93. Binomials, Multinomials and Expansions Solutions

1. \(\left(x + \frac{1}{x}\right)^5\)

   \(= {}^5C_0x^5 + {}^5C_1x^4\frac{1}{x} + {}^5C_2x^3\frac{1}{x^2} + {}^5C_3x^2\frac{1}{x^3} + {}^5C_4x\frac{1}{x^4} + {}^5C_5\frac{1}{x^5}\)

   \(= x^5 + 5x^3 + 10x + \frac{10}{x} + \frac{5}{x^3} + \frac{1}{x^5}\)

2. \((10.1)^5 = (10 + 0.1)^5\)

   \(= 10^5 + {}^5C_110^4(0.1) + {}^5C_210^3(0.1)^2 + {}^5C_310^2(0.1)^3 + {}^5C_410(0.1)^4 + (0.1)^5\)

   \(= 100000 + 5000 + 100 + 1 + 0.0005 + 0.00001\)

   \(= 105101.000501\)

3. Let \(\sqrt{x - 1} = a\)

   \((x + \sqrt{x - 1})^6 = (x + a)^6\)

   \(= x^6 + {}^6C_1x^5a + {}^6C_2x^4a^2 + {}^6C_3x^3a^3 + {}^6C_4x^2a^4 + {}^6C_5xa^5 + a^6\)

   \((x - \sqrt{x - 1})^6 = (x - a)^6\)

   \(= x^6 - {}^6C_1x^5a + {}^6C_2x^4a^2 - {}^6C_3x^3a^3 + {}^6C_4x^2a^4 - {}^6C_5xa^5 + a^6\)

   \(\therefore (x + \sqrt{x - 1})^6 + (x - \sqrt{x - 1})^6 = 2x^6 + 2{}^6C_2x^4a^2 + 2{}^6C_4x^2a^4 + 2a^6\)

   \(= 2[x^6 + 15x^5 - 29x^3 + 12x^2 + 3x - 1]\)

4. Clearly \((x + a)^n = A + B\) and \((x - a)^n = A - B\)

   Thus, \(A^2 - B^2 = (x^2 - a^2)^n\)

5. Let us solve these two parts one by one

   1. Given \((7 + 4\sqrt{3})^n = \alpha + \beta\)

      \(7 - 4\sqrt{3} = \frac{49 - 48}{7 + 4\sqrt{3}} = \frac{1}{7 + 4\sqrt{3}} < 1\)

      \(\therefore (7 - 4\sqrt{3})^n = \beta_1 < 1\)

      \(\alpha + \beta + \beta_1 = 2[7^n + {}^nC_27^{n - 2}48 + \ldots]\)

      \(=~\text{an even integer}\)

      \(\beta + \beta_1 =~\text{an even integer - a positive integer = an integer}\)

      \(\because 0 < \beta < 1\) and \(0 < \beta_1 < 1\)

      \(\therefore 0 < \beta + \beta_1 < 2\)

      \(\therefore \beta + \beta_1 = 1\)

      \(\alpha + 1 =~\text{an even number}~\therefore \alpha =~\text{an odd number}\)

   2. \((\alpha + \beta)(1 - \beta) = (\alpha + \beta)\beta_1\)

      \(= (7 + 4\sqrt{3})^n(7 - 4\sqrt{3})^n = 1\)

6. Let \(r^{th}\) term contain \(\frac{1}{y^2}\)

   \(t_r = {}^{10}C_{r - 1}y^{10 - r + 1}\left(\frac{c^3}{y^2}\right)^{r - 1}\)

   \(={}^{10}C_{r - 1}y^{13 - 3r}c^{3r - 3}\)

   \(\therefore 13 -3r - =2\Rightarrow r = 5\)

   \(\therefore\) Coefficient of \(y^-2 = {}^{210}c^{12}\)

7. \((1 + 3x + 3x^2 + x^3)^{15} = [(1 + x)^3]^{15} = (1 + x)^{45}\)

   \(\therefore\) Coefficient of \(x^9 = {}^{45}C_9 = \frac{45!}{9!36!}\)

8. Let \(r^{th}\) term be independent of \(x\)

   \(t_r = {}^9C_{r -1}\left(\frac{3}{2}x^2\right){10 - r}\left(-\frac{1}{3x}\right)^{r - 1}\)

   \(= (-1)^{r - 1}{}^9C_{r - 1}\left(\frac{3}{2}\right)^{10 - r}\frac{1}{3^{r - 1}}x^{21 - 3r}\)

   \(\therefore 21 - 3r = 0 \Rightarrow r = 7\)

   \(\therefore\) Coefficient \(= \frac{7}{18}\)

9. \((1 + x)^m\left(x + \frac{1}{x}\right)n = x^{-n}(1 + x)^{m + n}\)

   Thus, the term which will have \(x^n\) in \((1 + x)^{m + n}\) will be independent of \(x.\)

   \(\therefore\) Coefficient of \(x^n\) in \((1 + x)^{m + n} = {}^{m + n}C_n = \frac{(m + n)!}{m!n!}\)

10. Coefficient of \(x^{-1}\) in \((1 + 3x^2 + x^4)\left(1 + \frac{1}{x}\right)^8 =\) coefficient of \(x^{-1}\) in \(\left(1 + \frac{1}{x}\right)^8\) + \(3 *\) coefficient of \(x^{-3}\) in \(\left(1 + \frac{1}{x}\right)^8\) + coefficient of \(x^{-5}\) in \(\left(1 + \frac{1}{x}\right)^8\)

    \(r^{th}\) term in the expansion of \(\left(1 + \frac{1}{x}\right)^8\) is given by \(t_r = {}^8C_{r - 1}x^{1 - r}\)

    When \(1 - r = 1, r = 2,\) coefficient \(= {}^8C_1\)

    When \(1 - r = -3, r = 4,\) coefficient \(= {}^8C_3\)

    When \(1 - r = -5, r = 6\) coefficient \(= {}^8C_5\)

    Required coefficient \(= {}^8C_1 + {}^8C_3 + {}^8C_5 = 222\)

11. \(a_{r - 1} = {}^{2n - 1}C_{r - 1}(-1)^{r - 1}\)

    \(a_{2n - r} = {}^{2n - 1}C_{r - 1}(-1)^{r}\)

    \(a_{r - 1} + a_{2n - r} = 0\)

12. \(t_r = {}^{10}C_{r - 1}x^{\frac{15 - 5r}{2}}k^{r - 1}\)

    Since the term is independent of \(x, \therefore 15 - 5r = 0 \Rightarrow r = 3\)

    \(t_3 = {}^{10}C_2k^2 = 405 \therefore k = \pm 3\)

13. \(t_k = {}^{n - 3}C_{k - 1}x^{n - 3k}\)

    Since \(t_k\) should contains \(x^{2r}, \therefore 2k = n - 3k\)

    \(k = \frac{n - 2r}{3}\)

    If \(n - 2r\) is not divisible by \(3\) then the expansion will have no term with \(x^{2r}\)

14. \(t_r = {}^nC_{r - 1}x^{an - (a + b)(r - 1)}\)

    Since the term has to be independent of \(x, \Rightarrow an - (a + b)(r - 1) = 0\)

    \(r = 1 + \frac{an}{a + b}\)

    Thus, for \(r\) to be an integer \(an\) must be a multiple of \(a + b\)

15. \(\left(x + \frac{1}{x}\right)^7 = x^7 + {}^7C_1x^5 + {}^7C_2x^3 + {}^7C_3x + {}^7C_4x^{-1} + {}^7C_5x^{-3} + {}^7C_6x^{-5} + x^{-7}\)

    \(= x^7 + 7x^5 + 21x^3 + 35x + 35x^{-1} + 21x^{-3} + 7x^{-5} + x^{-7}\)

16. This problem is simple and has been left as an exercise.

17. \((1 + ax)^n = 1 + 8x + 24x^2 + \ldots = 1 + {}^nC_1ax + {}^nC_2a^2x^2 + \ldots\)

    Comparing coefficients of \(x, an = 8\)

    Comparing coefficients of \(x^2, \frac{n(n - 1)}{2}a^2 = 24\)

    \(\Rightarrow \frac{64 - 8a}{2} = 24, \Rightarrow a = 2, n = 4\)

18. This problem is simple and similar to 3 and has been left as an exercise.

19. \(t_7 = {}^9C_6\left(\frac{4x}{5}\right)^3\left(-\frac{5}{2x}\right)^6\)

    \(= \frac{10500}{x^3}\)

20. This problem is simple and similar to 3 and 18 and has been left as an exercise.

21. \((0.99)^{15} = (1 - 0.01)^15\)

    Since we have to evaluate only for \(4\) decimal places considering first four terms will do.

    First three terms \(= {}^{15}C_0 - {}^15C_1(.01) + {}^{15}C_2(.01)^2 + {}^{15}C_3(.01)^3 = 1 - .15 + .0105 - .000455 = 0.8600\)

22. \((999)^3 = (1000 - 1)^3 = 1000^3 - 3. \frac{1000^2}{2} + 3.\frac{1000}{2} - 1\)

    \(= 99700299\)

23. \((0.99)^{10} = (1 - 0.01)^{10}\)

    Since we have to evaluate only for \(4\) decimal places considering first four terms will do.

    First three terms \(= {}^{10}C_0 - {}^10C_1(.01) + {}^{10}C_2(.01)^2 - {}^{10}C_3(.01)^3 = 10 - .1 + 0.0045 - .000105 = .9044\)

24. This problem is simple and similar to 21 and 23 and has been left as an exercise.

25. \(A = {}^nC_0x^n + {}^nC_2x^{n - 2}a^2 + {}^nC_4x^{n - 4}a^4 + \ldots\)

    \(B = {}^nC_1x^{n - 1}a + {}^nC_3x^{n - 3}a^3 + {}^nC_5x^{n - 5}a^5 + \ldots\)

    \((x + a)^{2n} - (x - a)^{2n} = 2[{}^{2n}C_1x^{2n - 1}a + {}^2nC_3x^{2n - 3}a^3]\)

    \(4AB = 4{}^nC_0{}^nC_1x^{2n - 1}a + 4x^{2n - 3}a^3[{}^nC_0{}^nC_3 + {}^nC_1{}^nC_2 + \ldots]\)

    Thus, we see that \(4AM = (x + a)^n - (x - a)^n\)

26. Let \((5 + 2\sqrt{6})^n = \alpha + \beta\) where \(\alpha\) is a positive integer and \(beta\) is a proper fracttion.

    Also let, \(\gamma = (5 - 2\sqrt{6})^n\)

    Now, \(5 - 2\sqrt{6} = \frac{5^2 - 4*6}{5 + 2\sqrt{6}} = \frac{1}{5 + 2\sqrt{6}} < 1\)

    \(\therefore \gamma^n < 1\)

    \(\alpha + \beta + \gamma = 2[5^n + {}^nC_25^{n - 2}6^2 + \ldots] =\) An even number

    \(\beta + \gamma =\) An even number \(-\) An interger = An integer

    \(0 < \beta < 1\) and \(0 < \gamma < 1\)

    \(\therefore \beta + \gamma = 1\)

    \(\therefore \alpha\) is an off number.

27. This problem is simple and similar to 5. It has been left as an exercise.

28. \(t_r = {}^9C_{r - 1}(2x)^{9 - r + 1}\left(-\frac{3}{x}\right)^{r - 1}\)

    \(= {}^9C_{r - 1}2^{9 -r + 1}(-3)^{r - 1}x^{9 - r + 1 - r + 1}\)

    Now the power of \(x\) should be \(1, \therefore 11 - 2r = 1 \Rightarrow r = 5\)

    Coefficients of \(x\) is \(= {}^9C_42^{5}(-3)^{4} = 2592{}^9C_4\)

29. \(t_r = {}^{11}C_{r - 1}(3x^2)^{11 - r + 1}(5x)^{1 - r}\)

    \(= {}^{11}C_{r- 1}3^{12 - r}(5)^{r - 1}x^{24 - 2r + 1 - r}\)

    \(7 = 25 - 3r \Rightarrow r = 6\)

    \(\therefore\) Coefficients of \(t_6 = {}^{11}C_53^65^5\)

30. \(t_r = {}^{20}C_{r - 1}(2x^2)^{20 - r + 1}(-x)^{1 - r}\)

    \(= {-1}^{r - 1}{}^{20}C_{r - 1}2^{21 - r}x^{42 - 2r + 1 - r}\)

    Since we need coefficients of \(x^9, \therefore 9 = 43 - 3r, \Rightarrow r = \frac{34}{3}\)

    Since \(r\) is not an integer, there is no term containing \(x^9\) leading coefficient to be \(0.\)

31. \(t_r = {}^{15}C_{r - 1}(x^2)^{15 -r + 1}(3ax^{-1}){r - 1}\)

    \(t-r = {}^{15}C_{r - 1}.(3a)^{r - 1}x^{32 - 2r + 1 - r}\)

    Since we need coefficients of \(x^{24}, \therefore 24 = 33 - 3r\)

    \(\Rightarrow r = 3\)

    \(\therefore\) Coefficient of \(x^{24} = 9a^2{}^{15}C_2\)

32. \(t_r = {}^9C_{r - 1}(x^2)^{9 - r + 1}(-3x^{-1})^{r - 1}\)

    \(= {}^9C_{r - 1}(-3)^{r - 1}x^{20 - 2r + 1 - r}\)

    Since we need coefficients of \(x^{9}, \therefore 9 = 21 - 3r, \Rightarrow r = 4\)

    \(\therefore\) Coefficent of \(x^9, = 27.{}^9C_3\)

33. \(t_r = {}^{11}C_{r - 1}(2x)^{11 -r + 1}\left(\frac{1}{3x^2}\right)^{r - 1}\)

    \(= {}^{11}C_{r - 1}2^{12 - r}\frac{1}{3^{r - 1}}x^{12 - r + 2 - 2r}\)

    Since we need coefficients of \(x^{-7}, \therefore -7 = 14 - 3r, \Rightarrow r = 7\)

    \(\therefore\) Coefficients of \(x^{-7}, = {}^{11}C_62^6\frac{1}{3^5}\)

34. \(r^{th}\) term in the expansion of \(\left(ax^2 + \frac{1}{bx}\right)^{11}\) is given by \(t_r = {}^{11}C_{r - 1}(ax^2)^{12 - r}\frac{1}{(bx)^{r - 1}}\)

    \(t_r = {}^{11}C_{r - 1}a^{12 - r}b^{1 - r}x^{24 - 2r + 1 - r}\)

    Since we need coefficient of \(x^7, 7 = 25 -3r, \Rightarrow r = 6\)

    Coefficient of \(x^7, = {}^{11}C_5a^6b^{-5}\)

    \(r^{th}\) term in the expansion of \(\left(ax + \frac{1}{bx^2}\right)^{11}\) is given by \(t_r = (-1)^{r - 1}.{}^{11}C_{r - 1}(ax)^{12 - r}\frac{1}{(bx^2)^{r - 1}}\)

    \(t_r = (-1)^{r- 1}.{}^{11}C_{r - 1}a^{12 -r}b^{1 - r}x^{12 - r + 2 - 2r}\)

    Since we need coefficient of \(x^{-7}, -7 = 14 -3r, \Rightarrow r = 7\)

    Cofficient of \(x^{-7} = {}^{11}C_6a^5b^{-6}\)

    Equating the coefficients we get \(ab = 1 \because {}^{11}C_5 = {}^{11}C_6\)

35. \(t_r = {}^{2n}C_{r - 1}x^{4n - 2r + 2}\frac{1}{x^{r - 1}}\)

    \(t_r = {}^{2n}C_{r - 1}x^{4n - 3r + 3}\)

    Since it is the term containing \(x^p, p = 4n - 3r + 3, \Rightarrow r = \frac{4n - p + 3}{3}\)

    \(\therefore\) Coefficients of \(x^p, = {}^{2n}C_{\frac{4n - p}{3}}\)

    \(= \frac{2n!}{\left(\frac{4n -p}{3}\right)!\left(\frac{2n + p}{3}\right)}\)

36. Let us solve these one by one:

    1. \(t_r = {}^{2n}C_{r - 1}x^{2n - r + 1}\frac{1}{x^{r - 1}}\)

       Since the term has to be independent of \(x \Rightarrow 2n + 2 - 2r = 0 \Rightarrow r = n + 1\)

       \(t_{n + 1} = {}^{2n}C_n = \frac{2n!}{n!n!}\)

    2. \(t_r = {}^{15}C_{r - 1}(2x^2){16 - r}\frac{1}{x^{r - 1}}\)

       \(\Rightarrow 33 - 3r = 0 \Rightarrow r = 11\)

       \(t_{11} = {}^{15}C_10.2^5\)

    3. \(t_r = {}^{10}C_{r - 1}\left(\sqrt{\frac{x}{3}}\right)^{10 - r + 1}\left(\frac{3}{2x^2}\right)^{r - 1}\)

       \(\Rightarrow \frac{11 - r}{2} + 2 - 2r = 0 \Rightarrow r = 3\)

       \(t_3 = {}^{10}C_2\frac{1}{3^4}\frac{3^2}{2^2} = {}^{10}C_2\frac{1}{6^2}\)

    4. \(t_r = (-1)^{r - 1}.{}^{12}C_{r - 1}(2x^2)^{13 - r}\frac{1}{x^{r - 1}}\)

       \(= (-1)^{r - 1}.{}^{12}C_{r - 1}2^{13 - r}x^{27 - 3r}\)

       \(\Rightarrow 27 - 3r = 0 \Rightarrow r = 9\)

       \(t_9 = {}^12C_82^4\)

       5, 6, 7 and 8 are left as exercises.

37. \(t_r = {}^nC_{r - 1}x^{n - r + 1}\frac{1}{x^{2r - 21}}\)

    \(= {}^nC_{r - 1}x^{n - 3r + 3}\)

    For a term to be independent of \(x \Rightarrow r = \frac{n + 3}{3}\)

    \(\therefore\) Coefficient is \({}^nC_{\frac{n}{3}}\)

    \(= \frac{n!}{\left(\frac{n}{3}\right)!\left(\frac{2n}{3}\right)!}\)

38. Coeff. of \(x^m\) in \((1 + x)^{m + n} = {}^{m + n}C_m\)

    Coeff. of \(x^n\) in \((1 + x)^{m + n} = {}^{m + n}C_n\)

    Clearly, both the coefficients are equal.

39. \(t_4 = {}^nC_3(px)^{n - 3}\frac{1}{x^3} = \frac{5}{2}\)

    Since \(\frac{5}{2}\) independent of \(x, n - 6 = 0 \Rightarrow n = 6\)

    \({}^6C_3p^3 = \frac{5}{2} \Rightarrow p = \frac{1}{2}\)

40. Here \(n = 12,\) which is even, therefore, \(\frac{12}{2} + 1\) i.e. \(7{th}\) term will be middle term.

    \(t_7 = {}^{12}C_6x^6\left(-\frac{1}{2x}\right)^6 = \frac{231}{16}\)

41. Here \(n = 7,\) which is odd, therefore, \(\frac{7 + 1}{2}\) and \(\frac{7 + 3}{2}\) i.e. \(4^{th}\) and \(5^{th}\) terms will be middle terms.

    \(t_4 = {}^7C_3(2x^2){7 - 3}\left(-\frac{1}{x}\right)^3 = -560x^3\)

    \(t_5 = {}^7C_4(2x^2)^{7 - 4}\left(-\frac{1}{x}\right)^2 = 280x^2\)

42. \((1 - 2x + x^2)^n = (1 - x)^{2n}\)

    Since \(2n\) is even, therefore the middle term would be \((n + 1)^{th}\) term.

    \(t_{n + 1} = {}^{2n}C_n1^{2n - n}(-x)^n = \frac{2n!}{n!n!}(-1)^nx^n\)

43. \(\because 2n\) is even, the middle term would be \((n + 1)^{th}\) term.

    \(t_{n + 1} = {}^{2n}C_nx^{2n - n}\frac{1}{x^n} = \frac{2n!}{n!n!}\)

    \(= \frac{1.2.3.4\ldots 2n}{1.2.3\ldots n.(n!)}\)

    \(= \frac{1.3.5\ ldots (2n - 1).2^{n}.1.2.3\ldots n}{1.2.3\ldots n.(n!)}\)

    \(= \frac{1.3.5\ldots (2n - 1)}{n!}2^n\)

45. Clearly, middle term will have greatest coefficient which has been found in 44.

45. We have already found coefficient of middle term of \((1 + x)^{2n}\) which is \({}^{2n}C_n\frac{2n!}{n!n!}\)

    Since \(2n - 1\) is odd number we will have two middle terms for \((1 + x)^{2n - 1},\) which will be \(n^{th}\) and \((n + 1)^{th}\) terms of the expansion.

    Coefficient of \(t_n\) in \((1 + x)^{2n - 1} = {}^{2n - 1}C_{n - 1}\)

    Coefficient of \(t_{n + 1} = {}^{2n - 1}C_n\)

    Clearly, \({}^{2n - 1}C_n + {}^{2n - 1}C_{n - 1} = {}^{2n}C_n\) by invoking properties of combinations.

46. Let us find these:

    1. Since \(n = 20,\) which is an even number, the middle term would be \(11^{th}\) term.

       \(t_{11} = {}^{20}C_{10}\left(\frac{2x}{3}\right)^{10} \left(-\frac{3}{2x}\right)^{10}\)

       \(= {}^{20}C_{10}x^{10}y^{10}\)

    2. Since \(n = 6,\) an even number, the middle term would be \(4^{th}\) term.

       \(t_4 = {}^6C_3\left(\frac{2x}{3}\right)^3 \left(-\frac{3}{2x}\right)^3\)

       \(= -20\)

    3. Since \(n = 7,\) an odd number, the middle terms would be \(4^{th}\) and \(5^{th}\) terms.

       \(t_4 = {}^7C_3\frac{x^4}{y^4}.(-1)^3\frac{y^3}{x^3} = -35\frac{x}{y}\)

       \(t_5 = {}^7C_4\frac{x^3}{y^3}.(-1)^4\frac{y^4}{x^3} = 25\frac{y}{x}\)

    4. Since power of the expansion is \(2n,\) which is an even number so \((n + 1)^{th}\) term would be middle term.

       \(t_{n + 1} = {}^{2n}C_nx^n\)

    5. \((1 - 2x + x^2)^n = (1 - x)^{2n}\) and like previous exercise \(t_{n + 1} = (-1)^n{}^{2n}C_nx^n\) would be middle term.

47. Let \((r + 1)^{th}\) term be middle term then \(t_{r + 1} = {}^{2n + 1}C_r\frac{x^{2n - r + 1}}{y^{2n - r + 1}}\frac{y^r}{x^r}\)

    \(t_{r + 1} = {}^{2n + 1}C_r\frac{x^{2n - 2r + 1}}{y^{2n - 2r + 1}}\)

    Since \(2n + 1\) is an odd number, therefore there will be two middle terms, \((n + 1)^{th}\) and \((n + 2)^{th}\)

    \(t_{n + 1} = {}^{2n + 1}C_n\frac{x}{y}\)

    \(t_{n + 2} = {}^{2n + 1}C_n\frac{y}{x}\)

    \(2n - 2r + 1,\) which is power of \(x\) and \(y\) in general term, cannot be zero as both \(n\) and \(r\) are positive integers. Thus, no term is independent of \(x\) and \(y.\)

48. Since exponent of expansion is \(2n\) which is an even number, there will be one middle term and that term would be \((n + 1)^{th}\) term.

    \(t_{n + 1} = {}^{2n}C_n(-1)^{n}\frac{x^n}{x^n}\)

    \(= \frac{2n!}{n!n!} = \frac{2^n(1.2.3\ldots n)(1.3.5\ldots (2n - 1))}{n!n!}\)

    \(= (-2^n)\frac{1.3.5\ldots (2n- 1)}{n!}\)

49. Coefficient of \(t_{2n + 1} = {}^{43}C_{2r}\)

    Coefficient of \(t_{r + 2} = {}^{43}C_{r + 1}\)

    Given \({}^{43}C_{2r} = {}^{43}C_{r + 1}\)

    \(\therefore 2r + r + 1 = 43 \Rightarrow r = 14\)

50. Coefficient of \(r^{th}\) term \(t_r = {}^{2n}C_{r - 1}\)

    Coefficient of \((r + 4)^{th}\) term \(t_r = {}^{2n}C_{r + 3}\)

    \(\Rightarrow 2n = r - 1 + r + 3 \Rightarrow r = n - 1\)

    2nd Method Equal coefficient will be equidistant from mid term. In this case there is only one which is \((n + 1)^{th}\) term. So \(\frac{r + r + 4}{2} = n + 1 \Rightarrow r = n - 1\)

51. Following like previous exercise:

    \(\Rightarrow 18 = 2r + 3 + r - 3 \Rightarrow r = 6\)

52. Following like previous exercise:

    \(\Rightarrow 2r + 4 + r - 7 = 39 \Rightarrow r = 14\)

    \({}^rC_{12} = \frac{14 * 13}{2} = 91\)

53. Following like previous exercise:

    \(\Rightarrow 2n = 3r - 1 + r + 1 \Rightarrow r = \frac{n}{2}\)

54. Following like previous exercise:

    \(\Rightarrow 2n = p + p + 2 \Rightarrow p = n - 1\)

55. Let \(r^{th}\) and \((r + 1)^{th}\) have equal values for coefficients. Let \(C_r\) represent cofficient for \(r^{th}\) term and \(C_{r + 1}\) for \((r + 1)^{th}\) term.

    \(C_r = {}^{75}C_{r - 1}\)

    \(C_{r + 1} = {}^{75}C_r3\)

    Given \(C_r = C_{r + 1}\)

    \(\frac{1}{75 - r + 1} = \frac{1}{r}\)

    \(76 - r = r \Rightarrow r = 38\)

    \(\therefore {}^{75}C_{37}\) and \({}^{75}C_{38}\) are equal coefficients.

56. Coefficient of \((r + 1)^{th}\) term in \((1 + x)^{n + 1} = {}^{n + 1}C_r\)

    Coefficient of \(r^{th}\) term in \((1 + x)^n = {}^nC_{r - 1}\)

    Coefficient of \((r + 1)^{th}\) term in \((1 + x)^n = {}^nC_r\)

    Now we know that \({}^nC_{r - 1} + {}^nC_r = {}^{n + 1}C_{r + 1}\)

57. We have to find greatest term numerically.

    \(therefore\) greatest term in \(\left(7 -\frac{10}{3}\right)^{11} =\) greatest term in \(\left(7 - \frac{10}{3}\right)^{11}\)

    Let \(r^{th}\) term be the greatest term in the expansion of \(\left(7 + \frac{10}{3}\right)^{11}\)

    \(t_r = {}^{11}C_{r - 1}7^{11 -r + 1}\left(\frac{10}{3}\right)^{r - 1}\)

    \(t_{r + 1} = {}^{11}C_r7^{11 - r}\left(\frac{10}{3}\right)^r\)

    \(\frac{t_r}{t_{r + 1}} = \frac{21r}{(12 - r)}10\)

    Also, \(\frac{t_{r - 1}}{t_r} = \frac{21(r - 1)}{13 - r}10\)

    \(\because t_r\) is the greatest term \(\therefore t_r \geq t_{r + 1}\)

    \(\frac{21r}{(12 - r)10} \ geq 1 \Rightarrow r \geq \frac{120}{31}\)

    Also, \(\frac{t_{r - 1}}{t_r} \leq 1\)

    \(\Rightarrow r\leq \frac{151}{32}\)

    Thus, \(r = 4\)

    \(t_4 = \frac{440}{9}7^85^3\)

58. In any binomial expansion middle term has the greatest coefficient. Middle term in the expansion of \((1 + 2x)^{2n}\) is \((n + 1)^{th}\) term.

    \(t_{n + 1} = {}^{2n}C_nx^n, t_{n + 1} = {}^{2n}C_{n + 1}x^{n + 1}, t_{n} = {}^{2n}C_{n - 1}x^{n - 1}\)

    \(\frac{t_{n + 1}}{t_{n + 2}} = \frac{n + 1}{n}.\frac{1}{x} > 1 \Rightarrow x < \frac{n + 1}{n}\)

    Similalrly \(\frac{t_{n + 1}}{t_n} = \frac{(n + 1)x}{n} > 1 \Rightarrow x > \frac{n}{n + 1}\)

    Thus \(\frac{n}{n + 1} < x < \frac{n + 1}{n}\)

59. These are similar to problem 57 and have been left as an exercise.

60. This is similar to problem 58 and limit will be \(\frac{15}{16} < x < \frac{16}{15}\).

61. Given expression \(= 6^{2n} - 35n - 1 = (6^2)^n - 35n - 1 = 36^n - 35n - 1\)

    Clearly, \(35^2 = 1225\) so we rewrite \(36\) as \(1 + 35\)

    \((36)^n - 35n - 1 = (1 + 35)^n - 35n - 1\)

    \(= 1 + 35n + {}^{35}C_2.35^2 + {}^{35}C_3.35^3 + \ldots + {}^{35}C_{35}35^n - 35n - 1\)

    \(= {}^{35}C_2.35^2 + {}^{35}C_3.35^3 + \ldots + {}^{35}C_{35}35^n\)

    All the terms contain powers euqal or greater than \(2\) with base \(35\) therefore whole expression is divisible by \(35^2\) or \(1225\)

62. \(2^{4n} - 2^n(7n + 1) = (2^4)^n - 2^n.7n - 2^n\)

    \(= 16^n - 2^n.7n - 2^n = (2 + 14)^n - 2^n.7n - 2^n\)

    \(= {}^nC_02^n + {}^nC_12^{n - 1}14 + {}^nC_22^{n- 2}14^2 + {}^nC_32^{n - 3}14^3 + \ldots {}^nC_n14^n - 2^n.7n - 2^n\)

    \(= 14^2[{}^nC_22^{n - 2} + {}^nC_32^{n - 3} + \ldots + {}^nC_n14^{n - 2}]\)

    Thus, \(2^{4n} - 2^n(7n + 1)\) is divisible by \(14^2\)

63. \(3^{4n + 1} + 16n - 3 = 3[3^{4n}] + 16n - 3 = 3[(1 + 80)^n - 1] + 16n\)

    \(=3[{}^nC_0 + {}^nC_180 + {}^nC_280^2 + {}^nC_380^3 + \ldots + {}^nC_n80^n - 1] + 16n\)

    \(= 256n + {}^nC_280^2 + {}^nC_380^3 + \ldots + {}^nC_n80^n\)

    \(80^2\) is divisible by \(256,\) therefore above expression is divisible by \(256.\)

64. Let us solve these one by one:

    1. \(4^n - 3n - 1 = (1 + 3)^n - 3n - 1 = 1 + 3n + {}^nC_23^2 + {}^nC_33^3 + \ldots + {}^nC_n3^n - 3n -1\)

       \(= {}^nC_23^2 + {}^nC_33^3 + \ldots + {}^nC_n3^n\)

       \(= 3^2[{}^nC_2 + {}^nC_33 + \ldots + {}^nC_n3^{n - 2}]\)

       The above expression is divisible by \(9.\)

    2. \(2^{5n} - 31n - 1 = (2^5)^n - 31n - 1 = 32^n - 31n - 1\)

       \(= (1 + 31)^n - 31n - 1\)

       \(= 1 + 31n + {}^nC_231^2 + {}^nC_331^3 + \ldots + {}^nC_n31^n - 31 - 1\)

       \(= 31^2[{}^nC_2 + {}^nC_331 + \ldots + {}^nC_n31^{n - 2}]\)

       The above expression is divisible by \(961.\)

    3. \(3^{2n + 2} - 8n - 9 = 3^2.3^{2n} - 8n - 9 = 9(1 + 8)^n - 8n - 9\)

       \(= 9[1 + 8n + {}^nC_28^2 + {}^nC_38^3 + \ldots + {}^nC_n8^n] - 8n - 9\)

       \(= 64n + 9[{}^nC_28^2 + {}^nC_38^3 + \ldots + {}^nC_n8^n]\)

       The above expression is divisible by \(64\)

    4. \(2^{5n + 5} - 31n - 32 = 2^5.(2^5)^n - 31n - 31 = 32(1 + 31)^n - 31n - 32\)

       \(= 32[1 + 31n + {}^nC_231^2 + {}^nC_331^3 + \ldots + {}^nC_n31^n] - 31n - 32\)

       \(= 32[{}^nC_231^2 + {}^nC_331^3 + \ldots + {}^nC_n31^n]\)

       The above expression is divisible by \(31^2\) i.e. \(961\)

    5. \(3^{2n} - 1 + 24n - 32n^2 = (1 + 8)^n - 1 + 24n - 32n^2\)

       \(= 1 + 8n + {}nC_28^2 + {}nC_38^3 + \ldots + {}nC_n8^n - 1 + 24n - 32n^2\)

       \(= 32n + \frac{n(n - 1)}{2}8^2 + {}nC_38^3 + \ldots + {}nC_n8^n - 32n^2\)

       \(= 32n + 64n^2 - 32n + {}nC_38^3 + \ldots + {}nC_n8^n\)

       \(= 64n^2 + {}nC_38^3 + \ldots + {}nC_n8^n\)

       Above expression is divisible by \(256\) for \(n > 2\)

65. Let \(r^{th}, (r + 1)^{th}\) and \((r + 2)^{th}\) terms are the three consecutive terms given in the problem.

    \(t_r = {}^nC_{r - 1} = \frac{n!}{(r - 1)!(n - r + 1)!} = 165\)

    \(t_{r + 1} = {}^nC_r = \frac{n!}{r!(n - r)!} = 330\)

    \(t_{r + 2} \ {}^nC_{r + 1} = \frac{n!}{(r + 1)!(n - r - 1)!} = 462\)

    \(\frac{t_r}{t_{r + 1}} = \frac{r}{n - r + 1} = \frac{1}{2} \Rightarrow 3r = n + 1\)

    \(\frac{t_{r + 1}}{t_{r + 2}} = \frac{r + 1}{n - r} = \frac{330}{462}\)

    \(12r = 5n - 7\)

    Solving the two equations in \(n\) and \(r,\) we get, \(n = 11, r = 4.\)

66. Let \(a_1, a_2, a_3\) and \(a_4\) be coefficients of \(r^{th}, (r + 1)^{th}, (r + 2)^{th}\) and \((r + 3)^{th}\) terms of the expansion \((1 + x)^n\)

    \(a_1 = {}^nC_{r - 1}, a_2 = {}^nC_r, a_3 = {}^nC_{r + 1}, a_4 = {}^nC_{r + 2}\)

    \(\frac{a_2}{a_1} = \frac{n - r + 1}{r}\)

    \(\frac{a_1}{a_1 + a_2} = \frac{r}{n + 1}\)

    Substituting \(r + 1\) instead of \(r\) in previous equation, we get

    \(\frac{a_2}{a_2 + a_3} = \frac{r + 1}{n + 1}\)

    Similarly, \(\frac{a_3}{a_3 + a_4} = \frac{r + 2}{n + 1}\)

    \(\frac{a_1}{a_1 + a_2} + \frac{a_3}{a_3 + a_4} = \frac{2(r + 1)}{n + 1} = \frac{2a_2}{a_2 + a_3}\)

67. \(\frac{t_2}{t_3} = \frac{{}^nC_1x^{n - 1}y}{{}^nC_2x^{n - 2}y^2} = \frac{240}{720}\)

    \(\Rightarrow \frac{2x}{(n - 1)y} = \frac{1}{3}\)

    \(\frac{t_3}{t_4} = \frac{{}^nC_2x^{n - 2}y^2}{{}^nC_3x^{n - 3}y^3} = \frac{720}{1080}\)

    \(\Rightarrow \frac{3x}{(n - 2)y} = \frac{2}{3}\)

    Dividing the two obtained equations

    \(\frac{2x}{(n - 1)y}\frac{3x}{(n - 2)y} = \frac{1}{2}\frac{2}{3}\)

    \(\frac{2(n - 2)}{3(n - 1)} = \frac{1}{2}\)

    \(4n - 8 = 3n - 3 \Rightarrow n = 5\)

    Putting this value in \(\frac{t_r}{t_{r + 1}},\) we get

    \(\frac{2}{5 -1}.\frac{x}{y} = \frac{1}{3}\)

    \(y = \frac{3x}{2}\)

    \(t_r = {}^nC_1x^{n - 1}y = \frac{3}{2}5x^5 = 240 \Rightarrow x = 2\)

    \(\Rightarrow y = 3\)

68. Let \(n\) be the index of the expansion and \(a, b, c\) be the \(r^{th}, (r + 1)^{th}, (r + 2)^{th}\) term respectively.

    \(a = {}^nC_{r - 1}, b = {}^nC_r, c = {}^nC_{r + 1}\)

    \(\frac{a}{b} = \frac{r}{n - r + 1}, \frac{b}{c} = \frac{r + 1}{n - r}\)

    \(an + a = r(a + b), bn - c = r(b + c)\)

    \(\Rightarrow (b + c)(an + a) = (a + b)(bn - c)\)

    \(\Rightarrow n = \frac{2ac + b(a + c)}{b^2 - ac}\)

69. Coefficients are \(C_{14} = {}^nC_{13}, C_{15} = {}^nC_{14}, C_{16} = {}^nC_{15}\)

    These are in A.P., so we can write

    \(\frac{2.n!}{14!(n - 14)!} = \frac{n!}{13!(n - 13!)} + \frac{n!}{15!(n - 15)!}\)

    Multiplying both sides by \(15!(n - 13)!,\) we get

    \(2.15(n - 13) = 15.14 + (n - 13).(n - 14)\)

    \(\Rightarrow n^2 - 57n + 782 = 0\)

    \(n = 23, 34\)

70. Let those three terms are \(r^{th}, (r + 1)^{th}\) and \((r + 2)^{th}\) terms of the expansion.

    \(\frac{t_{r- 1}}{t_r} = \frac{r}{n - r + 1} = \frac{56}{70} = \frac{4}{5}\)

    \(9r - 4n = 4\)

    \(\frac{t_{r}}{t_{r + 1}} = \frac{r + 1}{n - r} = \frac{70}{56}\)

    \(9r - 5n = -4\)

    \(\Rightarrow n = 8, r = 4\)

71, 72 and 73 are similar problems like ones we have solved and has been left as exercises.

74. Let the binomial expansion be \((x + y)^n.\)

    \(a = {}^nC_5x^{n - 5}y^5, b = {}^nC_6x^{n - 6}y^6, c = {}^nC_7x^{n - 7}y^7, d = {}^nC_8x^{n - 8}y^8\)

    \(b^2 - ac = \left({}^nC_6x^{n - 6}y^6\right)^2 - {}^nC_5x^{n - 5}y^5{}^nC_7x^{n - 7}y^7\)

    \(= \left(\frac{n!}{6!(n - 6)!}x^{n - 6}y^6\right)^2 - \frac{n!}{5!(n - 5)!}x^{n - 5}y^5 \frac{n!}{7!(n - 6)!}x^{n - 7}y^7\)

    Similary \(c^2 - bd\) can be computed and it can be shown that \(\frac{b^2 - ac}{c^2 - bd} = \frac{4a}{3c}\)

75. Let us solve these one by one.

    1. We have to prove that \(\frac{a + b}{a}, \frac{b + c}{b}, \frac{c + d}{c}\) are in H.P.

       i.e. \(\frac{a}{a + b}, \frac{b}{b + c}, \frac{c}{c + d}\) are in A.P.

       Let \(a = {}^nC_r, b = {}^nC_{r + 1}, c = {}^nC_{r + 2}, d = {}^nC_{r + 3}\)

       \(\frac{a}{a + b} = \frac{n!}{r!(n - r!)}\left(\frac{1}{\frac{n!}{r!(n - r!)}} + \frac{n!}{(r + 1)!(n - r - 1)!}\right)\)

       \(= \frac{r + 1}{n + 1}\)

       Similarly \(\frac{b}{b + c} = \frac{r + 2}{n + 1}\)

       and, \(\frac{c}{c + d} = \frac{r + 3}{n + 1}\)

       Clearly, these are in A.P.

    2. Since \(\frac{a}{a + b}, \frac{b}{b + c}, \frac{c}{c + d}\) are in A.P. from the previous part.

       \(\frac{2b}{b + c} = \frac{a}{a + b} + \frac{c}{c + d}\)

       Solving this further leads to the equality \((bc + ad)(b - c) = 2(ac^2 - b^2d)\)

76. Since the coeficients of given terms are in A.P., we can write that

    \(\frac{n!}{n - 4!} - \frac{2.n!}{5!(n - 5)!} + \frac{n!}{6!(n - 6!)} = 0\)

    \(\frac{n!}{4!(n - 6!)}\left[\frac{1}{(n - 5)(n - 6)} -\frac{2}{5(n - 5) + \frac{1}{6.5}}\right]\)

    Clearly, \(\frac{n!}{(n - 6!)}\neq 0\)

    \(\Rightarrow n^2 - 23n + 132 = 0\Rightarrow n = 11, 12\)

77. Since the coeficients of given terms are in A.P., we can write that

    \(\frac{2n!}{(2n - 1)!} + \frac{2.2n!}{2!(2n - 2)!} + \frac{2n!}{3!(2n - 3)!} = 0\)

    \(2n - 2n(2n - 1) + \frac{n(2n - 1)(2n - 2)}{3} = 0\)

    \(4n^3 - 18^2 + 14n = 0\)

    \(\because n \neq 0, \Rightarrow 4n^2 - 9n + 7 = 0\)

78. Since the coeficients of given terms are in A.P., we can write that

    \({}^nC_{r - 1} - 2{}^nC_r + {}^nC_{r + 1} = 0\)

    \(\frac{n!}{(r - 1)!(n - r + 1)!} - 2\frac{n!}{r!(n - r)!} + \frac{n!}{(r + 1)!(n - r - 1)!} = 0\)

    \(\frac{n!}{(r - 1)!(n - r - 1)!}\left[\frac{1}{(n - r)(n - r + 1)} - 2\frac{1}{r(n - r)} + \frac{1}{r(r + 1)}\right] = 0\)

    Solving this gives the desired equation.

79. This problem is easy and similar to other problems and has been left as exercise.

80. Given series is \(C_1 + 2.C_2 + 3.C_3 + \ldots + n.C_n\)

    Its \(r^{th}\) term \(t_r = r.{}^nC_r = n.{}^{n - 1}C_{r -1}\)

    \(C_1 + 2.C_2 + 3.C_3 + \ldots + n.C_n = \sum_{r = 1}^nr.{}^nC_r\)

    \(= n\sum_{r=1}^n{}^{n - 1}C_{r - 1} = n.2^{n - 1}\)

    Calculus Method: This method requires knowledge of calculus.

    \((1 + x)^n = {}^nC_0 + {}^nC_1x + {}^nC_2x^2 + \ldots + {}^nC_nx^n\)

    Differentiating both sides w.r.t. \(x\)

    \(n.(1 + x)^{n - 1} = {}^nC_1 + 2{}^nC_2x^2 + \ldots + n{}^nCnx^{n - 1}\)

    Putting \(x = 1,\) we get

    \(n.2^{n - 1} = {}^nC_1 + 2{}^nC_2 + \ldots + n.{}^nC_n\)

81. Given series is \(C_0 + 2.C_1 + 3.C_2 + \ldots + (n + 1).C_n\)

    Its \(r^{th}\) term \(t_r = r{}^nC_{r - 1}\)

    \(= (r - 1 + 1){}^nC_{r - 1} = (r - 1){}^nC_{r - 1} + {}^nC_{r - 1}\)

    \(= n{}^{n - 1}C_{r - 2} + {}^nC_{r - 1}[\because (r - 1){}^nC_{r - 1} = n{}^{n - 1}C_{r - 2}]\)

    Thus, \(\sum_{r = 1}^{n + 1}t_r = n.2^{n - 1} + 2^n = (n + 2)2^{n - 1}\)

    Calculus Method:

    \((1 + x)^n = {}^nC_0 + {}^nC_1x + {}^nC_2x^2 + \ldots + {}^nC_nx^n\)

    Multiplying both sides by \(x,\) we get

    \(x(1 + x)^n = {}^nC_0x + {}^nC_1x^2 + {}^nC_2x^3 + \ldots + {}^nC_nx^{n + 1}\)

    Differentiating w.r.t. \(x,\) we get

    \([(1 + x)^n + nx(1 + x)^{n - 1}] = {}^nC_0 + 2{}^nC_1x + 3{}^nC_2x^2 + \ldots + (n + 1){}^nC_nx^n\)

    Substituting \(x = 1,\) we get

    \(C_0 + 2.C_1 + 3.C_2 + \ldots + (n + 1).C_n = (n + 2).2^{n - 1}\)

82. \(t_r = (2r - 1){}^nC_{r - 1} = 2(r - 1){}^nC_{r - 1} + {}^nC_{r - 1}\)

    \(=2n {}^{n - 1}C_{r - 2} + {}^nC_{r - 1}\)

    \(\sum_{r = 1}^{n + 1}t_r = \sum_{r = 1}^{n + 1}2n {}^{n - 1}C_{r - 2} + \sum_{r = 1}^{n + 1}{}^nC_{r - 1}\)

    \(= 2n({}^{n - 1}C_0 + {}^{n - 1}C_1 + {}^{n - 1}C_2 + \ldots + {}^{n - 1}C_{n - 1}) + ({}^nC_0 + {}^nC_1 + {}^nC_2 + \ldots + {}^nC_n)\)

    \(= 2n.2^{n - 1} + 2^n = 2^n(n + 1)\)

    Calculus Method:

    \((1 + x)^n = {}^nC_0 + {}^nC_1x + {}^nC_2x^2 + \ldots + {}^nC_nx^n\)

    Substituting \(x = x^2\) and multiplying both sides with \(x\)

    \(x(1 + x^2)^n = C_0x + C_1x^3 + C_2x^5 + \ldots + C_nx^{2n + 1}\)

    Differentiating both sides w.r.t \(x\) and substituting \(x = 1,\) we get the desired equation as earlier.

83. Given series is \(C_1 - 2.C_2 + 3.C_3 - 4.C_4 + \ldots + (-1)^nn.C_n = 0\)

    \(t_r = (-1)^{r - 1}r.{}^nC_r = (-1)^{r - 1}n.{}^{n - 1}C_{r - 1}\)

    \(\sum_{r = 1}^{n}t_r = \sum_{r = 1}^{n}(-1)^{r - 1}n.{}^{n - 1}C_{r - 1}\)

    \(n({}^{n - 1}C_0 - {}^{n - 1}C_1 + {}^{n - 1}C_2 + \ldots + (-1)^{n - 1}){}^{n - 1}C_{r - 1} = n(1 - 1)^{n - 1} = 0\)

    Calculus Method:

    \((1 + x)^n = {}^nC_0 + {}^nC_1x + {}^nC_2x^2 + \ldots + {}^nC_nx^n\)

    Differentiating w.r.t. \(x\) and substituting \(x = -1\)

    \(n({}^{n - 1}C_0 - {}^{n - 1}C_1 + {}^{n - 1}C_2 + \ldots + (-1)^{n - 1}){}^{n - 1}C_{r - 1} = n(1 - 1)^{n - 1} = 0\)

84. Given series is \(C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}\)

    \(t_r = \frac{{}^nC_{r - 1}}{r}\)

    \(C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n + 1} = \sum_{r = 1}^n \frac{{}^nC_{r - 1}}{r} = \sum_{r = 1}^n\frac{{}^{n + 1}C_r}{n + 1}\left[\because \frac{{}^nC_{r - 1}}{r} = \frac{{}^{n + 1}C_r}{n + 1}\right]\)

    \(= \frac{1}{n + 1}({}^{n + 1}C_1 + {}^{n + 1}C_2 + {}^{n + 1}C_{n + 1})\)

    \(=\frac{2^{n + 1} - 1}{n + 1}\)

    Calculus Method:

    \((1 + x)^n = {}^nC_0 + {}^nC_1x + {}^nC_2x^2 + \ldots + {}^nC_nx^n\)

    Integrating w.r.t \(x\) between limits \(0\) and \(1,\) we get

    \(\left[\frac{(1 + x)^{n + 1}}{n + 1}\right]_0^1 = \left[C_0x + C_1\frac{x^2}{2} + C_2\frac{x^3}{3} + \ldots + C_n\frac{x^{n + 1}}{n + 1}\right]_0^1\)

    \(\Rightarrow C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}\)

85. Given series is \(C_0 - \frac{C_1}{2} + \frac{C_2}{3} - \ldots + (-1)^n\frac{C_n}{n + 1}\)

    \(t_r = (-1)^{r - 1}\frac{{}^nC_{r - 1}}{r}\)

    \(\therefore C_0 - \frac{C_1}{2} + \frac{C_2}{3} - \ldots + (-1)^n\frac{C_n}{n + 1} = \sum_{r = 1}^{n + 1}(-1)^{r - 1}\frac{{}^nC_{r - 1}}{r} = \sum_{r = 1}^{n + 1}(-1)^{r - 1}\frac{{}^{n + 1}C_r}{n + 1}\left[\because \frac{{}^nC_{r - 1}}{r} = \frac{{}^{n + 1}C_r}{n + 1}\right]\)

    \(= \frac{1}{n + 1}[{}^{n + 1}C_1 - {}^{n + 1}C_2 + {}^{n + 1}C_3 - \ldots + (-1)^n.{}^{n + 1}C_{n + 1}]\)

    \(= \frac{1}{n + 1}[-({}^{n + 1}C_0 + {}^{n + 1}C_1 - {}^{n + 1}C_2 + {}^{n + 1}C_3 - \ldots + (-1)^{n+1}.{}^{n + 1}C_{n + 1}) + {}^{n + 1}C_0]\)

    \(= \frac{1}{n + 1}\left[-(1 - 1)^{n + 1} + {}^{n + 1}C_0\right] = \frac{1}{n + 1}\)

    Calculus Method:

    \((1 + x)^n = {}^nC_0 + {}^nC_1x + {}^nC_2x^2 + \ldots + {}^nC_nx^n\)

    Integrating w.r.t \(x\) between limits \(0\) and \(1,\) we get

    \(\left[\frac{(1 + x)^{n + 1}}{n + 1}\right]_0^{-1} = \left[C_0x + C_1\frac{x^2}{2} + C_2\frac{x^3}{3} + \ldots + C_n\frac{x^{n + 1}}{n + 1}\right]_0^{-1}\)

    \(0 - \frac{1}{n + 1} = -C_0 + \frac{C_1}{2} - \frac{C_3}{3} + \ldots + (-1)^{n + 1}\frac{C_n}{n + 1}\)

    \(\Rightarrow C_0 - \frac{C_1}{2} + \frac{C_2}{3} - \ldots + (-1)^n \frac{C_n}{n + 1} = \frac{1}{n + 1}\)

86. Given series is \(\frac{C_1}{2} + \frac{C_3}{4} + \frac{C_5}{6} + \ldots\)

    \(t_r = \frac{{}^nC_{2r - 1}}{23} = \frac{{}^nC_{2r - 1}}{(2r - 1) + 1} = \frac{{}^{n + 1}C_{2r}}{n + 1}\)

    \(\frac{C_1}{2} + \frac{C_3}{4} + \frac{C_5}{6} + \ldots = \sum_{r = 1} = \frac{1}{n + 1}\sum_{r = 1}{n + 1}C_{2r} = \frac{2^n - 1}{n + 1}\)

    Calculus Method:

    Adding the results of 84 and 85, we get

    \(2\left[\frac{C_1}{2} + \frac{C_3}{4} + \frac{C_5}{6} + \ldots\right] = \frac{2^{n + 1} - 1 - 1}{n + 1} = \frac{2(2^n - 1)}{n + 1}\)

    \(\Rightarrow \frac{C_1}{2} + \frac{C_3}{4} + \frac{C_5}{6} + \ldots = \frac{2^n - 1}{n + 1}\)

87. Given series is \(2.C_0 + 2^2\frac{C_1}{2} + 2^3\frac{C_2}{3} + \ldots + 2^{n + 1}\frac{C_n}{n + 1}\)

    \(t_r = 2^r\frac{{}^nC_{r - 1}}{r} = 2^r\frac{{}^{n + 1}C_r}{n + 1}\)

    Now \(2.C_0 + 2^2\frac{C_1}{2} + 2^3\frac{C_2}{3} + \ldots + 2^{n + 1}\frac{C_n}{n + 1} = \sum_{r = 1}^{n + 1} 2^r\frac{{}6{n + 1}C_r}{n + 1}\)

    \(= \frac{1}{n + 1}[\{{}^{n + 1}C_0 + 2.{}^{n + 1}C_1 + 2^2.{}^{n + 1}C_2 + \ldots + 2^{n + 1}{}^{n + 1}C_{n + 1}\} - {}^{n + 1}C_0]\)

    \(= \frac{1}{n + 1}[(1 + 2)^{n + 1} - 1] = \frac{3^{n + 1} - 1}{n + 1}\)

    Calculus Method:

    \((1 + x)^n = {}^nC_0 + {}^nC_1x + {}^nC_2x^2 + \ldots + {}^nC_nx^n\)

    Integrating between limits \(0\) and \(2,\) we get the desired result.

88. \((1 + x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n\)

    \((x + 1)^n = C_0x^n + C_1x^{n - 1} + C_2x^{n - 2} + \ldots + C_n\)

    Multiplying these two, we get

    \((1 + x)^{2n} = (C_0 + C_1x + C_2x^2 + \ldots + C_nx^n)(C_0x^n + C_1x^{n - 1} + C_2x^{n - 2} + \ldots + C_n)\)

    Coefficient of \(x^{n + r}\) on R.H.S \(= C_0C_r + C_1C_{r + 1} + \ldots + C_{n - r}C_n\)

    Coefficient of \(x^{n + r}\) on L.H.S. \(= {}^{2n}C_{n + r} = \frac{(2n)!}{(n + r)!(n - r)!}\)

    Equating coefficients we get desired result.

89. From 88 recall that \((1 + x)^{2n} = (C_0 + C_1x + C_2x^2 + \ldots + C_nx^n)(C_0x^n + C_1x^{n - 1} + C_2x^{n - 2} + \ldots + C_n)\)

    Equating coefficients of \(x^n,\) we get

    \(C_0^2 + C_1^2 + C_2^2 + \ldots + C_n^2 = \frac{(2n)!}{n!n!}\)

90. \(t_r = r.\frac{{}^nC_r}{{}^nC_{r - 1}} = n - r + 1\)

    \(\sum_{r=1}^n = \frac{n(n + 1)}{2}\)

91. \(\{(1 + x)^n\}^2 = (1 + x)^{2n}\)

    \(({}^nC_0x + {}^nC_1x + {}^nC_2 + \ldots + {}^nC_n)^2 = ({}^{2n}C_0 + {}^{2n}C_1x + {}^{2n}C_2x + \ldots + {}^{2n}C_{2n}x^{2n})\)

    Substituting \(x = 1,\) we get desired equation.

92. This problem is same as previous one, just that instead of \(2, 5\) has been used.

93. \(t_r = (4r + 1){}^nC_{r - 1} = 4n{}^{n - 1}C_{r - 1} + {}^nC_{r - 1}\)

    \(C_0 + 5.C_1 + 9.C_2 + \ldots + (4n + 1).C_n = \sum_{r = 0}^nt_r\)

    \(=4n \sum_{r = 0}^n{}^{n - 1}C_{r - 1} + \sum_{r = 0}^n{}^nC_{r - 1}\)

    \(= 4n.2^{n - 1} + 2^n = (2n + 1)2^n\)

94. L.H.S. \(1 - (1 + x)C_1 + (1 + 2x)C_2 - (1 + 3x)C_3 + \ldots\)

    \(= 1 - C_1 + C_2 - C_3 + \ldots -x(C_1 - 2C_2 + 3C_3 + \ldots)\)

    We know that \((1 + x)^n = C_0 + C_1x + C_2x^2 + C_3x^3 + \ldots\)

    Substituting \(x = -1,\) we get

    \(0 = = 1 - C_1 + C_2 - C_3 + \ldots\)

    Thus our original expression becomes \(-x(C_1 - 2C_2 + 3C_3 + \ldots) = 0\)

    \(C_1 - 2C_2 + 3C_3 + \ldots = 0\)

    \(t_r = (-1)^{r - 1}r.C_r\)

    \(C_1 - 2C_2 + 3C_3 + \ldots = \sum_{r = 1}(-1)^{r - 1}r.C_r = n\sum_{r = 1}{}^{n- 1}C_{r - 1}\)

    \(= n({}^{n - 1}C_0 - {}^{n - 1}C_1 + {}^{n - 1}C_2 + \ldots + (-1)^{n -1}{}^{n - 1}C_{n - 1})\)

    \(= n(1 - 1)^{n - 1} = 0\)

95. \(t_r = (4r - 1)C_{r - 1} = 4rC_{r - 1} - C_{r - 1} = 4n{}^{n - 1}C_{r - 1} - C_{r - 1}\)

    \(3.C_1 + 7.C_2 + 11.C_3 + \ldots + (4n - 1)C_n = 4n \sum_{r = 1}^{n}{}^{n - 1}C_{r - 1} - \sum_{r = 1}^n{}^nC_{r - 1}\)

    \(= (2n - 1)2^n + 1\)

Problem no. 96 and 97 are similar to what we have solved and have been left as exercise.

98. Let \((1 + x - 3x^2)^{2163} = a_0 + a_1x + a_2x^2 + \ldots +a_{6489}^{6489}\)

    Substituting \(x = 1,\) we get

    \(a_0 + a_1 + a_2 + \ldots + a_{6489} = (-1)^2163 = -1\)

99. This problem can be solved by substituting \(1, \omega, \omega^2\) for \(x\) and adding. it has been left as an exercise.

100. \(t_{r + 1} = {}^{10}C_r2^{\frac{10 - r}{2}}3^{\frac{r}{5}},\) where \(r = 0, 1, 2, \ldots, 10\)

     For rational terms \(r =\) a multiple of \(5 = 0, 5, 10\)

     \(10 - r =\) a multiple of \(2 = 0, 2, 4, 6, 8, 10\)

     For both the only common values are \(r = 0, 10\)

     \(\therefore\) sum of rational terms \(= t_1 + t_{11}\)

     \(= 41\)

101. \(\frac{2^{4n}}{15} = \frac{16^n}{15} = \frac{(1 + 15)^n}{15}\)

     \(= \frac{1 + {}^nC_1.15 + {}^nC_2.15^2 + \ldots + {}^nC_n.15^n}{15}\)

     Except first term all others are mutliple of \(15\) so clearly fractional part \(= \frac{1}{15}\)

102. Let \((\sqrt{3} + 1)^{2n} = p + f,\) where \(p\) is the integral part and \(f\) is the fractional part i.e. \(0<f<1\)

     Integer just above \((\sqrt{3} + 1)^{2n} = p + 1\)

     Now \((\sqrt{3} + 1)^{2n} = [(\sqrt{3} + 1)^2]^n = 2^n(2 + \sqrt{3})^n\)

     \(p + f = 2^n(2 + \sqrt{3})^n\)

     Also, \(0 < \sqrt{3} - 1 < 1 \Rightarrow 0 < (\sqrt{3} - 1)^{2n} < 1\)

     Let \(f_1 = \sqrt{3} - 1)^{2n} = 2^n(2 - \sqrt{3})^n, \Rightarrow 0 < f_1 < 0\)

     \(p + f + f_1 = 2^n.2[2^n + {}^nC_22^{n - 2}.3 + \dots]\)

     \(= 2^{n + 1}.\) an integer = an even integer

     \(f + f_1 =\) Even integer - p = an odd integer`

     Also, \(0 < f + f_1 < 2\)

     Clearly, \(f + f_1 = 1\)

     \(\Rightarrow p + 1 = 2^{n + 1}.\) an integer

     Thus, integer next to \((\sqrt{3} + 1)^{2n}\) is divisible by \(2^{n + 1}\)

103. Let \(p\) be the integral part of \(R,\) then \([R] = p\)

     Since \(f = R - [R] = R - p \therefore 0 < f < 1\)

     and \(R = p + f\)

     \(p + f = (5\sqrt{5} + 11)^{2n + 1}\)

     Let \(f_1 = (5\sqrt{5} - 11)^{2n + 1}\)

     We observe that \(5\sqrt{5} - 11 = \frac{4}{5\sqrt{5} + 11}\)

     \(\therefore 0 < f_1 < 1\)

     \(p + f - f_1 = 2[{2n + 1}C_1.(5\sqrt{5})^{2n}.11 + {2n + 1}C_3.(5\sqrt{5})^{2n - 2}.11^3 + \ldots]\)

     \(=\) an even number

     \(f - f_1 =\) en even number \(- p =\) an integer

     \(-1 < f - f_1 < 1\)

     Thus, we can say that \(f - f_1 = 0 \Rightarrow f = f_1\)

     \(\therefore Rf = Rf_1 = r^{2n + 1}\)

104. Let \(x = 101^{50}\) and \(y = 100^{50} + 99^{50}\)

     \(101^{50} - 99^{50} = 100^{50} + 2[{}^{50}C_3.100^{47} + \ldots + {}^{50}C_{49}100]\)

     \(= 100^{50} +\) a positive number

     \(101^{50} - 99^{50} > 100^{50}\)

     \(101^{50} > 100^{50} + 99^{50}\)

105. \(t_1 = \sum_{r = 0}^n (-1)^r{}^nC_r\left(\frac{1}{2}\right) = \left(1 - \frac{1}{2}\right)^n = \frac{1}{2^n}\)

     Similarly, \(t_2 = \frac{1}{2^{2n}}\)

     and \(t_r = \frac{1}{2^{3n}}\)

     \(\therefore\) required sum \(= \frac{1 - \frac{1}{2^{mn}}}{2^n - 1}\)

106. \(32^{32} = (2 + 30)^{32} = 2^{32} + 30k,\) where \(k \in N\)

     Therefore last digits in \(32^{32} =\) last digits in \(2^{32}\)

     \(2^{32} = (2^5)^6.2^2 = 32^6.4 = (2^6 + 30r)4, r\in N\)

     Last digit in \(2^6.4 =\) last digit in \(64.4 = 6\)

     \(\therefore\) last digit in \(32^{32} = 6\)

107. Let \(n = 2m,\) where \(m\) is a positive integer, then \(k = 3m\)

     L.H.S. \(= \sum_{r = 1}^{3m}(-3)^{r - 1}.{}^{6m}C_{2r - 1}\)

     \(={}^{6m}C_1 - {}^{6m}C_3.3 + {}^{6m}C_5.3^2 + \ldots + (-1)^{3m - 1}{}^{6m}C_{6m - 1}.3^{3m - 1}\)

     \(= \frac{1}{\sqrt{3}}[{}^{6m}C_1.\sqrt{3} - {}^{6m}C_3.(\sqrt{3})^3 + {}^{6m}C_5(\sqrt{5})^5 - \ldots + (-1)^{3m - 1}.{}^{6m}C_{6m - 1}(\sqrt{3})^{6m - 1}]\)

     Also, \((-1)^{3m - 1} = -i(i)^{6m - 1}\)

     \((1 + \sqrt{3}i)^{6m} = [1 - {}^6mC_2(\sqrt{3})^2 + {}^{6m}C4(\sqrt{3})^4 + \ldots] + i[{}^{6m}C_1.\sqrt{3} - {}^{6m}C_3(\sqrt{3})^3 + \ldots + {}^{6m}C_{6m -1}(i)^{6m - 2}(\sqrt{3})^{6m - 1}]\)

     However, \((1 + \sqrt{3}i)^6m = \left[2\left(\cos \frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\right]^{6m} = 2^{6m}[\cos 2m\pi + \sin 2m\pi]\)

     \(= 2^{6m}(1 + 0) = 2^{6m}\)

     Thus, coefficient of imaginary part is \(0,\) which proves the desired result.

108. \(t_0 = a^n, t_1 = {}^nC_1a^{n - 1}x, t_2 = {}^nC_2a^{n - 2}x^2, ...\)

     Substituting \(x = ix,\) we get

     \((a + ix)^n = a^n + {}^nC1a^{n - 1}.(ix) + {}nC_2a^{n - 2}(ix)^2 + \ldots + {}^nC_n(ix)^n\)

     \(=(t_0 - t_2 + t_4 - \ldots) + i(t_1 - t_3 + t_5 - \ldots)\)

     Taking modulus and then squaring, we get

     \((t_0 - t_2 + t_4 - \ldots)^2 + (t_1 - t_3 + t_5 - \ldots)^2 = (a^2 + x^2)^n\)

109. Putting \(x = 1,\) we get

     \(3^n = a_0 + a_1 + a_2 + a_3 + \ldots + a_{2n}\)

110. Putting \(x = -1,\) we get

     \(1^n = a_0 - a_1 + a_2 - a_3 + \ldots +(-1)^{2n}a_2n\)

     \(1^n = a_0 - a_1 + a_2 - a_3 + \ldots +a_2n\)

111. Substituting \(x = 1, \omega, \omega^2\) and adding them we get the desired result.

112. \(S_n = \frac{1 - q^{n + 1}}{1 - q}\)

     \(S_n^{'} = \frac{2^{n + 1} - (q + 1)^{n + 1}}{(1 - q).2^n}\)

     Now, \({}^{n + 1}C_1 + {}^{n + 1}C_2.S_1 + {}^{n + 1}C_3.S_2 + \ldots + {}^{n + 1}C_{n + 1}.S_n\)

     \(= {}^{n + 1}C_1\frac{1 - q}{1 - q} + {}^{n + 1}C_2\left(\frac{1 - q^2}{1 - q}\right) + {}^{n + 1}C_3\left(\frac{1 - q^3}{1 - q}\right) + \ldots + {}^{n + 1}C_{n + 1}\left(\frac{1 - q^{n + 1}}{1 - q}\right)\)

     \(= \frac{1}{1 - q}[({}^{n + 1}C_1 + {}^{n + 1}C_2 + \ldots + {}^{n + 1}C_{n + 1}) -({}^{n + 1}C_1.q + {}^{n + 1}C_2.q^2 + \ldots + {}^{n + 1}C_{n + 1}.q^{n + 1})]\)

     \(= \frac{1}{1 - q}[2^{n + 1}- (1 + q)^{n + 1}] = 2^nS_n^{'}\)

113. \((\sqrt[4]{9} + \sqrt[6]{8})^{1000} = (\sqrt{3} + \sqrt{2})^{1000}\)

     Clearly, terms \(0, 2, 4, \ldots, 1000\) will be rational. Thus, total no. of rational terms is \(501.\)

114. Clearly, terms which are divisible by L.C.M. of \(3\) and \(5\) i.e. \(15\) will be rational numbers. Such terms are first and last term.

     Sum of rational terms \(= {}^{15}C_02^5 + {}^{15}C_03^3 = 59\)

115. \(t_3 = {}^5C_2x^4(\log_{10}x)^2 = 1000,000\)

     Clearly \(x = 10\)

116. \(x^3 + 3x^2 - 5 + \frac{3}{x^2} - \frac{1}{x^3}\)

117. Since coefficients of second, third and fourth terms are in A.P., we can write

     \(2{}^mC_2 = {}^mC_1 + {}^mC_3\)

     \(m(m - 1) = m + \frac{m(m - 1)(m - 2)}{6}\)

     \(\Rightarrow m^2 - 9m + 14 = 0 \Rightarrow m = 2, 7\) but since we are dealing with sixth term \(m\) will be 7 and we will need to discard \(2.\)

     \(t_6 = {}7C_5(2^{\log(10 - 3^x)}).2^{(x - 2)\log 3} = 21\)

     \(\Rightarrow 2^{\log(10 - 3^x) + (x - 2)\log 3} = 1\)

     \(\log [10 - 3^x + (x - 2)\log 3] = 2\)

     \(10 - 3^x + (x - 2)\log 3 = 100\)

     Solving this yields

     \(\Rightarrow x = 0, 2\)

118. This problem is similar to previous one and has been left as an exercise.

119. Given \(\frac{1}{(81)^n} - \frac{10}{(81)^n}{}^{2n}C_1 + \frac{10^2}{(81)^n}{}^{2n}C_2 - \frac{10^3}{(81)^n}{}^{2n}C_3 + \ldots + \frac{10^{2n}}{(81)^n} = 1\)

     L.H.S. \(= \frac{1}{81^n}(1 - 10)^{2n}\)

     \(= \frac{1}{81^n}(-9^2)^n = \frac{1}{81^n}81^n = 1\)

120. \(\because {}^nC_r = {}^nC){n - r}\)

     \(\therefore S_n = {}^nC_0 - {}^nC_1\frac{2}{3} + {}^nC_2\left(\frac{2}{3}\right)^2 - \ldots + (-1)^n\left(\frac{2}{3}\right){}^nC_n\)

     \(= \left(1 - \frac{2}{3}\right)^n = \frac{1}{3^n}\)

     \(\therefore \lim_{n \to \infty}S_n = 0\)

121. This problem can be solved like 103 and has been left as an exercise.

122. \(\sum_{r = 0}^n(-1)^r.{}^nC_r\left[\frac{1}{2^r} + \frac{3^r}{2^{2r}} + \frac{7^r}{w^{3r}} + \ldots~\text{up to}~n~\text{terms}\right]\)

     \(= \left(1 - \frac{1}{2}\right)^n + \left(1 - \frac{1}{4}\right)^n + \left(1 - \frac{1}{8}\right)^n + \ldots n\) terms

     \(= \frac{1}{2^n} + \frac{1}{2^{2n}} + \frac{1}{2^{3n}} + \ldots n\) terms

     Summing this G.P. yields

     \(= \frac{1}{2^n - 1} - \frac{1}{2^{n^2}(2^n - 1)}\)

123. \(17^{256} = 289^{128} = (290)^{128}\)

     \(= 1000m + {}^{128}C_{126}290^2 - {}^{128}C_{127}290 + 1 [m \in I]\)

     \(= 1000(m + 683527) + 681\)

     Thus the digits are \(6, 8\) and \(1.\)

124. \((n + 1)^n = n^n\left(1 + \frac{1}{n}\right)^n\)

     Upon expansion(as shown in following problem) you will find that \(\left(1 + \frac{1}{n}\right)^n < n\)

     Thus, \(n^{n + 1} > (n + 1)^n\)

125. \(\left(1 + \frac{1}{n}\right)^n = 1 + 1 + \frac{1}{2!}\left(1 - \frac{1}{n}\right) + \frac{1}{3!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) + \ldots\)

     \(< 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots + \frac{1}{n!}\)

     \(< 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^{n - 1}}\)

     \(= 3 - \frac{1}{2^{n - 1}}\)

     Thus, we have proven the desired inequality.

126. We will make use of the fact that \(x^n - y^n\) is divisible by \(x - y\)

     \((1992^{1998} - 1955^{1998}) - (1938^{1998} - 1901^{1998})\) is divisible by \(37\)

     \((1992^{1998} - 1938^{1998}) - (1955^{1998} - 1901^{1998})\) is divisible by \(54\)

     \(\therefore 1992^{1998} - 1955^{1998} - 1938^{1998} + 1901^{1998}\) is divisible by \(37*54\) i.e. \(1998\)

127. Given expression is \((50 + 3)^{53} - {30+ 3}^{33}\)

     \(= 50p + {}^{53}C_{53}3^{53} - 30q - {}^{33}C_{33}3^{33}, p, q \in I\)

     Thus, now we have to prove that \(3^{53} - 3^{33}\) is divisible by \(10\)

     \(3^{33}(3^{20} - 1)\) now if you see carefully \(3^{20} = 81^{5}\) which will awlays have last digit as 1. Thus, \(3^{20} - 1\) will be always divisible by \(10\) making out original expression also divisible by 10.

128. \((1 + x)^{m + 1} = {}^{m + 1}C_0 + {}^{m + 1}C_1x + {}^{m + 1}C_2x^2 + \ldots + {}^{m + 1}C_mx^m + {}{m + 1}C_{m + 1}x^{m + 1}\)

     \((1 + x)^{m + 1} - 1 - x^{m + 1} = {}^{m + 1}C_1x + {}^{m + 1}C_2x^2 + \ldots + {}^{m + 1}C_mx^m\)

     Substituting \(x = 1, 2, 3, 4, \ldots, n\) in the above expression and adding, we get

     \(= (n + 1)^{m + 1} - (n + 1)\)

129. \(\sum_{i = 1}^k\sum_{k = 1}^n{}^nC_k{}^kC_i\)

     \(= \sum_{k = 1}^n({}^nC_k).({}^kC_1) + \sum_{k = 1}^n({}^nC_k).({}^kC_2) + \ldots + {}^nC_n.{}^nC_n\)

     \(= {}^nC_1.{}^1C_1 + {}^nC_2({}^2C_1 + {}^2C_2) + \ldots + {}^nC_n({}^nC_1 + {}^nC_2 + \ldots + {}^nc_n)\)

     \(= {}^nC_1(2 - 1) + {}^nC_2.(2^2 - 1) + \ldots + {}^nC_n(2^n - 1)\)

     \(= {2 + 1}^n - 1 + {1 + 1}^n - 1 = 3^n - 2^n\)

130. Though it may appear that this problem is dependent on logarithmic manipulation but sych is not the case. We will prove it for general value \(z\) rather than \(10\)

     \(\sum_{r=0}^n{(-1)^r\binom{n}{r}\frac{1+r\log z}{(1+\log z^n)^r}}\)

     \(=\sum_{r=0}^n(-1)^r\binom{n}{r}\frac{1}{\left(1+\log z^n\right)^r} +\sum_{r=1}^n(-1)^r\binom{n-1}{r-1}\frac{n\log z}{\left(1+\log z^n\right)^r}\)

     \(=1+\sum_{r=1}^{n-1}(-1)^r\left(\binom{n-1}{r}+\binom{n-1}{r-1}\right)\frac{1}{\left(1+\log z^n\right)^r} +(-1)^n\frac{1}{(1+\log z^n)^r}-\sum_{r=0}^{n-1}(-1)^r\binom{n-1}{r}\frac{\log z^n}{\left(1+\log z^n\right)^{r+1}}\)

     \(=1+\sum_{r=1}^{n-1}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^r} -\sum_{r=0}^{n-2}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^{r+1}} +(-1)^n\frac{1}{(1+\log z^n)^r}-\sum_{r=0}^{n-1}(-1)^r\binom{n-1}{r}\frac{\log z^n}{\left(1+\log z^n\right)^{r+1}}\)

     \(=1+\sum_{r=1}^{n-1}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^r}+(-1)^n\frac{1}{(1+\log z^n)^r} -\sum_{r=0}^{n-2}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^{r}}+(-1)^n\frac{\log z^n}{(1+\log z^n)^n}\)

     \(=1+(-1)^{n-1}\frac{1}{(1+\log z^n)^{n-1}}+(-1)^n\frac{1}{(1+\log z^n)^{n-1}}-1+(-1)^n\frac{\log z}{(1+\log z^n)^n}\)

     \({=0}\)

131. \(32^{32} = (2^5)^{32} = 2^{160} = {3 - 1}^{160} = 3m + 1, m \in N\)

     \({32}^{32^{32}} = 32^{3m + 1} = 2^{15m + 5} = 2^{3{5m + 1}}.2^2\)

     \(=8^{5m + 1}.4 = 32.8^{5m} = 32(1 + 7)^{5m}\)

     \(=32(1 + 7k), k \in N = 4 + 28 + 7(32k) = 4 + 7r, r \in N\)

     Thus, remainder is \(4\) when divided by \(7\)

132. Let \(t = x - 3,\) then \(x - 2 = 1 + t\)

     \(\sum_{r=0}^{2n}a_r(x - 2)^r = \sum_{r=0}^{2n}b_r(x - r)^r\)

     \(\Rightarrow \sum_{r=0}^{2n}a_r(1 + t)^r = \sum_{r=0}^{2n}b_rt^r\)

     Equating coefficients of \(t^n\) we obtain desired result.

133. Given expression is \(= (1 + x)^{1000}\left[1 + 2\frac{x}{1 + x} + 3\left(\frac{x}{1 + x}\right)^2 + \ldots + 1001\left(\frac{x}{1 + x}\right)^{1000}\right]\)

     The series is arithmetico-geometric series. Solving it yields

     \(= (1 + x)^1002 - x^{1001}(1 + x) - 1001.x^{1001}\)

     Required coefficient of \(x^{50} = {}^{1000}C_50\)

134. L.H.S. = coeff. of \(x^n\) in \((1 + x)^n + (1 + x)^{n + 1} + \ldots + (1 + x)^{n + k}\)

     \((1 + x)^n + (1 + x)^{n + 1} + \ldots + (1 + x)^{n + k}\)

     \(= (1 + x)^n\left[\frac{1 + x}^{k + 1 - 1}{x}\right]\)

     \(= \frac{1}{x}(1 + x)^{n + k + 1} - \frac{1}{x}(1 + x)^n\)

     Coeff. of \(x^n = {}^{n + k + 1}C_{n + 1}\)

     Hence, we have proved the desired equation.

135. Let \(S = (x + 2x^2 + 3x^3 + \ldots + nx^n)\)

     \(xS = x^2 + 2x^3 + \ldots + (n - 1)x^n + nx^{n + 1}\)

     \(S = x\frac{1 - x^n}{(1 - x)^2} - n\frac{x^{n + 1}}{1 - x}\)

     \((1 + x + 2x^2 + 3x^3 + \ldots + nx^n)^2 = \left[1 + \frac{x(1 - x^n)}{1 - x}^2 - \frac{nx^{n + 1}}{1 - x}\right]^2\)

     Required coeff. of \(x^n\) = coeff. of \(x^n\) in \(\left(1 + \frac{x}{(1 - x)^2}\right)^2\) [leaving terms containing powers of \(x\) greater than \(n\)]

     \(=\) coeff. of \(x^n\) in \(\left[1 + \frac{2x}{(1 - x)^2} + \frac{x^2}{(1 - x)^4}\right]\)

     Solving this yields answer as \(\frac{n(n ^2 + 11)}{6}\)

136. Let \(S = 1 + (1 + x) + (1 + x)^2 + \ldots + (1 + x)^n\)

     \((1 + x)S = (1 + x) + (1 + x)^2 + \ldots + (1 + x)^n + (1 + x)^{n + 1}\)

     Subtracting, we get

     \(xS = (1 + x)^{n + 1} - 1\)

     \(\therefore ` coff. of :math:`x^k\) in \(S = {}^{n + 1}C_{k + 1}\)

137. Let the expression be \(E = (x + 1)^n + (x + 1)^{n - 1}(x + 2) + (x + 1)^{n - 2}(x + 2)^2 + \ldots + (x + 2)^n\)

     We know that \((x - y)(x^{n - 1} + x^{n - 2}y + \ldots + y) = x^n - y^n\)

     Thus, \(E(x + 2 - x - 1) = (x + 2)^{n + 1} - (x + 1)^{n + 1}\)

     Coeff. of \(x^3\) in \((x + 2)^{n + 1} - (x + 1)^{n + 1}\) is

     \(= {}^{n + 1}C_3.2^{n - 2} - {}^{n + 1}C_3\)

138. \(\left(\frac{a + 1}{a^{\frac{2}{3}} - a^{\frac{1}{3}} + 1} - \frac{a - 1}{a - a^{\frac{1}{2}}}\right)^{10}\)

     \(= (\sqrt[3]{a} - \frac{1}{\sqrt{a}})^{10}\)

     \(t_{r + 1} = {}^{10}C_r a^{\frac{10 - r}{3}}a^{-\frac{r}{2}}\)

     Since the term has to be independent of \(a, \Rightarrow \frac{10 - r}{3} - \frac{r}{2} \Rightarrow r = 4\)

     Thus \(t_5 = {}^{10}C_4 = 210\)

139. Coeff. of \(x^2\) in \(\left(x + \frac{1}{x}\right)^{10}(1 - x + 2x^2) =\) coeff. of \(x^2\) in \(\left(x + \frac{1}{x}\right)^{10}\) - coeff. of \(x\) in \(\left(x + \frac{1}{x}\right)^{10}\) - 2 * coeff. of term independent of \(x\) in \(\left(x + \frac{1}{x}\right)^{10}\)

     \((r + 1)^{th}\) term in \(\left(x + \frac{1}{x}\right)^{10} = {}^{10}C_rx^{10 - r}x^{-r} = {}^{10}C_rx^{10 - 2r}\)

     Coeff. of \(x^2\) means \(10 - 2r = 2 \Rightarrow r = 4.\) Thus, coeff. \(= {}^10C_4 = 210\)

     Coeff. of \(x\) means \(10 - 2r = 1\) which makes \(r\) a fraction. Thus, coeff. \(= 0\)

     Coeff. of term indepdent of \(x\) means \(10 -2r = 0 \Rightarrow r = 5.\) Thus, coeff. \(= 2.{}^10C_5 = 504\)

     Thus, final coeff. \(= 210 + 504 = 714\)

140. Coeff. of \(x^4\) in \((1 + x - 2x^2)^6 =\) coeff. of \(x^r\) in \((1 + x(1 - 2x))^6\)

     Thus coefficient of \(x^4\) will occur in \(3^{rd}\) terms onward.

     Adding coefficients we get \(-45\) as our answer.

141. We have \((1 + x + 2x^3)\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9\)

     \(= (1 + x + 2x^3)\left[\left(\frac{3}{2}x^2\right)^9 - {}^9C_1\left(\frac{2}{2}x^2\right)^8\frac{1}{3x} + \ldots + (-1)^9\left(\frac{1}{3x}\right)^9\right]\)

     Thus, the term independent of \(x\) in the expansion is \(1a_0 + 1a_1 + 2a_3\) where \(a_m\) is the coefficient of \(x^m\) in the second bracket \([]\) of previous equation. Now, \((r + 1)^{th}\) term in \([]\) of previous equation is

     \({}^9C_r\left(\frac{3}{2}x^2\right)^{9 - r}\left(-\frac{1}{3x}\right)^r = (-1)^r{}^9C_r\left(\frac{3}{2}\right)^{9 - r}\left(\frac{1}{3^r}\right)x^{18 -r}\)

     \(\therefore a_{18 - 3r} =\) coeff. of \(x^{18 - 3r}\)

     For \(a_0, 18 - 3r = 0 \Rightarrow r = 6 \Rightarrow a_0 = {}^9C_6 \frac{3^3}{2^3}\frac{1}{3^6} = \frac{7}{18}\)

     For \(a_1, 18 - 3s = 1, \Rightarrow r = \frac{19}{3}\) which is fractional. \(\therefore a_1 = 0\)

     For \(a_3, 18 - 3r = 3 \Rightarrow r - 7 \Rightarrow a_3 = -{}^9C_7\left(\frac{3}{2}\right)^2\frac{1}{3^7} = -\frac{1}{27}\)

     Thus, required term \(= 1.\frac{7}{18} + .0 + 2.\frac{-1}{27} = \frac{17}{52}\)

142. Given \(\left(x^2 + \frac{1}{x^3}\right)^7(2 - x)^{10}\) and we have to find term independent of \(x\) in this.

     Coeff of term independent of \(x\) to be found in \(\frac{1}{x^{21}}(x^5 + 1)^7(2 - x)^{10}\) i.e. coeff. of \(x^{21}\) in \((x^5 + 1)^7(2 - x)^{10}\)

     \((x^5 + 1)^7\) will have coeff. of powers of \(x\) where powers will be \(35, 30, 25, 20, \ldots, 0\) while \((2 - x)^{10}\) will have powers of \(1, 2, 3, \ldots, 10\)

     Clearly, combinations of \(20\) and \(1\) and \(15\) and \(6\) satisfy our needs.

     Thus, computing these terms leads to answer of \(-61600\)

143. We have to find term independent of \(x\) in \((1 + x + x^{-2} + x^{-3})^{10}\)

     Coeff. of term independent of \(x\) in \(\frac{1}{x^30}(1 + x)^{10}(1 + x^3)^{10}\)

     \((1 + x)^{10}\) will have powers of \(x\) in \(0, 1, 2, 3, \ldots, 10\) while \((1 + x^3)^{10}\) will have powers of \(x\) in \(0, 3, 6, 9, \ldots, 30\)

     Thus, combinaions of \((0, 30), (3, 27), (6, 24), (9, 21)\) are the combinations which will satisfy our needs, where first number is power of \(x\) in \((1 + x)^{10}\) and second number is power of \(x\) in \((1 + x^3)^{10}\)

     Thus coeff. is \(= {}^{10}C_1.{}^{30}C_{30} + {}^{10}C_3.{}^{30}C_{27} + {}^{10}C_6.{}^{30}C_{24} + {}^{10}C_9.{}^{30}C_{21}\)

144. Given \((1 + x^2)^2(1 + x)^n = \sum_{k = 0}^{n + 4}a_kx^k\)

     \(a_1, a_2, a_3\) are coefficients of \(x, x^2, x^3\) respectively.

     Thus, we can find \(a_1, a_2, a_3\) like we did in 143 and then apply arithmetic progression confition \(2a_2 = a_1 + a_3\) to find \(n\)

145. We have to prove that \({}^mC_1 + {}^{m + 1}C_2 + {}^{m + 2}C_3 + \ldots + {}^{m + n - 1}C_n = {}^nC_1 + {}^{n + 1}C_2 + {}^{n + 2}C_3 + \ldots + {}^{n + m - 1}C_n\)

     Keep in mind that \({}^mC_0 = {}^nC_0\)

     Adding \({}^mC_0\) to L.H.S., we get \({}^mC_0 + {}^mC_1 + {}^{m + 1}C_2 + {}^{m + 2}C_3 + \ldots + {}^{m + n - 1}C_n\)

     Now we know that \({}^nC_r + {}^nC_{r - 1} = {}^{n + 1}C_r\) applying the repeatedly we obtain L.H.S. as \({}^{m + n}C_n\)

     Similarly adding \({}^nC_0\) to R.H.S. and applying above formula repeatedly we obtain R.H.S. as \({}^{m + n}C_m\)

     Clearly, \({}^{m + n}C_m = {}^{m + n}C_n\)

146. Let us solve these one by one.

     1. We observe that \(1 + x + x^2 = (x + \omega)(x + \overline{\omega})\) where \(\omega\) and \(\overline{\omega}\) are cube root of unity not equal to \(-1\)

        \(\sum_{r=0}^{2n}c_rx^r=(x^2+x+1)^n=(x+w)^n(x+\overline{w})^n=\sum_{k=0}^n\binom{n}{k}x^kw^{n-k}\cdot \sum_{l=0}^n\binom{n}{\ell}x^l \overline{w}^{n-l}\)

        Thus, \(a_r=\sum_{j=0}^r\binom{n}{j}\binom{n}{r-j}w^{n-j}\overline{w}^{\,n-(r-j)}=\sum_{j=0}^r\binom{n}{j}\binom{n}{r-j}w^{r-2j}\)

        Thus, \(a_r = a_{2n - r}\)

2. \(a_r = a_{2n - r}\) from previous part.

   Substituting \(x = 1\)

   \(a_0 + a_1 + \ldots + a_2n = 2(a_0) + 2(a _1) + \ldots + 2a_{n - 1} + a_n = 3^n\)

   \(\Rightarrow a_0 + a_1 + a_2 + \ldots + a_{n - 1} = \frac{1}{2}(3^n - a_n)\)

3. Differentiating and using result obtained in part 1 we can prove this.

147. Given \(\frac{(1 - x^3)^n}{(1 - x)^{3n}} = \sum_{r = 0}^na_r\frac{x^r}{(1 - x)^{2r}}\)

     \(\Rightarrow \left(\frac{1 + x + x^2}{(1 - x)^2}\right)^n = \sum_{r = 0}^na_r\alpha^r,\) where \(\alpha = \frac{x}{(1 - x)^2}\)

     \((1 + 3\alpha)^n = \sum_{r = 0}^na_r\alpha^r\)

     Coeff of \(\alpha^r = a_r = {}^nC_r3^r\)

148. Coeff. of middle term of \((1+ x)^{2n} = {}^{2n}C_n\)

     Coeff. of \(x^n\) in \((1 + x)^{2n - 1} = {}^{2n - 1}C_n\)

     Clearly, \(2.{}^{2n - 1}C_n = {}^{2n}C_n\)

149. Coeff. of terms in the expansion of \((x + y)^{200}\) are \({}^{200}C_1, {}^{200}C_2, {}^{200}C_3, \ldots, {}^{200}C_{200}\)

     Since middle term has greatest coefficient, therefore \(r = 100\)

150. Let committees of \(r\) persons be made out of \(20\) persons. Then number of committees \(= {}^20C_r\)

     Since middle term has greatest coefficient therefore \({}^{20}C_r\) will be maximum when \(r = 10\)

     Therefore, \(10\) persons should be chosen for maximum no. of committees.

151. This problem is similar to 149 and 150 and has been left as an exercise.

152. \((3 + 2x)^7\) will have \(8\) terms with \(4^{th}\) and \(5^{th}\) terms as middle terms. We know that when there are two middle terms coefficients are equal. Therefore, these two middle terms are consecutive terms which have equal coefficient.

153. Let \((1 + 5x^2 - 7x^3)^{2000} = a_0 + a_1x + a_2x + \ldots + a_{6000}x^{6000}\)

     Substituting \(x = 1,\) we get

     \((1 + 5 - 7)^{2000} =\) sum of coefficients \(= 1\)

154. Substitutin \(x^{-\frac{x}{4}} = 1\) and \(x^{\frac{5x}{4}} = 1\)

     \((1 + 1)^n = 64 \Rightarrow n = 6\)

     Hence greatest term = middle term = \(4^{th}\) term

     According to question \(t_4 = (n - 1)+ t_3\)

     Solving this gives us, \([\alpha] = 0\)

155. Let \(S\) be the sum of coeff. in \((5p - 4q)^n\). Substituting \(p =1, q = 1,\) we get

     \(S = (5 - 4)^n = 1\)

156. Let \(S\) be the sum of coeff. in \((5p - 4q)^n\). Substituting \(x = 1,\) we get

     \(S = (1 - 3 + 1)^{201}(1 + 5 - 5)^{503} = -1\)

157. Substituing \(x =1\) reduces the two expansions as \((t - 1)^n\) and \((1 - t)^n\)

     Clearly, if \(n\) is odd then sign will differe except when \(t = 1.\) However, if \(n\) is even both expansions will be positive and \(t\) can assume any real value.

158. Substituting \(x = 1, i, -i\) and then multiplying when \(x = i, -i\) we obtain the desired result.

159. Let \(r^{th}\) term be the greatest term.

     \(t_r = \sqrt{3}\left[{}^{20}C_{r - 1}\left(\frac{1}{\sqrt{3}}\right)^{r - 1}\right]\)

     \(t_{r + 1} = \sqrt{3}\left[{}^{20}C_r\left(\frac{1}{\sqrt{3}}\right)^r\right]\)

     \(\frac{t_r}{t_{r + 1}} = \frac{{}^{20}C_{r - 1}}{{}^{20}C_r}\sqrt{r}\)

     \(= \frac{r}{21 - r}\sqrt{3} \geq 1 \Rightarrow r = 7.69\)

     Also, \(\frac{t_{r -1}}{t_r} \leq 1\) gives us \(r = 8.5\)

     Hence, \(8^{th}\) term will be greatest term and \(t_8 = \frac{25840}{9}\)

160. Since \(t_{11}\) is G.M. of \(t_8\) and \(t_{12}\)

     \(t_{11} = \sqrt{t_8.t_{12}}\)

     \(\left(\frac{15!}{10!5!}x^5a^{10}\right)^2 = \frac{15!15!}{7!8!11!4!}x^{12}a^{18}\)

     \(\Rightarrow \frac{x}{a} = \sqrt{\frac{77}{75}}\)

     Let \(r^{th}\) term be the greatest term. Now,

     \(t_r = {}^{15}C_{r - 1}x^{16 - r}a^{r - 1}\)

     \(t_{r + 1} = {}^{15C_r}x^{15 - r}a^r\)

     \(\frac{t_r}{t_{r + 1}} = \frac{r}{16 - r}\frac{x}{a} \geq 1\)

     \(r \geq 7.947\)

     Hence, \(8^{th}\) term is the greatest term.

161. \(t_{n + 1} = {}^{2n}C_nx^n, t_{n + 2} = {}^{2n}C_{n + 1}x^{n + 1}, t_n = {}^{2n}C_{n - 1}x^{n - 1}\)

     \(\frac{t_{n + 1}}}{t_{n + 2}} = \frac{n + 1}{n}.\frac{1}{x} > 1 \Rightarrow x < \frac{n + 1}{n}\)

     Also, \(t_{n + 1} = \frac{n + 1}{n}x > 1 \Rightarrow x > \frac{n}{n + 1}\)

     Thus, greatest term will have greatest coefficient if and only if \(x \in \left(\frac{n}{n + 1}, \frac{n + 1}{n}\right)\)

     Given \(x \in \left(\frac{10}{11}, \frac{11}{10}\right)\)

     Thus, \(n = 10\)

     \(t_4 = \frac{n}{4} = \frac{5}{2}\)

     \(\Rightarrow {}^{m}C_3(kx)^{m - 3}\frac{1}{x^3} = \frac{5}{2}\)

     R.H.S. is independent of \(x, \Rightarrow m - 6 = 0, m = 6\)

     \({}^{6}C_3k^3 = \frac{5}{2}\Rightarrow k = \frac{1}{2}\)

     \(\Rightarrow mk = 3\)

162. This problem is similar to last problem and has been left as an exercise. The range is \(2 < x < \frac{64}{21}\)

163. \(9^n + 7 = (8 + 1)^n + 7 = {}^nC_0.8^n + {}^nC_18{n - 1} + \ldots + {}^nC_{n - 1}8 + {}^nC_n + 7\)

     \(= 8k + 1 + 7, k \in N = 8(k + 1)\)

     Thus, the number is divisible by \(8\)

164. Given expression is \(3^{2n + 1} + 2^{n + 2}\) which can be rewritten as \(3.(7 + 2)^n + 4.2^n,\) which upon expansion yields

     \(3({}^nC_07^n + {}^nC_17^{n - 1}2 + \ldots + {}^nC_n2^n) + 4.2^n\)

     \(= 7k + 2^n(3 + 4), k \in N\)

     The above expression is divisible by \(7.\)

165. Let the binomial expansion be \((x + y)^n\) and \(a, b, c\) be the coefficients of \(r^{th}, (r + 1)^{th}, (r + 2)^{th}\) terms respectively.

     Then, \(a = {}^nC_{r - 1}, b = {}^nC_r, c = {}^nC_{r + 1}\)

     The descriminant of given quadratic equation is \(D = b^2 - 4ac\)

     Substituting values of \(a, b, c\) and simplifying we obtain

     \(D = 4({}^nC_r)^2 \frac{n + 1}{(n - r + 1)(r + 1)}\) where \(r\) is non-negative integer.

     Clearly \(D > 0,\) hence roots of given equation are real and unequal.

166. This problem is similar to previous one and has been left as an exercise.

167. Calculus Method:

     \((1 + x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n\)

     Multiplying with \(x\) and differentiating w.r.t. \(x,\) we get

     \((1 + x)^n + nx(1 + x)^{n - 1} = C_0 + 2xC_1 + 3x^2C_2 + (n + 1)x^nC_n\)

     Substituting \(x = -1,\) we obtain

     \(C_n - 2.C_1 + 3.C_2 - \ldots +(-1)^n(n + 1)C_n = 0\)

     Second Method:

     \(t_r = (-1)^{r - 1}r.{}^nC_{r - 1} = (-1)^{r - 1}(r- 1 + ).{}^nC_{r - 1} = (-1)^{r - 1}n.{}^{n - 1}C_{r - 2} + (-1)^{r - 1}.{}^nC_{r - 1}\)

     \(\sum_{r = 1}^{n + 1} = -n({}^{n - 1}C_0 - {}^{n - 1}C_1 + {}^{n - 1}C_2 - \ldots + {-1}^{n - 1}.{}^{n - 1}C_{n - 1}) + ({}^nC_0 - {}^nC_1 + \ldots + (-1)^n.{}^nC_n)\)

     \(= -n(1 - 1)^{n - 1} + (1 - 1)^n = 0\)

168. Calculus Method:

     We know that \((1 + x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n\)

     Substituting \(x = x^2\) and multiplying with \(x,\) we get

     \(x(1 + x^2)^n = C_0x + C_1.x^3 + C_2.x^5 + \ldots + (2n + 1)x^{2n}.C_n\)

     Differentiating both sides w.r.t \(x,\) and substituting \(x = i,\) we get

     \(C_0 -3.C_1 + 5.C_2 - \ldots + (-1)^n(2n + 1)C_n = (1 - 1)^n + i.n.(1 - 1)^{n - 1}.2i = 0\)

     Second Method:

     \(t_r = = (-1)^{r - 1}(2r - 1).C_{r - 1} = (-1)^r[2(r - 1) + 1]{}^nC_{r - 1}\)

     \(= 2(-1)^{r - 1}.n{}^{n - 1}C_{r - 1} + (-1)^{r - 1}{}^nC_{r - 1}\)

     \(\sum_{r = 1}^{n + 1}t_r = -2n[{}^{n - 1}C_0 - {}^{n - 1}C_1 + \ldots + (-1)^{n - 1}{}^{n - 1}C_{n - 1}] + [{}^{n}C_0 - {}^nC_1 + \ldots + (-1)^n{}^nC_n]\)

     \(= -2n(1 - 1)^{n - 1} + (1 - 1)^n = 0\)

169. Calculus Method:

     L.H.S. = \(a[C_0 - C_1 + C_2 - \ldots + (-1)^n.C_n] + [1.C_1 - 2C_2 + 3.C_3 - \ldots + (-1)^n(-n)C_n]\)

     \(= a(1 - 1)^n + 1.C_1 - 2C_2 + 3.C_3 - \ldots + (-1)^n(-n)C_n\)

     Given, \((1 + x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n\)

     Differentiating w.r.t. \(x\) and substuting \(x = -1,\) we get

     \(1.C_1 - 2C_2 + 3.C_3 - \ldots + (-1)^n(-n)C_n = (1 - 1)^n = 0\)

     Hence, desired equality is proved.

     Second Method:

     \(t_r = (-1)^{r - 1}[a - (r - 1)]{}^nC_{r -1}\)

     \(= a(-1)^{r - 1}.{}^nC_{r - 1} - (-1)^{r - 1}n.{}^{n - 1}C_{r -1}\)

     This can be proven to be \(0\) like previous problems.

170. \(t_{r + 1} = r^r.{}^nC_rp^rq^{n - r}\)

     \(= r.n.{}^{n - 1}C_{r - 1}p^rq^{n - r}\)

     \(= n(r - 1 + 1){}^{n - 1}C_{r - 1}p^rq^{n - r}\)

     \(= n[(n - 1).{}^{n - 2}C_{r - 2} + {}^{n - 1}C_{r - 1}]p^rq^{n - r}\)

     L.H.S. \(= \sum_{r = 0}^n t_{r + 1}\)

     \(= n(n - 1)p^2\sum_{r = 0}^n {}^{n - 2}C_{r - 2}q^{n -2 - (r - 2)} + np\sum_{r = 0}^{n}{}^{n - 1}C_{r - 1}p^{r - 1}q^{n - 1 - (r - 1)}\)

     \(= n(n - 1)p^2(p + q)^{n - 2} + np(p + q)^{n - 1}\)

     \(= n(n - 1)p^2 + np [\because p + q = 1]\)

     \(= n^2p^2 + npq\)

171. \((1 + x)^{10} = C_0 + C_1x + C_2x^2 + \ldots + C_10x^{10}\)

     Integrating between limits \(0\) and \(2\) gives the desired result.

172. Calculus Method:

     \((1 - x)^n = 1 - {}^nC_1x + {}^nC_2x^2 - \ldots + (-1)^{n - 1}{}^nC_nx^n\)

     \(\frac{1 - (1 - x)^n}{x} = {}^nC_1 - {}^nC_2x + \ldots + (-1)^{n - 1}C_{n}x^{n - 1} = \frac{x - (1 - x)^n}{x}\)

     Integrating between limits \(0\) and \(1,\) we get

     \(\left[{}^nC_1x - {}^nC_2\frac{x^2}{2} + \ldots + (-1)^{n - 1}\frac{x^n}{n}\right]_0^1 = \int_{0}^1\frac{1 - (1 - x)^n}{x}dx\)

     Let \(z = 1 - x\) then R.H.S. becomes \(\int_1^0 -\frac{1 - z^n}{1 - z}dz\)

     \(= \int_0^1(1 + z + z^2 + \ldots + z^{n - 1})dz\)

     \(= \left[z + \frac{z^2}{2} + \ldots + \frac{z^n}{n}\right]_0^1\)

     \(= 1 + \frac{1}{2} +\frac{1}{3} + \ldots + \frac{1}{n}\)

     Seccond Method:

     Let \(S_n = C_1 - \frac{1}{2}C_2 + \frac{1}{3}C_3 - \ldots + \frac{(1)^nC_n}{n}\)

     \(S_n = n - \frac{1}{2}\frac{n(n - 1)}{2!} + \frac{1}{3}\frac{n(n - 1)(n - 2)}{3!} + \ldots\)

     \(= (n - 1 + 1) - \frac{1}{2}\frac{(n - 1)(n - 2 + 2)}{2!} + \frac{(n - 1)(n - 2)(n - 3 + 3)}{3!} + \ldots\)

     \(= S_{n - 1} + \frac{1}{n}\left[n - \frac{n(n - 1)}{2!} + \frac{n(n - 1)(n - 2)}{3!} + \ldots\right]\)

     \(= S_{n - 1} - \frac{1}{n}[C_0 - C1 + C_2 + \ldots -1] = S_{n - 1} - \frac{1}{n}[(1 - 1)^n - 1]\)

     \(S_{n - 1} + \frac{1}{n} \therefore S_n - S_{n - 1} = 1\)

     Similarly \(S_{n - 1} - S_{n - 2} = \frac{1}{n - 2}\)

     \(S_{n - 2} - S_{n - 3} = \frac{1}{n - 3}\)

     \(\ldots\)

     \(S_2 - S_1 = \frac{1}{3}\)

     \(S_1 = 1\)

     Adding we get \(S_n = 1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{n}\)

173. \((1 - x)^n = 1 - {}^nC_1x + {}^nC_2x^2 - \ldots + (-1)^{n - 1}{}^nC_nx^n\)

     Putting \(x = x^4,\) and integrating between limits \(0\) and \(1,\) we get

     \(C_0 - \frac{C_1}{5} + \frac{C_2}{9} - \ldots + (-1)^n\frac{C_n}{4n + 1} = \int_{0}^1(1 - x^4)^ndx\)

     It can be proven that \(\int_0^1(1 - x^4)^ndx = \frac{4n.n!}{1.5.9\ldots (4n + 1)}\) [Refer to any book on calculus]

174. \((1 - x)^n = 1 - {}^nC_1x + {}^nC_2x^2 - \ldots + (-1)^{n - 1}{}^nC_nx^n\)

     Multiplying both sides with \(x^{n - 1},\) and integrating with limit \(0\) and \(1,\) we get

     \(\frac{C_0}{n} - \frac{C_1}{n + 1} + \frac{C_2}{n + 2} - \ldots + (-1)^n\frac{C_n}{2n} = \int_0^1 x^{n - 1}(1 - x)^ndx\)

     Integrating R.H.S. by parts we obtain the desired result.

175. L.H.S. \(= \left(\frac{C_0}{n} - \frac{C_0}{n + 1}\right) - \left(\frac{C_1}{n + 1} - \frac{C_1}{n + 2}\right) + \ldots + (-1)^{n - 1}\left(\frac{C_n}{2n} - \frac{C_n}{2n + 1}\right)\)

     \(= \left(\frac{C_0}{n} - \frac{C_1}{n + 1} + \frac{C_2}{n + 2} - \ldots + (-1)^n\frac{C_n}{2n}\right) - \left(\frac{C_0}{n + 1} - \frac{C_1}{n + 2} + \ldots + (-1)^n\frac{C_n}{2n + 1}\right)\)

     Now this can be solved like 174.

176. This problem is similar to 174 and has been left as an exercise.

177. \((1 + x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n\)

     \((1 - x)^n = 1 - {}^nC_1x + {}^nC_2x^2 - \ldots + (-1)^{n - 1}{}^nC_nx^n\)

     Multiplying these two we get

     \((C_0 + C_1x + C_2x^2 + \ldots + C_nx^n)(1 - {}^nC_1x + {}^nC_2x^2 - \ldots + (-1)^{n - 1}{}^nC_nx^n) = (1 - x^2)^n\)

     Coeff. of \(x^n\) on L.H.S. \(= C_0^2 - C_1^2 + C_2^2 - \ldots + (-1)^nC_n^2\)

     R.H.S. \(= C_0 + (-1)^1C_1x^2 + C_2x^2 + \ldots\)

     \(=\) an expression having even powers of \(x\)

     Thus, if \(n\) is odd, coeff. of \(x^n\) on R.H.S \(= 0\)

     If \(n\) is even, coeff. of \(x^n\) on R.H.S. \(= (-1)^{\frac{n}{2}}{}^nC_{\frac{n}{2}}\)

     \(= \frac{n!}{\left(\frac{n!}{2}\right)^2}\)

178. :math:`(1 + x)^n = C_0 + C_1x + C_2x^2 + ldots + C_nx^n`X

     \((1 + x)^m = C_0 + C_1x + C_2x^2 + \ldots + C_mx^m\)

     Multiplying these two, we get

     \((C_0 + C_1x + C_2x^2 + \ldots + C_nx^n)(C_0 + C_1x + C_2x^2 + \ldots + C_mx^m) = (1 + x)^{m + n} = {}^{m + n}C_0 + {}^{m + n}C_1x + \ldots + {}^{m + n}C_rx^r + \ldots + {}^{m + n}C_{m + n}x^{m + n}\)

     Equating the coefficient of \(x^r\) we get the desired result.

179. \((1 + x)^{2n} = C_0 + C_1x + C_2x^2 + \ldots + C_{2n}x^{2n}\)

     \((x + 1)^{2n} = C_0x^{2n} + C_1x^{2n - 1} + C_2x^{2n - 2} + \ldots + C_2n\)

     Multiplying we get

     \((C_0 + C_1x + C_2x^2 + \ldots + C_{2n}x^{2n})(C_0x^{2n} + C_1x^{2n - 1} + C_2x^{2n - 2} + \ldots + C_{2n}) = (1 - x^2)^{n}\)

     Equating the coefficients of \(x^{2n}\) we get the desired result.

180. \((1 + x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n\)

     Differentaiating w.r.t \(x,\) we obtain

     \(n(1 + x)^{n - 1} = C_1 + 2C_2x + \ldots + nC_nx^{n -1}\)

     Also, \((x + 1)^n = C_0x^n + C_1x^{n - 1} + C_2x^{n - 2} + \ldots + C_n\)

     Multiplying two previous equations, we obtain

     \(n(1 + x)^{2n - 1} = (C_1 + 2C_2x + \ldots + nC_nx^{n -1})(C_0x^n + C_1x^{n - 1} + C_2x^{n - 2} + \ldots + C_n)\)

     Equating coefficients of \(x^{n - 1},\) we obtain

     \(C_1^2 + 2.C_2^2 + 2.C_3^3 + \ldots + n.C_n^2 = n.{}^{2n - 1}C_{n - 1} = \frac{(2n - 1)!}{[(n - 1)!]^2}\)

181. \((1 + x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n\)

     Integrating with limits \(0\) and \(x,\) we obtain

     \(\left[\frac{(1 + x)^{n + 1}}{n + 1}\right]_0^x = \left[C_0x + C_1\frac{x^2}{2} + C_2\frac{x^3}{3} + \ldots + C_n\frac{x^{n + 1}}{n + 1}\right]_0^x\)

     Also, \((x + 1)^n = C_0x^n + C_1x^{n - 1} + C_2x^{n - 2} + \ldots + C_n\)

     Multiplying and equating coefficients of \(x^{n + 1},\) we obtain

     \(C_0^2 + \frac{C_1^2}{2} + \frac{C_2^2}{3} + \ldots + \frac{C_n^2}{n + 1} = \frac{{}^{2n + 1}C_{n + 1} - 0}{n + 1} = \frac{(2n + 1)!}{[(n + 1)!]^2}\)

182. \((1 - x)^n = C_0 - C_1x + C_2x^2 - \ldots + (-1)^nC_nx^n\)

     Multiplying with \(x,\) we obtain

     \(x(1 - x)^n = C_0x + C_1x^2 + \ldots + (-1)^nC_nx^n\)

     Differentiating w.r.t \(x,\) we get

     \((1 - x)^n - nx(1 - x)^{n - 1} = C_0 -2xC_1 + \ldots + (-1)^n(n + 1)x^nC_n\)