# 33. Summation of Series Solutions¶

We have studied the formula for A.P http://10hash.com/algebra/arithmetic_progressions/#sum-of-squares-and-cubes-and-more. So, we have, \(t_n = 12n^2 - 6n + 5\)

Thus,

\[\sum_{i=1}^n t_n = 12\sum_{i=1}^n 12i^2 - 6 \sum_{i=1}^n i + \sum_{i=1}^n 5\]\[\sum_{i=1}^n t_n = 12 \frac{n(n + 1)}{2n + 1} - 6\frac{n()n +1}{2} + 5n\]Now, it is trivial to solve above expression.

Clearly, nth term = \(2n -1\) for A. P. 1, 3, 5, 7 …

Thus, nth term of given series = \(t_n = (2n - 1)^2 = 4n^2 - 4n + 1\)

Thus, \(S_n = \sum t_n = \sum (4n^2 - 4n + 1) = \sum 4n^2 - \sum 4n + \sum 1\)

\(= \frac{4n(n + 1)(2n + 1)}{6} + \frac{4n(n + 1)}{2} + n\)

\(= \frac{n(4n^2)}{3}\)

Clearly \(t_n = n(n + 1)(n + 2) = n^3 + 3n^3 + 2n\)

\(S_n = \sum t_n = \sum n^3 + 3n^2 + 2n\)

\(= \left\{\frac{n(n + 1)}{2}\right\}^2 + \frac{3n(n + 1)(2n + 1)}{6} + \frac{2n(n + 1)}{2}\)

Now, it is trivial to evaluate above expression.

Clearly, rth term is \(t_r = r*(r - r + 1) = nr - r^2 + r\)

\(S_n = \sum_{r=1}^n (nr - r^2 + r\)

= \(\frac{n(n + 1)(n + 2)}{6}\)

nth term of the series \(t_n = (1 + 2 + 3 + ... + n) = \frac{n(n + 1)}{2}\)

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

\(= \frac{n(n + 1)(n + 2)}{6}\)

As we can see that nth term will have \(n\) numbers. Thus, when brackets are removed, we have follwoing number of terms

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

\(\therefore S_n = 1 + 2 + 3 + ... + \frac{n(n + 1)}{2}\)

\(= \frac{1}{2}\frac{2(n + 1)(n^2 + n + 2)}{8}\)

nth term of the given series

\[t_n = \frac{1^3 + 2^3 + 3^3 + ... + n^3}{1 + 3 + 5 + ... ~\text{to n terms}}\]\[= \frac{\left\{\frac{n(n + 1)}{2}\right\}^2}{\frac{n}{2}*[2*1 + (n - 1)2]}\]\[= \frac{\left\{\frac{n(n + 1)}{2}\right\}^2}{n^2} = \frac{n^2 + 2n + 1}{4}\]\(\therefore\) Sum to \(n\) terms, \(S_n = \sum_{n = 1}^n t_n\)

\[= \frac{1}{4}\sum_{n = 1}^n (n^2 + 2n + 1)\]\[= \frac{1}{4}\left\{\frac{n(n + 1)(2n + 1)}{6} + \frac{2n(n + 1)}{2} + n\right\}\]\(\therefore\) sum to 16 terms

\(S_{16} = 446\)

nth term of the seuqence 3, 5, 7, … is \(t_n = 3 + (n - 1)2 = 2n + 1\)

nth term of the seuquence 2, 4, 6, … is \(t_n = 2 + (n - 1)2 = 2n\)

Thus, nth term of the given seuqnce is \((2n + 1)^3 - (2n)^3 = 12n^2 + 6n + 1\)

Hence, \(S_n\) can be easily found and is \(S_n = 4n^3 + 9n^2 + 6n\) and calculation for this has been left to reader as an exercise. Substituting \(n = 10\) will yield the required result.

As shown below:

\[t_1 = \frac{1}{1*2} = \frac{1}{1} - \frac{1}{2}\]\[t_2 = \frac{1}{2*3} = \frac{1}{2} - \frac{1}{3}\]\[t_3 = \frac{1}{3*4} = \frac{1}{3} - \frac{1}{4}\]…

\[t_n = \frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}\]we get our sum by adding all the terms

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

Let \(t_n = \frac{A}{n} + \frac{B}{n + 1} + \frac{C}{n + 2}\).

Then by partial fractions, \(A = \text{value of }\frac{1}{(n + 1)(n + 2)\text{when }n = 0}\)

\(\therefore A = \frac{1}{(0 + 1)() + 2)} = \frac{1}{2}\)

Similarly, \(B = -1\) and \(C = \frac{1}{2}\).

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

Now, it is trivial to add these and get our result.

\(S_n = 1 + 5 + 11 + 19 + 29 +~...~+ t_n~~~~~~~~~~~~~(i)\)

\(S_n = ~~~~~~1 + ~~5 + 11 + 19 + ~...~+ t_{n - 1}+ t_{n}~~~(ii)\)

Subtracting, we get

\(t_n = 1 + [4 + 6 + 8 + ... \text{to (n - 1) terms}\)

\(t_n = 1 + \frac{n - 1}{2}[2*4 + (n - 1 - 1)2] = n^2 + n 1 - 1\)

Now it is trivial to find the sum of the series.

Given, first person gets 1 ruppee.

Second person gets (1 + 1) = 2 rupees.

Third person gets (2 + 2) = 4 rupees.

Fourth person gets (4 + 3) = 7 rupees and so on.

Let there be \(n\) persons and nth person gets \(t_n\) rupees

Let \(S_n = 1 + 2 + 4 + 7 + ... + t_n\)

Proceeding like in previous exercise, we have

\(t_n = 1 + (1 + 2 + 3 + ... \text{to (n - 1) terms})\)

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

Substituting \(t_n = 67\) will yield the desired result which is 12.

Since 1st group contains 1 number, 2nd group contains 2 numbers, 3rd group contains 3 numbers, and so on, therefore, nth group will contain \(n\) numbers.

Here, we observe that numbers in each group are in A. P. whose c. d. is 1. (n - 1)th group will have (n - 1) numbers. Total numbers of numbers till the end of (n - 1)th group is

\(N = 1 + 2 + 3 + ... \text{for (n - 1) terms}\)

\(N = \frac{n - 1}{2}\left\{2 + (n - 2)1\right\}\)

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

Thus, first term of nth group will be \(N + 1\) i.e. \(\frac{n^2 - n + 2}{2}\)

Sum of numbers in nth group is:

\(S_n = \frac{n}{2}\left\{2.\frac{(n^2 - n + 2)}{2} + (n - 1)\right\}\)

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

Let \(S_n = 1 + 3 + 7 + 15 + ... + t_{n - 1} + t_n\)

Rearranging and rewriting like exercise 11 we have

\(t_n = 1 + [2 + 4 + 8 + ... \text{to (n - 1) terms}]\)

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

Now, \(S_n = t_1 + t_2 + t_3 + ... + t_n\)

\(S_n = (2^1 - 1) + (2^2 - 1) + (2^3 - 1) + ... + (2^n - 1)\)

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

Let \(S_n = 1 + 2x + 3x^2 + ... + t_n\)

\(\therefore~xS_n = ~~~~~~~x + 2x^2 + ... + xt_{n-1} + xt_n\)

Subtracting, we get

\((1 - x)S_n = 1 + [(2 - 1)x + (3 - 2)x^2 + ... \text{to (n - 1) terms}] - xt_n\)

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

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

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

This problem is same as previous with \(x = 2\) and \(n = 99\).

Proceeding as exercise 15, we have

\((1 - x)S = 1 + (2^2 - 1)x + (3^2 - 2^2)x^2 + ...~\text{to }\infty\)

\((1 - x)S = 1 + 3x + 5x + 7x^2 + ...\)

Repeating the process again,

\((1 - x)^2S = 1 + 2x(1 + x + x^2 + ...~\text{to }\infty)\)

\(S = \frac{1 + x}{(1 - x)^3}\)

Given, \(S_n = 2n^2 + 4\) therefore \(S_{n - 1} = 2(n - 1)^2 + 4 = 2n^2 - 4n + 6\)

\(t_n = S_n - S_{n - 1} = 4n - 2\)

Since n is a first order term the sequence is in A. P. which can be found out by finding common difference which is c.d. = \(t_n - t_{n - 1} = 4\). Now, 4 is a constant with no variable multiplicand. Thus, series is in A. P.

Given, \(t_n = n(n - 1)(n + 1) = n^3 - n\)

\(S_n = \sum t_n = \left\{\frac{n(n+1)}{2}\right\}^2 - \frac{n(n+1)}{2}\)

The nth term is same as previous exercise. Thus, substituting 80 for \(n\) we get

\(S_80 = \left\{\frac{80(80+1)}{2}\right\}^2 - \frac{80(80+1)}{2} = 10494360\)

nth term = \((2n - 1)^3 = 8n^3 - 12n^2 + 6n -1\)

Now it is trivial to find the sum and has been left as an exercise.

\(t_n = (3n - 2)^2 = 9n^2 - 12n + 4\)

Now it is trivial to find the sum and has been left as an exercise.

\(S_{2n} = 1^2 + 3^2 + 5^2 + ... \text{ to n terms} + 2 + 4 + 6 + ... \text{ to n terms}\)

\(S_{2n} = S1_n + S2_n\)

\(t1_n = (2n - 1)^2 = 4n^2 - 4n + 1\)

\(t2_n = 2n\)

Now I trust the reader to find out the sum.

\(S_n = 1^2 + 3^2 + 5^2 + ... \text{ to n/2 terms } - (2^2 + 4^2 + 6^2 + ... \text{to n/2 terms})\)

\(S_n = S1_n - S2_n\)

\(t1_n = (2n - 1)^2 = 4n^2 - 4n + 1\)

\(t2_n = 4n^2\)

As usual find out the final answer.

\(t_n = (2n - 1)*(2n + 1) = 4n^2 - 1\)

Please find sum as I am tired.

\(t_n = n*(n + 1)\)

Not another sum.

\(t_n = n*(n + 1)^2\)

By now you know the drill.

\(t_n = (n + 1)*n^2\)

Silence.

\(t_n = 1 + 3 + 5 + .. \text{ to n terms}\)

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

Now it is as easy as putting a hot knife in butter.

\(t_n = 1^2 + 2^2 + 3^2 + ... \text{ to n terms}\)

\(t_n = \frac{n(n + 1)(2n + 1)}{6}\)

Grrr. Find the sum.

\(t_n = n*(n + 1)*(2n + 1)\)

Please find the final answer. I do not know that why I am giving partial answer.

\(t_n = n*(n + 1)*(n + 2)\)

It is better if you calculate the final answer.

\(t_n = n*(2n + 1)^2\)

Please do the needful.

\(S_n = n^2 + 2n^2 + 3n^2 + ... \text{ to n terms} - (1^3 + 2^3 + 3^3 + ... \text{ to n terms})\)

Now split it in two and find out nth term and then the sum

\(t_n = (1^2 + 2^2 + 3^2 + ... \text{ to n terms}) = \frac{n(n + 1)(2n + 1)}{6}\)

Now find \(S_n\) and then \(S_{10}\).

\(t_n = (2n + 1)^3 - (2n)^3\)

\(t_n = \frac{1}{1 + 2 + 3 + ... \text{ to n terms}}\)

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

\(t_1 = 2\left[\frac{1}{1} - \frac{1}{2}\right]\)

\(t_2 = 2\left[\frac{1}{2} - \frac{1}{3}\right]\)

\(t_3 = 2\left[\frac{1}{3} - \frac{1}{4}\right]\)

…

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

Thus, \(S_n = 2\left[\frac{1}{1} - \frac{1}{n + 1}\right]\)

\(S_n = \frac{2n}{n + 1}\)

\(t_n = \frac{1}{2n * 2(n + 1)}\)

\(t_n = \frac{1}{4}\left[\frac{1}{n} - \frac{1}{n + 1}\right]\)

Proceed like previous problem to find out the sum.

\(S_n = 2 + 6 + 12 + 20 +~...~+ t_n~~~~~~~~~~~~~(i)\)

\(S_n = ~~~~~~~2 + ~~6 + 12 + 20 + ~...~+ t_{n - 1}+ t_{n}~~~(ii)\)

Subtracting, we get

\(t_n = 2 + [4 + 6 + 8 + ... \text{to (n - 1) terms}\)

\(t_n = 2 + \frac{n - 1}{2}[2*4 + (n - 1 - 1)2] = n^2 + n 1 - 1\)

Now it is trivial to find the sum of the series.

\(S_n = 3 + 6 + 11 + 18 +~...~+ t_n~~~~~~~~~~~~~(i)\)

\(S_n = ~~~~~~~3 + ~~6 + 11 + 18 + ~...~+ t_{n - 1}+ t_{n}~~~(ii)\)

Subtracting, we get

\(t_n = 3 + [3 + 5 + 7 + ... \text{to (n - 1) terms}\)

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

Now it is trivial to find the sum of the series.

\(S_n = 1 + 9 + 24 + 46 + 75 + ~...~+ t_n~~~~~~~~~~~~~(i)\)

\(S_n = ~~~~~~~1 + ~~9 + 24 + 46 + ~...~+ t_{n - 1}+ t_{n}~~~(ii)\)

Subtracting, we get

\(t_n = 1 + [8 + 15 + 22 + ... \text{to (n - 1) terms}\)

\(t_n = 1 + \frac{n - 1}{2}[2*8 + (n - 1 - 1)*7]\)

Now it is trivial to find the sum of the series.

\(S_n = 2 + 4 + 7 + 11 + 16 + ~...~+ t_n~~~~~~~~~~~~~(i)\)

\(S_n = ~~~~~~~2 + ~~4 + 7 + 11 + ~...~+ t_{n - 1}+ t_{n}~~~(ii)\)

Subtracting, we get

\(t_n = 2 + [2 + 3 + 4 + ... \text{to (n - 1) terms}\)

\(t_n = 2 + \frac{n - 1}{2}[2*2 + (n - 1 - 1)]\)

\(S_n = 1 + 3 + 6 + 10 + ~...~+ t_n~~~~~~~~~~~~~(i)\)

\(S_n = ~~~~~~~1 + ~~3 + 6 + 10 + ~...~+ t_{n - 1}+ t_{n}~~~(ii)\)

Subtracting, we get

\(t_n = 1 + [2 + 3 + 4 + ... \text{to (n - 1) terms}\)

\(t_n = 1 + \frac{n - 1}{2}[2*2 + (n - 1 - 1)]\)

Now you can find \(S_n\) and then \(S_{10}\).

This is same as previous problem just substitute 10 with 30.

1st group has 2 terms, 2nd has 4 and 3rd has 6. Thus, (n - 1)th group will have 2(n - 1) terms.

Total no. of terms till (n - 1)th group = 2 + 4 + 6 + … to (n - 1) terms

\(N = \frac{n - 1}{2}[2*2 + (n - 1 - 1)*2] = \frac{n(n - 1)}{2}\)

Numbers are 1, 3, 5, 7 … thus, \(\frac{n(n - 1)}{2} + 1 \text{th}\) term will be

\(t_{n + 1} = \frac{n^2 - n + 2}{4}\left[2*1 + \left(\frac{n^2 - n + 2}{2} - 1\right)*2\right]\)

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

Now the nth group will start with \(t_{n + 1}\) for 2n terms with common difference 2 which can be easily evaluated.

This is same problem as previous and can be solved easily.

47 to 52 are similar problems as we have solved earlier and can be done with ease so I am leaving them as exercise to the reader.