39. Summation of Series Problems#
Find the sum of \(n\) terms of the series whose nth term is \(12n^2 - 6n +5\).
Find the sum to \(n\) terms of the series \(1^2 + 3^2 + 5^2 + 7^2 + ...\)
Find the sum to \(n\) terms of the series \(1.2.3 + 2.3.4 + 3.4.5 + ...\)
Find the sum of the series \(1.2 + 2.(n - 1) + 3.(n - 2) + ... + n.1\)
Find the sum to \(n\) terms of the series \(1 + (1 + 2) + (1 + 2 + 3) + ...\)
Find the sum to \(n\) terms of the series \(1 + (2 + 3) + (4 + 5 + 6) + ...\)
Find the sum of the series
\[\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + ... ~\text{to 16 terms}\]Find \((3^3 - 2^3) + (5^3 - 4^3) + (7^3 - 6^3) + ...\) to 10 terms.
Find \(\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} ...\) to \(n\) terms.
Find the sum of the follwing series to infinity:
\[\frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{3.4.5} + ...\]Find the sum of \(n\) terms of the series \(1 + 5 + 11 + 19 + 29 ...\)
A sum is distributed among certain number of persons. Second person gets one rupee more than the first, third person gets two rupees more than the second, fourth person gets three rupees more than the third and so on. If the first person gets one rupee and the last person get 67 rupees, find the number of person.
Natural numbers have been grouped in the following way 1, (2, 3), (4, 5, 6), (7, 8, 9, 10), … Show that the sum of the numbers in the nth group is \(\frac{n(n^2 + 1)}{2}\).
Find \(1 + 3 + 7 + 15 + ...\) to \(n\) terms.
Find \(1 + 2x + 3x^2 + 4x^3 + ...\) to \(n\) terms.
Find \(1 + 2.2 + 3.2^2 + 4.2^3 + ... + 100.2^{99}\)
Find \(1 + 2^2x + 3^2x^2 + 4^2x^3 + ...\) to \(\infty, |x| < 1\)
If the sum of \(n\) terms of a sequence be \(2n^2 + 4\), find its n th term. Is this sequence in A. P.?
Find the sum of \(n\) terms of the series whose nth term is \(n(n - 1)(n + 1)\).
Find the sum of 80 terms of the series whose nth term is \(n(n^2 - 1)\)
Find the sum of the following series:
\(1^3 + 3^3 + 5^3 + ...\) to \(n\) terms.
\(1^2 + 4^2 + 7^2 + 10^2 + ...\) to \(n\) terms.
\(1^2 + 2 + 3^2 + 4 + 5^2 + 6 + ...\) to \(2n\) terms.
\(1^2 - 2^2 + 3^3 - 4^2 + ...\) to \(n\) terms.
\(1.3 + 3.5 + 5.7 + ...\) to \(n\) terms.
\(1.2 + 2.3 + 3.4 + ...\) to \(n\) terms.
\(1.2^2 + 2.3^2 + 3.4^2 + ...\) to \(n\) terms.
\(2.1^2 + 3.2^2 + 4.3^2 + ...\) to \(n\) terms.
\(1 + (1 + 3) + (1 + 3 + 5) + ...\) to \(n\) terms.
\(1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ...\) to \(n\) terms.
\(1.2.3 + 2.3.5 + 3.4.7 + ...\) to \(n\) terms.
\(1.2.3 + 2.3.4 + 3.4.5 + ...\) to \(n\) terms.
\(1.3^2 + 2.5^2 + 3.7^2 + ...\) to \(20\) terms.
\((n^2 - 1^2) + 2(n^2 - 2^2) + 3(n^2 - 3^2) + ...\) to \(n\) terms.
\(1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ...\) to \(10\) terms.
\((3^3 - 2^3) + (5^3 - 4^3) + (7^3 - 6^3) + ...\) to \(10\) terms.
\(1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + ...\) to \(n\) terms.
Find the sum to infinity of the series
\[\frac{1}{2.4} + \frac{1}{4.6} + \frac{1}{6.8} + \frac{1}{8.10} + ... \]
Find the nth term and sum to \(n\) terms of the following series:
\(2 + 6 + 12 + 20 + ...\)
\(3 + 6 + 11 + 18 + ...\)
\(1 + 9 + 24 + 46 + 75 + ...\)
Find the nth term of the series \(2 + 4 + 7 + 11 + 16 ...\)
Find the sum to 10 terms of the series \(1 + 3 + 6 + 10 + ...\)
A man collects Rs 1 on first day, Rs. 3 on the second day, Rs. 6 on third day, Rs. 10 on fourth day and so on in the month of April. What will be his collection on 30th of April and what will be his total collection in the month of April?
The odd natural numbers have been divided in groups as: (1, 3); (5, 7, 9, 11); (13, 15, 17, 19, 21, 23); … Show that the sum of numbers in the nth group is \(n^3\).
Show that the sum of the numbers in each of the following groups is an sqaure of an odd positive integer. (1); (2, 3, 4); (3, 4, 5, 6, 7,); …
Find the sum to \(n\) terms of the following series:
\(2 + 5 + 14 + 41 + ...\)
\(1 + 5 + 13 + 29 + 61 + ...\)
\(3 + 5 + 9 + 17 + 33 + ...\)
\(1.1 + 2.3 + 4.5 + 8.7 + ...\)
- \[1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + ... \]
- \[1 - \frac{4}{2} + \frac{4}{2^2} - \frac{4}{2^3} + ... \]