Find the sum of the n terms of the series fra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The $n^{th}$ term of the series is given by $T_n = \frac{2n + 1}{\sum_{k=1}^{n} k^2}$.Using the formula for the sum of squares,$\sum_{k=1}^{n} k^2

Vedclass · Accepted Answer

$\frac{6n}{n + 1}$

Vedclass · Answer

(C) The $n^{th}$ term of the series is given by $T_n = \frac{2n + 1}{\sum_{k=1}^{n} k^2}$.Using the formula for the sum of squares,$\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}$.Thus,$T_n = \frac{2n + 1}{\frac{n(n + 1)(2n + 1)}{6}} = \frac{6}{n(n + 1)}$.Using partial fractions,$T_n = 6 \left( \frac{1}{n} - \frac{1}{n + 1} \right)$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = 6 \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k + 1} \right)$.This is a telescoping series: $S_n = 6 \left( (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + ... + (\frac{1}{n} - \frac{1}{n + 1}) \right)$.$S_n = 6 \left( 1 - \frac{1}{n + 1} \right) = 6 \left( \frac{n + 1 - 1}{n + 1} \right) = \frac{6n}{n + 1}$.

Find the sum of the $n$ terms of the series $\frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + ...$

Similar Questions

$1 + \sum\limits_{r = 0}^{22} {\left\{ {r\left( {r + 2} \right) + 1} \right\}} \cdot r! = k!$,then the number of divisors of $k$ is

The sum of $n$ terms of the series $\frac{1}{1 + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + \dots$ is

The sum of the series $\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots$ up to $n$ terms is equal to:

The value of the expression $3(1!) - 4(2!) + 5(3!) - 6(4!) + \dots - 2008(2006)! + (2007)!$ is

The sum of the series $1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots$ up to $10$ terms is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module