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

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

The sum of the infinite series $\frac{1}{3 \times 7} + \frac{1}{7 \times 11} + \frac{1}{11 \times 15} + \dots$ is

If $\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+\frac{1}{(180-a)(200-a)}=\frac{1}{256}$,then the maximum value of $a$ is.

$\frac{1}{3 \times 7} + \frac{1}{7 \times 11} + \frac{1}{11 \times 15} + \ldots$ to $50$ terms $=$

If ${t_n} = \frac{1}{4}(n + 2)(n + 3)$ for $n = 1, 2, 3, \dots$,then $\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \dots + \frac{1}{t_{2003}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module