Sum of the n terms of the series frac{3}{1^2} | 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 of the first $n$ na

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 of the first $n$ natural numbers,$\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}$.Substituting this into the expression for $T_n$:$T_n = \frac{2n + 1}{\frac{n(n + 1)(2n + 1)}{6}} = \frac{6}{n(n + 1)}$.Using partial fractions,we can write $T_n = 6 \left( \frac{1}{n} - \frac{1}{n + 1} \right)$.The sum of $n$ terms $S_n = \sum_{k=1}^{n} T_k = 6 \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k + 1} \right)$.This is a telescoping series:$S_n = 6 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + ... + (\frac{1}{n} - \frac{1}{n + 1}) \right]$.$S_n = 6 \left( 1 - \frac{1}{n + 1} \right) = 6 \left( \frac{n + 1 - 1}{n + 1} \right) = \frac{6n}{n + 1}$.

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

Similar Questions

If the sum of three consecutive terms of an $A.P.$ is $51$ and the product of the last and first term is $273$,then the numbers are

Let $S_1, S_2, \dots, S_{101}$ be the consecutive terms of an $A.P.$ If $\frac{1}{S_1 S_2} + \frac{1}{S_2 S_3} + \dots + \frac{1}{S_{100} S_{101}} = \frac{1}{6}$ and $S_1 + S_{101} = 50$, then $|S_1 - S_{101}|$ is equal to

Three numbers are in $G.P.$ such that their sum is $38$ and their product is $1728$. The greatest number among them is

If $1+(1-2^{2} \cdot 1)+(1-4^{2} \cdot 3)+(1-6^{2} \cdot 5)+\ldots+(1-20^{2} \cdot 19) = \alpha - 220 \beta$,then an ordered pair $(\alpha, \beta)$ is equal to

The first term of an $A.P.$ of consecutive integers is $p^2 + 1$. The sum of $(2p + 1)$ terms of this series can be expressed as:

Vedclass Test Series

Exam Paper Generator

Online Exam Module