Question

Let $T_r$ and $S_r$ be the $r^{th}$ term and sum up to $r^{th}$ term of a series respectively. If for an odd natural number $n,S_n=n$ and $T_n=\frac{T_{n-1}}{n^2}$, then $T_m$ ($m$ being even) is:

• A. $\frac{2}{1+m^2}$
• B. $\frac{2m^2}{1+m^2}$
• C. $\frac{{(m+1)}^2}{2+{(m+1)}^2}$
• D. $\frac{2{(m+1)}^2}{1+{(m+1)}^2}$

Answer 1

The correct option is D $\frac{2{(m+1)}^2}{1+{(m+1)}^2}$

Given,

$S_n=n$ and $T_n=\frac{T_{n-1}}{n^2},$ $n$ is an odd natural number.

$\therefore S_n-S_{n-2}=\left(n\right)-(n-2)=2\Rightarrow T_n+T_{n-1}=2\Rightarrow \frac{T_{n-1}}{n^2}+T_{n-1}=2\Rightarrow (\frac{1}{n^2}+1)T_{n-1}=2\Rightarrow \left(\frac{1+n^2}{n^2}\right)T_{n-1}=2\Rightarrow T_{n-1}=\frac{2n^2}{1+n^2}$

Since, $n$ is odd $\Rightarrow (n-1)$ is odd.

Replacing $(n-1)$ by $m$ we get,

$T_m=\frac{2(m+1{)}^2}{1+(m+1{)}^2}$