Question

If the sum to $n$ terms of the series $\frac{3}{1^2\times 2^2}+\frac{5}{2^2\times 3^2}+\frac{7}{3^2\times 4^2}+\dots \dots$ is $0.99$, then the value of $n$ is

Answer 1

​​​​​​Let $T_r$ be the $r^{\text{th}}$ term of the given series. Then,
$T_r=\frac{(2r+1)}{r^2(r+1{)}^2},$
$=[\frac{1}{r^2}-\frac{1}{(r+1{)}^2}]$
Hence, the required sum is
$\sum _{r=1}^n{T}_r=\sum _{r=1}^n[\frac{1}{r^2}-\frac{1}{(r+1{)}^2}]$
$=(\frac{1}{1^2}-\frac{1}{2^2})+(\frac{1}{2^2}-\frac{1}{3^2})+(⋮⋮)+(\frac{1}{n^2}-\frac{1}{(n+1{)}^2})$
$=1-\frac{1}{(n+1{)}^2}$
$=\frac{2n+n^2}{(n+1{)}^2}$
Now,
$0.99=\frac{2n+n^2}{(n+1{)}^2}$
$\Rightarrow \frac{99}{100}=\frac{2n+n^2}{(n+1{)}^2}\Rightarrow 99n^2+99+198n=200n+100n^2\Rightarrow n^2+2n-99=0\Rightarrow n=-11,9$
But $n\in N$
$\therefore n=9$