Question

The sum of $n$ terms of the series $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots$ is

(A) $\frac{6 n}{n+1}$

(B) $\frac{9 n}{n+1}$

(C) $\frac{12 n}{n+1}$

(D) $\frac{3 n}{n+1}$

Answer 1

Here, $T_r=\frac{2 r+1}{1^2+2^2+\ldots+r^2}$

$\Rightarrow\frac{(2 r+1)}{\frac{1}{6} r(r+1)(2 r+1)}=6\left(\frac{1}{r}-\frac{1}{r+1}\right)$

$\therefore \text { Required sum }=\sum_{r=1}^n T_r $

$\Rightarrow6 \sum_{r=1}^n\left(\frac{1}{r}-\frac{1}{r+1}\right)$

$\Rightarrow6\left(1-\frac{1}{n+1}\right)=\frac{6 n}{n+1}$

So, The correct option will be (A).