Question

The sum to $50$ terms of the series $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\dots$ is

• A. $\frac{100}{17}$
• B. $\frac{150}{17}$
• C. $\frac{200}{51}$
• D. $\frac{50}{17}$

Answer 1

The correct option is A $\frac{100}{17}$

Let $T_r$ be the $r^{th}$ term of the given series. Then,
$T_r=\frac{2r+1}{1^2+2^2+\dots \dots +r^2}$
$=\frac{6(2r+1)}{\left(r\right)(r+1)(2r+1)}=\frac{6}{r(r+1)}$
$=6(\frac{1}{r}-\frac{1}{r+1})$
So, sum is given by
$\sum _{r=1}^{50}{T}_r=6\sum _{r=1}^{50}(\frac{1}{r}-\frac{1}{r+1})$
$=6[(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\dots \dots +(\frac{1}{50}-\frac{1}{51})]$
$=6[1-\frac{1}{51}]$
$=\frac{100}{17}$