Question

The sum to 50 term of the series $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}$ +. 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,

$T_r=\frac{2r+1}{1^2+2^2+\cdots +r^2}$

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

$=6(\frac{1}{r}-\frac{1}{r+1})$

So, sum is given by

$\sum _{r=1}^{50}{T}_r=6\sum _{r=1}^{50}(\frac{1}{r}-\frac{1}{r+1})$

$=6[(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots +(\frac{1}{50}-\frac{1}{51})]$

$=6[1-\frac{1}{51}]$

$=\frac{100}{17}$