Question

The sum $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+....$ upto 11 terms is:

• A. $\frac{7}{2}$
• B. $\frac{11}{4}$
• C. $\frac{11}{2}$
• D. $\frac{60}{4}$

Answer 1

Answer: (C)

Given sum is
$\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+....$
$nth$ term $=T_n=\frac{2n+1}{\frac{n\left(n+1\right)\left(2n+1\right)}{6}}=\frac{6}{n\left(n+1\right)}$
or $T_n=6\left[\frac{1}{n}-\frac{1}{n+1}\right]$
$\therefore S_n=\sum T_n=6\sum \frac{1}{n}-6\sum \frac{1}{n+1}=\frac{6n}{n}-\frac{6}{n+1}$
$=6-\frac{6}{n+1}=\frac{6n}{n+1}$
So, sum upto 11 terms means
$S_{11}=\frac{6\times 11}{11+1}=\frac{66}{12}=\frac{33}{6}=\frac{11}{2}$