Question

If $S=\frac{2^2-1}{2}+\frac{3^2-2}{6}+\frac{4^2-3}{12}+\dots .$. upto 10 terms, then $S$ is equal to

• A. $\frac{123}{11}$
• B. $\frac{10}{11}$
• C. $\frac{13}{11}$
• D. $\frac{120}{11}$

Answer 1

Answer: (D)

$T_{10}=\frac{(n+1{)}^2-n}{n(n+1)}=1+\left(\frac{1}{n}-\frac{1}{n+1}\right)$
$\therefore S_{10}+\left(1-\frac{1}{11}\right)=\frac{120}{11}$