Thank you for reporting, we will resolve it shortly
Q.
If $S=\frac{2^2-1}{2}+\frac{3^2-2}{6}+\frac{4^2-3}{12}+\ldots .$. upto 10 terms, then $S$ is equal to
Sequences and Series
Solution:
$ 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} $