Question

If the $n^{th}$ term of the sequence is given by $\frac{(n^2-1)}{(n+1)}$ , then the sum of first n terms of the sequence is given by

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$n^{th}$ term of the sequence = $\frac{(n^2-1)}{(n+1)}=n-1$

Sum of n terms =$\left[\frac{n(n+1)}{2}\right]-n=\left[\frac{n(n-1)}{2}\right]$