Question

Suppose a series of $n$ terms is given by $S_n=a_1+a_2+...+a_n$ Then $S_{n-1}=a_1+a_2+...+a_{n-1}(\forall n>1)$ and then we can write $t_n=S_n-S_{n-1}\forall n\ge 2$ for $n=1$ we have $S_1=t_1$
On the basis of above information answer the following questionsThe sum to $n$ terms of a series is $\frac{n(n+1)(n+2)}{3}$ then its $n^{th}$ term is given by

• A. $n(n+2)$
• B. $(n+1)(n+2)$
• C. $n(n+1)$
• D. $\frac{n(n+1)}{3}$

Answer 1

The correct option is C $n(n+1)$

$S_n=\frac{n(n+1)(n+2)}{3}$
$\therefore S_{n-1}=\frac{(n+1)\left(n\right)(n+2)}{3}$
$\therefore t_n=S_n-S_{n-1}=\frac{n(n+1)(n+2)-(n-1)\left(n\right)(n+1)}{3}$
$=\frac{n(n+1)[n+2-n+1]}{3}=n(n+1)$