Question

$1\times 2+2\times 3+3\times 4+....nterms=$

• A. $\frac{n(n+1)(n+2)}{3}$
• B. $\frac{n(n+1)(n-2)}{3}$
• C. $\frac{n(n+1)(n+2)}{6}$
• D. $\frac{n(n-1)(n-2)}{6}$

Answer 1

The correct option is A $\frac{n(n+1)(n+2)}{3}$

$n^{th}$ term of sequence $=n(n+1)$
Then sum of $n$ terms $=\sum _{r=1}^{n}n(n+1)=\sum _{r=1}^n(n^2+n)$
$=\sum _{r=1}^n{n}^2+\sum _{r=1}^{n}n=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$
$=\frac{n(n+1)(2n+1)+3n(n+1)}{6}$
$=\frac{n(n+1)[2n+1+3]}{6}$
$=\frac{n(n+1)(2n+4)}{6}$
$=\frac{n(n+1)(n+2)}{3}$.