Question

The value of $1.3+2.4+3.5+...+n(n+2)$ is equal to

• A. $\frac{n(n+1)(2n+7)}{6}$
• B. $\frac{2(n+2)(2n+3)}{6}$
• C. $\frac{n(n+1)(2n+5)}{6}$
• D. $\frac{n(n+1)(2n+1)}{6}$

Answer 1

Answer: (A)

$\frac{n(n+1)(2n+7)}{6}$

${\sum }_{k=1}^{n}k(k+2)={\sum }_{k=1}^n{k}^2+2{\sum }_{k=1}^{n}k$
$=\frac{n(n+1)(2n+1)}{6}+2\times \frac{n(n+1)}{2}$
$=n(n+1)(\frac{2n+1}{6}+1)$
$=\frac{n(n+1)(2n+7)}{6}$