Question

For any integer $n\ge 1,{\sum }_{K=1}^{n}K(K+2)=$

• A. $\frac{n(n+1)(n+2)}{6}$
• B. $\frac{n(n+1)(2n+7)}{6}$
• C. $\frac{n(n+1)(2n+1)}{6}$
• D. $\frac{n(n-1)(2n+8)}{6}$

Answer 1

The correct option is B $\frac{n(n+1)(2n+7)}{6}$

We need to find value of ${\sum }_{K=1}^{n}K(K+2)$

$={\sum }_{K=1}^n{K}^2+2{\sum }_{K=1}^n\left(K\right)$
$=\frac{n(n+1)(2n+1)}{6}+2\frac{n(n+1)}{2}$

$=\frac{(n+1)n}{6}[2n+1+6]$

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