Question

For any integer $n\ge 1$, the sum $\sum _{k=1}^{n}k(k+1)$ is equal to

• A. $\frac{n(n+1)(n+2)}{3}$
• B. $\frac{n(n+1)(2n+1)}{3}$
• C. $\frac{n(n+1)(2n+7)}{6}$
• D. $\frac{n(n+1)(2n+9)}{6}$

Answer 1

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

The value of ${\sum }_{k=1}^{n}k(k+1)={\sum }_{k=1}^n{k}^2+{\sum }_{k=1}^{n}k$
$=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$
$=\frac{n(n+1)(n+2)}{3}$