For any integer $n \geq 1$, the sum $\sum_{k=1}^n k(k+2)$ is equal to

Question

Aicharya · Accepted Answer

Now, $\begin{aligned} & \sum_{k=1}^n k(k+2) \ & =\sum_{k=1}^n\left(k^2+2 kight)=\sum_{k=1}^n k^2+2 \sum_{k=1}^n k \ & =\frac{n(n+1)(2 n+1)}{6}+\frac{2 \cdot n(n+1)}{2} \ & =n(n+1)\left(\frac{2 n+1}{6}+1ight) \ & =\frac{n(n+1)(2 n+7)}{6}\end{aligned}$

For any integer $n \geq 1$, the sum $\sum_{k=1}^n k(k+2)$ is equal to

Solution