# For any integer $n \leq 1$, the sum $\sum \limits _{k=1}^n$. $k(k+2)$ is equal to

$\begin {array} {1 1} (1)\;\frac{n(n+1)(n+2)}{6} & \quad (2)\;\frac{n(n+1)(2n+1)}{6} \\ (3)\;\frac{n(n+1)(2n+7)}{6} & \quad (4)\;\frac{n(n+1)(2n+9)}{6} \end {array}$

$(3)\;\frac{n(n+1)(2n+7)}{6}$