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

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

Can you answer this question?

$(3)\;\frac{n(n+1)(2n+7)}{6}$
answered Nov 7, 2013 by