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

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

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1 Answer

$ (3)\;\frac{n(n+1)(2n+7)}{6}$
answered Nov 7, 2013 by pady_1
 
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