Discussion Forum



1)

For an integer  $n\geq1,\sum_{k=1}^{n}K(K+2)=$


A) $\frac{n(n+1)(n+2)}{6}$

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

C) $\frac{n(n+1)(2n+1)}{6}$

D) $\frac{n(n-1)(2n+8)}{6}$

Answer:

Option B

Explanation:

Consider,

$\sum_{k=1}^{n}k(k+2)=\sum_{k=1}^{n}(k^{2}+2k)=\sum_{k=1}^{n}k^{2}+2\sum_{k=1}^{n}k$

    =$\frac{n(n+1)(2n+1)}{6}+2\frac{n(n+1)}{2}$

      $=n(n+1)\left[\frac{(2n+1)}{6}+1\right]=n(n+1)\left[\frac{2n+7}{6}\right]$

  =$\frac{n(n+1)(2n+7)}{6}$