Mathematics | TS EAMCET 2017 | Discussion Page

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

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

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

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

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

Option B

Consider,

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

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

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

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

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