Q106 / 160 · TS EAMCET 2017 · Mathematics

For any integer $n \geq 1, \sum_{K=1}^n K(K+2)=$

  • A
    $\frac{n(n+1)(n+2)}{6}$
  • Correct answer
    B
    $\frac{n(n+1)(2 n+7)}{6}$
  • C
    $\frac{n(n+1)(2 n+1)}{6}$
  • D
    $\frac{n(n-1)(2 n+8)}{6}$

Solution

Consider, $$ \begin{aligned} & \sum_{k=1}^n k(k+2)=\sum_{k=1}^n\left(k^2+2 k\right)=\sum_{k=1}^n k^2+2 \sum_{k=1}^n k \\ & =\frac{n(n+1)(2 n+1)}{6}+2 \frac{n(n+1)}{2} \\ & =n(n+1)\left[\frac{(2 n+1)}{6}+1\right]=n(n+1)\left[\frac{2 n+7}{6}\right] \\ & =\frac{n(n+1)(2 n+7)}{6} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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