Q46 / 160 · AP EAMCET 2008 · Mathematics

For any integer $n \geq 1$, the sum $\sum_{k=1}^n k(k+2)$ is equal to

  • A
    $\frac{n(n+1)(n+2)}{6}$
  • B
    $\frac{n(n+1)(2n+1)}{6}$
  • Correct answer
    C
    $\frac{n(n+1)(2n+7)}{6}$
  • D
    $\frac{n(n+1)(2n+9)}{6}$

Solution

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

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

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