Q80 / 160 · AP EAMCET 2014 · Mathematics

$\sum_{k=1}^{2 n+1}(-1)^{k-1} \cdot k^2$ equals to

  • A
    $(n-1)(2 n-1)$
  • Correct answer
    B
    $(n+1)(2 n+1)$
  • C
    $(n+1)(2 n-1)$
  • D
    $(n-1)(2 n+1)$

Solution

$\begin{aligned} & \sum_{k=1}^{2 n+1}(-1)^{n-1} k^2 \\ & =\left[1^2-2^2+3^2-4^2+\ldots-(2 n)^2+(2 n+1)^2\right] \\ & =\left[1^2+3^2+5^2+\ldots+(2 n+1)^2\right] \\ & -\left[2^2+4^2+6^2+\ldots+(2 n)^2\right] \\ & =\left[1^2+2^2+3^2+4^2+\ldots+(2 n+1)^2\right] \\ & -2\left[2^2+4^2+6^2+\ldots+(2 n)^2\right] \\ & =\sum(2 n+1)^2-2 \times 2^2\left[1+2^2+3^2+\ldots+n^2\right] \\ & =\frac{(2 n+1)(2 n+1+1)\{2(2 n+1)+1\}}{6}-\sum n^2 \\ & {\left[\because \sum n^2=\frac{n(n+1)(2 n+1)}{6}\right]} \\ & =\frac{(2 n+1)(2 n+2)(4 n+3)}{6} \\ & -8 \frac{n(n+1)(2 n+1)}{6} \\ & =\frac{(n+1)(2 n+1)}{6}[2(4 n+3)-8 n] \\ & =\frac{(n+1)(2 n+1)}{6}[8 n+6-8 n] \\ & =(n+1)(2 n+1) \\ & \end{aligned}$

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

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