Q91 / 155 · WBJEE 2016 · Mathematics

The sum of $n$ terms of the following series $1^{3}+3^{3}+5^{3}+7^{3}+\ldots$ is

  • Correct answer
    A
    $n^{2}\left(2 n^{2}-1\right)$
  • B
    $n^{3}(n-1)$
  • C
    $n^{3}+8 n+4$
  • D
    $2 n^{4}+3 n^{2}$

Solution

We have. $\mathrm{1}^{3}+3^{3}+5^{3}+7^{3}+\ldots$ $\begin{aligned} \text { Now, } T_{n} &=(2 n-1)^{3} \\ &=8 n^{3}-3(2 n)^{2}(1)+3(2 n)\left(0^{2}-(\pi)^{3}\right.\\ &=8 n^{3}-12 n^{2}+6 n-1 \end{aligned}$ $\therefore \quad S_{n}=\Sigma T_{n}$ $=\frac{8 n^{2}(n+1)^{2}}{4}-\frac{12 n(n+0) 2 n+1}{6}$$+\frac{6 n(n+1)}{2}-n$ $=2 n^{2}(n+1)^{2}-2 n(n+1)(2 n+1)+3 n(n+1)-n$ $=n\left[2 n(n+1)^{2}-2(n+1)(2 n+1)+3(n+1)-1\right]$ $=n\left[2 n\left(n^{2}+1+2 n\right)-2\left(2 n^{2}+3 n+1\right)+3n+3-1\right].$ $=n\left[2 n^{3}+2 n+4 n^{2}-4 n^{2}-6 n-2+3 n+2\right]$ $=n\left[2 n^{3}-n\right]=n^{2}\left(2 n^{2}-1\right).$

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

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