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The sum of $n$ terms of the following series $1^{3}+3^{3}+5^{3}+7^{3}+\ldots$ is
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Verified Answer
The correct answer is:
$n^{2}\left(2 n^{2}-1\right)$
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}$$+\frac{6 n(n+1)}{2}-n$
$=2 n^{2}(n+1)^{2}-2 n(n+1)(2 n+1)+3 n(n+1)-n$
$=n\left[2 n(n+1)^{2}-2(n+1)(2 n+1)+3(n+1)-1\right]$
$=n\left[2 n\left(n^{2}+1+2 n\right)-2\left(2 n^{2}+3 n+1\right)+3n+3-1\right].$
$=n\left[2 n^{3}+2 n+4 n^{2}-4 n^{2}-6 n-2+3 n+2\right]$
$=n\left[2 n^{3}-n\right]=n^{2}\left(2 n^{2}-1\right).$
$\begin{aligned} \text { Now, } T_{n} &=(2 n-1)^{3} \\ &=8 n^{3}-3(2 n)^{2}(1)+3(2 n)\left(0^{2}-(\pi)^{3}\right.\\ &=8 n^{3}-12 n^{2}+6 n-1 \end{aligned}$
$\therefore \quad S_{n}=\Sigma T_{n}$
$=\frac{8 n^{2}(n+1)^{2}}{4}-\frac{12 n(n+0) 2 n+1}{6}$$+\frac{6 n(n+1)}{2}-n$
$=2 n^{2}(n+1)^{2}-2 n(n+1)(2 n+1)+3 n(n+1)-n$
$=n\left[2 n(n+1)^{2}-2(n+1)(2 n+1)+3(n+1)-1\right]$
$=n\left[2 n\left(n^{2}+1+2 n\right)-2\left(2 n^{2}+3 n+1\right)+3n+3-1\right].$
$=n\left[2 n^{3}+2 n+4 n^{2}-4 n^{2}-6 n-2+3 n+2\right]$
$=n\left[2 n^{3}-n\right]=n^{2}\left(2 n^{2}-1\right).$
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