Solve the following

Solution:

Let $T_{n}$ be the $n$th term of the given series.

Thus, we have:

Now, let $S_{n}$ be the sum of $n$ terms of the given series.

Thus, we have:

$S_{n}=\sum_{k=1}^{n} T_{k}$

$=\sum_{k=1}^{n}[2 k-1]^{3}$

$=\sum_{k=1}^{n}\left[8 k^{3}-1-6 k(2 k-1)\right]$

$=\sum_{k=1}^{n}\left[8 k^{3}-1-12 k^{2}+6 k\right]$

$=\sum_{k=1}^{n}\left[8 k^{3}-1-12 k^{2}+6 k\right]$

$=8 \sum_{k=1}^{n} k^{3}-\sum_{k=1}^{n} 1-12 \sum_{k=1}^{n} k^{2}+6 \sum_{k=1}^{n} k$

$=\frac{8 n^{2}(n+1)^{2}}{4}-n-\frac{12 n(n+1)(2 n+1)}{6}+\frac{6 n(n+1)}{2}$

$=2 n^{2}(n+1)^{2}-n-2 n(n+1)(2 n+1)+3 n(n+1)$

$=n(n+1)[2 n(n+1)-2(2 n+1)+3]-n$

$=n(n+1)\left[2 n^{2}-2 n+1\right]-n$

$=n\left[2 n^{3}-2 n^{2}+n+2 n^{2}-2 n+1-1\right]$

$=n\left[2 n^{3}-n\right]$

$=n^{2}\left[2 n^{2}-1\right]$