The sum of the series

Solution:

(b) $\frac{n(2 n-1)(2 n+1)}{3}$

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

Thus, we have:

$T_{n}=(2 n-1)^{2}=4 n^{2}+1-4 n$

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

Thus, we have:

 

$S_{n}=\sum_{k=1}^{n}\left(4 k^{2}+1-4 k\right)$

$\Rightarrow S_{n}=4 \sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} 1-4 \sum_{k=1}^{n} k$

$\Rightarrow S_{n}=\frac{4 n(n+1)(2 n+1)}{6}+n-\frac{4 n(n+1)}{2}$

$\Rightarrow S_{n}=\frac{2 n(n+1)(2 n+1)}{3}+n-2 n(n+1)$

$\Rightarrow S_{n}=n\left[\frac{2(n+1)(2 n+1)}{3}+1-2(n+1)\right]$

$\Rightarrow S_{n}=\frac{n}{3}[(2 n+2)(2 n+1)+3-6(n+1)]$

$\Rightarrow S_{n}=\frac{n}{3}\left[\left(4 n^{2}-1\right)\right]$

$\Rightarrow S_{n}=\frac{n(2 n-1)(2 n+1)}{3}$