Find the sum of the series whose nth term is:
(2n − 1)2
Solution
Let \[T_n\] be the nth term of the given series.
Thus, we have:
\[T_n = \left( 2n - 1 \right)^2\]
Let \[S_n\] be the sum of n terms of the given series.
Now,
\[S_n = \sum^n_{k = 1} T_k\]
\[\Rightarrow S_n = \sum^n_{k = 1} \left( 2k - 1 \right)^2 \]
\[ \Rightarrow S_n = \sum^n_{k = 1} \left( 4 k^2 + 1 - 4k \right)\]
\[ \Rightarrow S_n = {4\sum}^n_{k = 1} k^2 + \sum 1^n_{k = 1} - 4 \sum^n_{k = 1} k \]
\[ \Rightarrow S_n = \frac{4n\left( n + 1 \right)\left( 2n + 1 \right)}{6} + n - \frac{4n\left( n + 1 \right)}{2}\]
\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left[ \frac{4\left( 2n + 1 \right)}{3} - 4 \right] + n\]
\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{8n + 4 - 12}{3} \right) + n\]
\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{8n - 8}{3} \right) + n\]
\[ \Rightarrow S_n = 4n\left( n + 1 \right)\left( \frac{n - 1}{3} \right) + n\]
\[ \Rightarrow S_n = \frac{n\left( 4n + 4 \right)\left( n - 1 \right) + 3n}{3}\]
\[ \Rightarrow S_n = \frac{n}{3}\left( 4 n^2 + 4n - 4n - 4 + 3 \right)\]
\[ \Rightarrow S_n = \frac{n}{3}\left( 4 n^2 - 1 \right)\]
\[ \Rightarrow S_n = \frac{n}{3}\left( 2n - 1 \right)\left( 2n + 1 \right)\]
