Question

The sum of the series 1² + 3² + 5² + ... to n terms is 

पर्याय

• \[\frac{n (n + 1) (2n + 1)}{2}\]

• \[\frac{n (2n - 1) (2n + 1)}{3}\]

• \[\frac{(n - 1 )^2 (2n + 1)}{6}\]

• \[\frac{(2n + 1 )^3}{3}\]

Answer 1

\[\frac{n (2n - 1) (2n + 1)}{3}\]

Let \[T_n\]  be the nth term of the given series.
Thus, we have:

\[T_n = \left( 2n - 1 \right)^2 = 4 n^2 + 1 - 4n\]

Now, let

\[S_n\] be the sum of n terms of the given series.
Thus, we have:

\[S_n = \sum^n_{k = 1} \left( 4 k^2 + 1 - 4k \right)\]

\[ \Rightarrow S_n = {4\sum}^n_{k = 1} k^2 + \sum 1^n_{k = 1} - 4 \sum^n_{k = 1} k\]

\[ \Rightarrow S_n = \frac{4n\left( n + 1 \right)\left( 2n + 1 \right)}{6} + n - \frac{4n\left( n + 1 \right)}{2}\]

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

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

\[ \Rightarrow S_n = \frac{n}{3}\left[ \left( 2n + 2 \right)\left( 2n + 1 \right) + 3 - 6\left( n + 1 \right) \right]\]

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

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