Question

The sum of the series 1² + 3² + 5² + ... to n terms is
(a) $\frac{n(n+1)(2n+1)}{2}$

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

(c) $\frac{(n-1{)}^2(2n+1)}{6}$

(d) $\frac{(2n+1{)}^3}{3}$

Answer 1

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

Let $T_n$ be the nth term of the given series.

Thus, we have:

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

Now, let $S_n$ be the sum of n terms of the given series.

Thus, we have:

$$\begin{gathered} S_n=\sum _{k=1}^n\left(4k^2+1-4k\right) \\ \Rightarrow S_n=\underset{k=1}{\overset{n}{4\sum }}{k}^2+\underset{k=1}{\overset{n}{\sum 1}}-4\sum _{k=1}^{n}k \\ \Rightarrow S_n=\frac{4n\left(n+1\right)\left(2n+1\right)}{6}+n-\frac{4n\left(n+1\right)}{2} \\ \Rightarrow S_n=\frac{2n\left(n+1\right)\left(2n+1\right)}{3}+n-2n\left(n+1\right) \\ \Rightarrow S_n=n\left[\frac{2\left(n+1\right)\left(2n+1\right)}{3}+1-2\left(n+1\right)\right] \\ \Rightarrow S_n=\frac{n}{3}\left[\left(2n+2\right)\left(2n+1\right)+3-6\left(n+1\right)\right] \\ \Rightarrow S_n=\frac{n}{3}\left[\left(4n^2-1\right)\right] \\ \Rightarrow S_n=\frac{n\left(2n-1\right)\left(2n+1\right)}{3} \end{gathered}$$