Question

The sum of the first $n$ terms of the series $1^2+{2.2}^2+3^2+{2.4}^2+5^2+{2.6}^2+\dots ..$ is $\frac{n(n+1{)}^2}{2}$ when $n$ is even. When $n$ is odd, the sum is -

• A. $\frac{3n(n+1)}{2}$
• B. $\frac{n^2(n+1)}{2}$
• C. $\frac{n(n+1{)}^2}{4}$
• D. ${\left[\frac{n(n+1)}{4}\right]}^2$

Answer 1

The correct option is B $\frac{n^2(n+1)}{2}$

Let $n=Even$ then,

$\begin{array}{c}{S}_n=1^2+{2.2}^2+3^2+{2.4}^2+.......+2.n^2\\ =1^2+{2.2}^2+3^2+{2.4}^2+5^2+{2.6}^2+.......+{\left(n-1\right)}^2+2.n^2\\ =\frac{n{\left(n+1\right)}^2}{2}\end{array}$

If $n=$odd then, $n-1$ is even then,

$\begin{array}{c}{T}_n=1^2+{2.2}^2+3^2+{2.4}^2+5^2+{2.6}^2+.......+2{\left(n-1\right)}^2+n^2\\ =S_{n-1}+n^2=\frac{\left(n-1\right)\left(n-1+2\right)}{2}+n^2\\ i.e\frac{\left(n-1\right){\left(n\right)}^2}{2}+n^2\\ \frac{n^2\left(n-1\right)+2n^2}{2}=\frac{n^2\left(n-1+2\right)}{2}=\frac{n^2\left(n+1\right)}{2}\end{array}$

Hence, this is the answer.