Question

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

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

Answer 1

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

$S_n=1^2+2.2^2+3^2+2.4^2+5^2+...$

$=(1^2+3^2+5^2+...)+(2.2^2+2.4^2+2.6^2+...)+{(n-1)}^2+2n^2$

$=(1^2+3^2+5^2+...+{(n-1)}^2)+(2.2^2+2.4^2+2.6^2+...+2n^2)$

When $n$ is even,

$S_n=\frac{n{(n+1)}^2}{2}$

When $n$ is odd,

$S_n=1+2.2^2+3^2+2.4^2+...+2{(n-1)}^2+n^2$

When $n-1$ is even,

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

$=\frac{(n-1)n^2}{2}$

When $n$ is odd,

$S_n=S_{n-1}+n^2$

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

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