Question

The sum of first $n$ terms of the series $1^2+2.2^2+3^2+2.4^2+5^2+5.6^2+...$ is $\frac{n{(n+1)}^2}{2}$ where $n$ is even. When $n$ is odd the sum is

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

Answer 1

Answer: (A)

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

When $n$ is odd, last term will be $n^2$,

$\therefore$ Then the sum is

$1^2+2.2^2+3^2+2.4^2+5^2+5.6^2+...+2{(n-1)}^2+n^2$

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

$[$ Replacing $n$ by $n-1$ in $\frac{n{(n+1)}^2}{2}]$

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