Question

The sum of first n terms of the series $1^2+2\times 2^2+3^2+2\times 4^2+5^2+2\times 6^2+.....$ is $\frac{n(n+1{)}^2}{2}$ when $n$ is even. When $n$ is odd the sum of the series is

• A. $\frac{n^2(3n+1)}{4}$
• B. $\frac{n^2(n+1)}{2}$
• C. $\frac{n^3(n-1)}{2}$
• D. None of these.

Answer 1

Answer: (B)

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

Let $n=2m$, then
$S_{2m}=1^2+2\times 2^2+3^2+2\times 4^2+.....+(2m-1{)}^2+2(2m{)}^2$
$=2m(2m+1{)}^2/2=m(2m+1{)}^2$
When $n=2m-1$
$1^2+2\times 2^2+3^2+2\times 4^2+.....+(2m-1{)}^2$
$=S_{2m}-2(2m{)}^2=m(2m+1{)}^2-2(2m{)}^2$
$=m[4m^2+4m+1-8m]=m(2m-1{)}^2$
$=n^2(n+1)/2$
Hence, option 'B' is correct.