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, when $n$ is even. When $n$ is odd, the sum is

• A. $\frac{n{(n+1)}^2}{4}$
• B. $\frac{n^2(n+2)}{4}$
• C. $\frac{n^2(n+1)}{4}$
• D. $\frac{n{(n+2)}^2}{4}$

Answer 1

The correct option is A $\frac{n{(n+1)}^2}{4}$

$S=1^2+2^2+3^2+{2.4}^2+....+n^2,nisoddS=(1^2+2^2+3^2+....+n^2)+(2^2+4^2+....(n-1{)}^2=\frac{n(n+1)(2n+1)}{6}+2^2(1^2+2^2+3^2+....+{\left(\frac{n-1}{2}\right)}^2=\frac{n(2n^2+3n+1)}{6}+4\frac{\left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right)n}{6}S=\frac{n^2(n+1)}{2},nisoddIfniseven,S=(1^2+2^2+3^2+....+n^2)+(2^2+4^2+....+n^2)=\frac{n(n+1)(2n+1)}{6}+\frac{4\left(\frac{n}{2}\right)(\frac{n}{2}+1)(n+1)}{6}=\frac{n(n+1)(2n+1+n+2)}{6}=\frac{n(n+1{)}^2}{2}$