Question

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

When $n$ is odd, then the sum is

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

Answer 1

Answer: (D)

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

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

$(n-1{)}^{th}$ term is $2(n-1{)}^2$ the $n^{th}$ term is $n^2$
Hence, using the given formula is valid if $n$ is even, so substitute $n-1$ in that formula in place of $n$ and add $n^2$
We get,

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