Question

Find the sum of first $n$ terms of the series.
$1^3+3\times 2^2+3^3+3\times 4^2+5^3+3\times 6^2+....$ If $n$ is odd.

• A. $\frac{1}{8}(n+1)[n^3+7n^2-3n-1]$
• B. $\frac{n}{8}(n^2+4n^2+10n+6)$
• C. $\frac{1}{8}(n+1)[n^2+7n^2-3n-1]$
• D. $\frac{1}{8}(n+1)[n^2+7n^2-3n]$

Answer 1

The correct option is A $\frac{1}{8}(n+1)[n^3+7n^2-3n-1]$

If $n$ is odd.
Then $(n+1)$ is even in the case
Sum of first $n$ terms $=$ Sum of first $(n+1)$ terms $-$ $(n-1{)}^{th}$ term.
$=\frac{(n+1)}{8}\left[\right(n+1{)}^3+4(n+1{)}^2+10(n+1)+8]-3(n+1{)}^2$
$=\frac{1}{8}(n+1)[n^3+3n^2+3n+1+4n^2+8n+4+10n+10+8-24n-24]$
$=\frac{1}{8}(n+1)[n^3+7n^2-3n-1]$

Hence option ${}^{\prime }{A}^{\prime }$ is the answer.