Question

The sum of the series up to $n$ term $1.3 .5+3.5 .7+5.7 .9+\ldots \ldots \ldots .$. is

• A. $8 n^3+12 n^2-2 n-3$
• B. $n\left(8 n^3+11 n^2-n-3\right)$
• C. $n\left(2 n^3+8 n^2+7 n-2\right)$
• D. $n^4+6 n^3+7 n^2+n$

Answer 1

Answer: (C)

The sum of the series $1.3 .5+3.5 .7+5.7 .9+$ up to $n$ terms here $T_n=(2 n-1)(2 n+1)(2 n+3)$
$T_n=8 n^3+12 n^2-2 n-3$
$S_n=8 \sum n^3+12 \sum n^2-2 \sum n-3 n$
$=8\left[\frac{n(n+1)}{2}\right]^2+\frac{12 n(n+1)(2 n+1)}{6}-\frac{2 n(n+1)}{2}-3 n$
$=2 n^2(n+1)^2+2 n(n+1)(2 n+1)-n(n+1)-3 n$
$=n(n+1)[2 n(n+1)+2(2 n+1)-1]-3 n$
$=n(n+1)\left[2 n^2+6 n+1\right]-3 n$
$=2 n^4+8 n^3+7 n^2-2 n$
$=n\left(2 n^3+8 n^2+7 n-2\right)$