Question

The sum of the series up to $n$ term $\mathrm{1.3.5}+\mathrm{3.5.7}+\mathrm{5.7.9}+\dots \dots \dots .$. is

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

Answer 1

Answer: (C)

The sum of the series $\mathrm{1.3.5}+\mathrm{3.5.7}+\mathrm{5.7.9}+$ up to $n$ terms here $T_n=(2n-1)(2n+1)(2n+3)$
$T_n=8n^3+12n^2-2n-3$
$S_n=8\sum n^3+12\sum n^2-2\sum n-3n$
$=8{\left[\frac{n(n+1)}{2}\right]}^2+\frac{12n(n+1)(2n+1)}{6}-\frac{2n(n+1)}{2}-3n$
$=2n^2(n+1{)}^2+2n(n+1)(2n+1)-n(n+1)-3n$
$=n(n+1)[2n(n+1)+2(2n+1)-1]-3n$
$=n(n+1)\left[2n^2+6n+1\right]-3n$
$=2n^4+8n^3+7n^2-2n$
$=n\left(2n^3+8n^2+7n-2\right)$