Question

Find the sum of the series up to $n$ terms $\mathrm{1.3.5}+\mathrm{3.5.7}+\mathrm{5.7.9}+.....$

• A. $8n^3+12n^{2}2n-3$
• B. $n(8n^3+11n^2-n-3)$
• C. $n(2n^3+8n^2+7n-2)$
• D. None of these

Answer 1

The correct option is C $n(2n^3+8n^2+7n-2)$

$T_n=[1+(n-1\left)2\right][3+(n-1\left)2\right][5+(n-1\left)2\right]=(2n-1)(2n+1)(2n+3)=8n^3+12n^2-2n-3$

$\therefore S_n=\sum T_n=8\sum n^3+12\sum n^2-2\sum n-3\sum 1$

$=8\frac{\left(n\right(n+1){)}^2}{4}+12\frac{n(n+1)(2n+1)}{6}-2\frac{n(n+1)}{2}-3n$

$=2n^2(n+1{)}^2+2n(2n^2+3n+1)-n(n+1)-3n$

$=2n^4+8n^3+7n^2-2n=n(2n^3+8n^2+7n-2)$

Hence, option 'C' is correct.