Answer:
Option B
Explanation:
Central Idea. Write the nth term of the given series and simplify it to get its lowest forth. Then , apply , Sn = ∑ Tπ
Given series is
$\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+...\infty$
Let Tn be the nth term of the given series
$\therefore$ $\frac{1^{3}+2^{3}+3^{3}+....+n^{3}}{1+3+5+.....+n terms}$
= $\frac{\left\{\frac{n(n+1)}{2}\right\}^{2}}{n^{2}}$
= $\frac{(n+1)^{2}}{4}$
$S_{9}=\sum_{n=1}^{9}\frac{(n+1)^{2}}{4}=\frac{1}{4}[(2^{2}+3^{2}+.....+10^{2})$
+ $1^{2}-1^{2}]$
$=\frac{1}{4}\left[\frac{10(10+1)(20+1)}{6}-1\right]=\frac{384}{4}=96$
