The sum of first 9 terms of the series 1 3 1 + 1 3 + 2 3 1 + 3 + 1 3 + 2 3 + 3 3 1 + 3 + 5 + . . . is

Given, $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+...$

$\therefore T_n=\frac{1^3+2^3+...+n^3}{1+3+...+\left(2n-1\right)}$

$=\frac{\frac{n^2{\left(n+1\right)}^2}{4}}{n^2}$

$=\frac{{\left(n+1\right)}^2}{4}$

Sum of $9$ terms

$=\sum _{n=1}^9\frac{{\left(n+1\right)}^2}{4}$

$=\frac{1}{4}\left[2^2+3^2+...+10^2\right]$
$=\frac{1}{4}\left[\left(1^2+2^2+...+10^2\right)-1^2\right]$   

$=\frac{1}{4}\left[\frac{10·11·21}{6}-1\right]$     

$=\frac{1}{4}\left[384\right]=96$