The sum to 16 terms of the series

$1^3+\frac{1+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\frac{1^3+2^3+3^3+4^3}{1+3+5+7}+\dots$

$t_r=\frac{1^3+2^3+\dots +r^3}{1+3+\dots (2r-1)}=\frac{r^2(r+1{)}^2}{4r^2}=\frac{(r+1{)}^2}{4}$

Thus, $\sum _{r=1}^{16} t_r=\frac{1}{4}\left(\sum _{r=1}^{17} r^2-1\right)$ 

$=\frac{1}{4}\left(\frac{1}{6}\left(17\right)\left(18\right)\left(35\right)-1\right)=446$