The sum $1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+...+\frac{1^3+2^3+3^3+...+{15}^3}{1+2+3+...+15}-\frac{1}{2}\left(1+2+3+...+15\right)$ is equal to

Given series is $\left[1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+...+\frac{1^3+2^3+3^3+...+{15}^3}{1+2+3+...+15}\right]-\frac{1}{2}\left(1+2+3+...+15\right)$

First, we find the sum $1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+...+\frac{1^3+2^3+3^3+...+{15}^3}{1+2+3+...+15}$

General term for the series, is $T_r=\frac{1^3+2^3+3^3+...+r^3}{1+2+3+...+r}$

Using, the sum $1+2+3+...+n=\frac{n\left(n+1\right)}{2}$ and $1^3+2^3+3^3+...+n^3={\left(\frac{n\left(n+1\right)}{2}\right)}^2,$ we get

$T_r=\frac{{\left({\displaystyle \frac{r\left(r+1\right)}{2}}\right)}^2}{\left({\displaystyle \frac{r\left(r+1\right)}{2}}\right)}$

$\Rightarrow T_r=\frac{r\left(r+1\right)}{2}$

$\Rightarrow T_r=\frac{1}{2}(r^2+r)$

$\Rightarrow \sum _{r=1}^{15}{T}_r=\frac{1}{2}\left(\sum _{r=1}^{15}{r}^2+\sum _{r=1}^{15}r\right)$

Now, using $1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6},$ we get

$\sum _{r=1}^{15}{T}_r=\frac{1}{2}\left(\frac{15\times 16\times 31}{6}+\frac{15\times 16}{2}\right)$

$\Rightarrow \sum _{r=1}^{15}{T}_r=680$

Now, we will find the sum $\frac{1}{2}\left(1+2+3+\dots +15\right)=\frac{1}{2}\times \frac{15\times 16}{2}=60$

Hence, the required sum$=680-60=620.$