$\frac{2^3-1^3}{1\times 7}+\frac{4^3-3^3+2^3-1^3}{2\times 11}+\frac{6^3-5^3+4^3-3^3+2^3-1^3}{3\times 15}+\dots ..+\frac{{30}^3-{29}^3+{28}^3-{27}^3+\dots +2^3-1^3}{15\times 63}$ is equal to ______.

Given series, $\frac{2^3-1^3}{1\times 7}+\frac{4^3-3^3+2^3-1^3}{2\times 11}+\frac{6^3-5^3+4^3-3^3+2^3-1^3}{3\times 15}+\dots ..+\frac{{30}^3-{29}^3+{28}^3-{27}^3+\dots +2^3-1^3}{15\times 63}$ 

Now finding $T_n=\frac{{\sum }_{k=1}^n\left[{\left(2k\right)}^3-{\left(2k-1\right)}^3\right]}{n\left(4n+3\right)}$

$=\frac{{\sum }_{k=1}^{n}4k^2+{\left(2k-1\right)}^2+2k\left(2k-1\right)}{n\left(4n+3\right)}$

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

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

$=\frac{n^2\left(4n+3\right)}{n\left(4n+3\right)}=n$

So,$T_n=n$

Now finding $S_n=\sum _{n=1}^{15}{T}_n=\frac{15\times 16}{2}=120$