Finding coefficient of $x^{15}$ $$(x+x^2+x^3+x^4+x^5)\cdot (x^2+x^3+x^4+…)^5$$
I have done a little :
$$x(1 + x+x^2+x^3+x^4)\cdot x^{10}(1 + x^2+x^3+…)^5$$
By generating functions:
$$\begin{align}&x^{11}\cdot\frac{1 - x^5}{1-x}\cdot\frac{1}{(1-x)^5}\\[1ex] 
\implies &x^{11}(1 - x^5)\cdot\frac{1}{(1-x)^6}\\[1.5ex]
\implies &x^{11}(1 - x^5) \cdot\sum_{n=0}^{\infty}\binom{n+5}{5} x^n\\
\end{align}$$
How am I supposed to find $x^{15}$or any other one like $x^{18} , x^{19}$ 
Any clues? Thanks.

You are most of the way home with what you have done already, the last step involves manipulating the summation  by multiplying through by $x^{11}(1-x^5)=x^{11}-x^{16}$ like so
$$\begin{align}(x^{11}-x^{16})\sum_{n=0}^{\infty}\binom{n+5}{5}x^n&=\sum_{n=0}^{\infty}\binom{n+5}{5}(x^{n+11}-x^{n+16})\\&=\sum_{n=0}^{\infty}\binom{n+5}{5}x^{n+11}-\sum_{n=0}^{\infty}\binom{n+5}{5}x^{n+16}\end{align}$$
relabelling summation indices $n\rightarrow n-11$ and $n\rightarrow n-16$ for the first and second summations respectively gives
$$\sum_{n=11}^{\infty}\binom{n-6}{5}x^{n}-\sum_{n=16}^{\infty}\binom{n-11}{5}x^{n}$$
but if we define $\binom{a}{b}=0$ for $a\lt b$ such that $a,b\in \mathbb{Z}$ then we can write this as
$$\sum_{n=0}^{\infty}\binom{n-6}{5}x^{n}-\sum_{n=0}^{\infty}\binom{n-11}{5}x^{n}$$
or simply
$$\sum_{n=0}^{\infty}\left(\binom{n-6}{5}-\binom{n-11}{5}\right)x^n$$
hence your coefficients $c_n$ in
$$\frac{x^{11}-x^{16}}{(1-x)^6}=\sum_{n=0}^{\infty}c_nx^n$$
are
$$c_n=\binom{n-6}{5}-\binom{n-11}{5}\tag{Answer}$$
e.g. To find the coefficient $x^{15}$ plug $n=15$ into that formula.
