$$(x+x^2+x^3+x^4+x^5)\cdot (x^2+x^3+x^4+…)^5$$
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.
On
It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.
We obtain \begin{align*} \color{blue}{[x^{15}]}&\color{blue}{x^{11}(1-x^5)\sum_{n=0}^\infty \binom{n+5}{5}x^n}\\ &=[x^4](1-x^5)\sum_{n=0}^\infty\binom{n+5}{5}x^n\tag{1}\\ &=[x^4]\sum_{n=0}^\infty\binom{n+5}{5}x^n\tag{2}\\ &=\binom{9}{5}\tag{3}\\ &\color{blue}{=126} \end{align*}
Comment:
In (1) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
In (2) we note the term $x^5$ does not contribute to the coefficient of $x^4$ and can be skipped.
In (3) we select the coefficient of $x^4$.
In general we obtain for $k\geq 0$: \begin{align*} \color{blue}{[x^{k}]}&\color{blue}{x^{11}(1-x^5)\sum_{n=0}^\infty \binom{n+5}{5}x^n}\\ &=[x^{k-11}](1-x^5)\sum_{n=0}^\infty\binom{n+5}{5}x^n\tag{4}\\ &=\left([x^{k-11}]-[x^{k-16}]\right)\sum_{n=0}^\infty\binom{n+5}{5}x^n\tag{5}\\ &\color{blue}{=\binom{k-6}{5}-\binom{k-11}{5}}\tag{6}\\ \end{align*}
Comment:
In (4) we apply the same rule as in (1).
In (5) we use the linearity of the coefficient of operator and apply the same rule as in (1).
In (6) we select the coefficients accordingly. Note that here we set $\binom{n}{k}=0$ if $k>n$ or $n<0$.
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.