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.
 A: 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.
A: 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 $n<0$.
