Find the coefficient of $x^{30}$. Find the coefficient of $x^{30}$ in the given polynomial
$$
\left(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}+x^{11}+x^{12}\right)^5
$$
I don't know how to solve problems with such high degree.
 A: $$[x^{30}]\left(\frac{1-x^{13}}{1-x}\right)^5 = [x^{30}]\sum_{k=0}^{5}\binom{5}{k}(-1)^k x^{13k}\sum_{n\geq 0}\binom{n+4}{4}x^n \tag{1}$$
hence the LHS of $(1)$ equals:
$$\binom{5}{0}\binom{34}{4}-\binom{5}{1}\binom{21}{4}+\binom{5}{2}\binom{8}{4}=\color{red}{17151.}\tag{2}$$
A: The coefficient attached to $x^{30}$ will be the number of ways you can add up to $30$ by using the numbers $0$-$12$ up to five times. (Here order matters)
For instance $1+1+2+10+6=30$ is one way.
$10+10+10+0+0=30$ is another and so is $0+10+10+10+0 = 30$.
The reason for this is more apparent for smaller polynomials.
For instance let's calculate the coefficient of $x^4$ in $(1+x+x^2)^3$.
$$(1+x+x^2)(1+x+x^2)(1+x+x^2)$$
Eventually we will have every combination of products between the three terms. We can achieve $x^4$ is several ways:
$$1\cdot x^2 \cdot x^2$$
$$x^2 \cdot 1 \cdot x^2$$
$$x^2 \cdot x^2 \cdot 1$$
$$x \cdot x \cdot x^2$$
$$x \cdot x^2 \cdot x$$
$$x^2 \cdot x \cdot x$$
Thus we have six ways of achieving $x^4$, and the powers of $x$ in each of these cases add up to $4$. Therefore the coefficient of $x^4$ is six.
A: This turns out to be a detailed explanation of Jack's answer.
$$
\begin{align}
\left(\frac{1-x^{13}}{1-x}\right)^5
&=(1-x^{13})^5(1-x)^{-5}\\
&=\sum_{j=0}^5(-1)^j\binom{5}{j}x^{13j}\sum_{k=0}^\infty(-1)^k\binom{-5}{k}x^k\tag{1}\\
&=\sum_{j=0}^5(-1)^j\binom{5}{j}x^{13j}\sum_{k=0}^\infty\binom{k+4}{k}x^k\tag{2}\\
&=\sum_{m=0}^\infty\sum_{j=0}^5(-1)^j\binom{5}{j}\binom{m-13j+4}{m-13j}x^m\tag{3}\\
&=\sum_{m=0}^{60}\sum_{j=0}^5(-1)^j\binom{5}{j}\binom{m-13j+4}{m-13j}x^m\tag{4}\\
&=\sum_{m=0}^{60}\sum_{j=0}^{\lfloor m/13\rfloor}(-1)^j\binom{5}{j}\binom{m-13j+4}{4}x^m\tag{5}
\end{align}
$$
Explanation:
$(1)$: Binomial Theorem
$(2)$: $\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}$
$(3)$: change variables $k\mapsto m-13j$
$(4)$: if $m\gt60$, then the sum in $j$ is an order $5$ difference of a degree $4$ polynomial
$(5)$: if $k\lt0$, then $\binom{n}{k}=0$; if $0\le k\le n$, then $\binom{n}{k}=\binom{n}{n-k}$
Plugging in $m=30$, we get the coefficient of $x^{30}$ to be
$$
\begin{align}
\sum_{j=0}^2(-1)^j\binom{5}{j}\binom{34-13j}{4}
&=\binom{5}{0}\binom{34}{4}-\binom{5}{1}\binom{21}{4}+\binom{5}{2}\binom{8}{4}\\
&=17151
\end{align}
$$
