<s>[INST] 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.
 [/INST] 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
$$
\begin{align}
\sum_{j=0}^2(-1)^j\binom{5}{j}\binom{34-13j}{4}
&=\binom{5}{0}\binom{34}{4}-\binom{5}{1}\binom{21}{4}+\binom{5}{2}\binom{8}{4}\\
&=17151
\end{align}
$$
</s>