Coefficients of $(1 + x^2 + x^4 + x^7)^{10}$ I've been asked to find the coefficient of $x^{25}$ in $(1 + x^2 + x^4 + x^7)^{10}$. While reasoning through the problem, I was thinking that this seems equivalent to solving the system of equations
$$
\begin{cases}
0\cdot x_1 + 2\cdot x_2 + 4\cdot x_3 + 7\cdot x_4 = 25 \\
x_1 + x_2 + x_3 + x_4 = 10 \\
x_i \geq 0.
\end{cases}
$$
While at first, this seemed helpful to me, I am not sure where to proceed from here. I have a feeling that either a clever use of generating functions or the multinomial theorem would be helpful, but I can't figure out how to go about it. Thanks in advance for any help.
 A: I think it's easiest if you break it up into cases.  Since 25 is odd and of the exponents only $x^7$ is an odd power, it follows that the only terms contributing to $x^{25}$ are those that use one or three factors of $x^7$.  
In essence, the desired coefficient is the same as taking the sum of the coefficients of $x^{18}y$ and $x^{4}y^3$ in the expansion of $(1+x^2+x^4+y)^{10}$.
This reduces the computation to the following pieces:


*

*You need to count the number of ways you can select one or three of the ten factors to correspond to the chosen factors of $x^7$, depending on the case.

*You need to determine the coefficient of $x^{18}$ in $(1+x^2+x^4)^{9}$.

*You need to determine the coefficient of $x^{4}$ in $(1+x^2+x^4)^7$.


The first of these should be routine, the second is trivial, and since the power of $x$ is low enough, the coefficient for the third is the same as if you were expanding $(1+x^2+x^4+x^6+\ldots)^7$ instead.
A: The following approach is rather unusual but might be interesting. We consider the polynomial
\begin{align*}
p(x)=\left(1+x^2+x^4+x^7\right)^{10}=\sum_{k=0}^{70}a_kx^k
\end{align*}
and are looking for $a_{25}$. We do so by using the technique of series multisection. Usually this technique is used to filter out elements in a periodical manner and obtain for instance a polynomial containing every third element in $p(x)$:
\begin{align*}
\sum_{k=0}^{\lfloor 70/3\rfloor} a_{3k}x^{3k}
\end{align*}

To get $a_{25}$ we use this technique to filter out each element but the first and apply a shift by $25$. In order to do so, we take the smallest prime $p$ greater than $70$, i.e. $p=71$ and consider a primitive $p$-th root of unity:
\begin{align*}
\omega=e^{2\pi i /71}
\end{align*}
We obtain
  \begin{align*}
\color{blue}{a_{25}}&\color{blue}{=\frac{1}{71}\sum_{k=0}^{70}e^{-50k\pi i/71}\left(1+e^{4k\pi i/71}+e^{8k\pi i/71}+e^{14k\pi i/71}\right)^{10}}\\
&\color{blue}{\,\,=34\,750}
\end{align*}
  where the value was found with some help of Wolfram Alpha.

