<s>[INST] Number of Non negative integer solutions of $x+2y+5z=100$ Find Number of Non negative integer solutions of $x+2y+5z=100$
My attempt: 
we have $x+2y=100-5z$ 
Considering the polynomial $$f(u)=(1-u)^{-1}\times (1-u^2)^{-1}$$
$\implies$
$$f(u)=\frac{1}{(1-u)(1+u)}\times \frac{1}{1-u}=\frac{1}{2} \left(\frac{1}{1-u}+\frac{1}{1+u}\right)\frac{1}{1-u}=\frac{1}{2}\left((1-u)^{-2}+(1-u^2)^{-1}\right)$$ 
we need to collect coefficient of $100-5z$ in the above given by
$$C(z)=\frac{1}{2} \left((101-5z)+odd(z)\right)$$
Total number of solutions is
$$S(z)=\frac{1}{2} \sum_{z=0}^{20} 101-5z+\frac{1}{2} \sum_{z \in odd}1$$
$$S(z)=540.5$$
what went wrong in my analysis?
 [/INST] Given: $x+2y=100-5z$, tabulate:
$$\begin{array}{c|c|c}
z&x&\text{count}\\
\hline
0&100,98,\cdots, 0&\color{red}{51}\\
1&\ \ 95,93,\cdots, 1&\color{blue}{48}\\
2&\ \ 90,88,\cdots, 0&\color{red}{46}\\
3&\ \ 85,83,\cdots, 1&\color{blue}{43}\\
4&\ \ 80,78,\cdots, 0&\color{red}{41}\\
\vdots&\vdots&\vdots\\
17&15,13,\cdots,1&\color{blue}{8}\\
18&10,8,\cdots,0&\color{red}{6}\\
19&5,3,1&\color{blue}{3}\\
20&0&\color{red}{1}\\
\hline
&&\color{red}{286}+\color{blue}{255}=541
\end{array}$$
</s>