$a_n=$ $(-2)^{2n-1}$ for $n\geq2$, $a_0=0$, $a_1=10$
Solution I found:
$A(z)=0*z^0+10*z^1+\sum_{n=\geq2}^{\infty}(-2)^{2n-1}*z^n=$
$$A(z)=10 + (-2)^3z^2+(-2)^5z^3+...+(-2)^{2n-1}z^n$$
$$(-2)^{2n-1}z^n=b_n$$
$$\frac{b^{n+1}}{b^n}$$
$$\frac{(-2)^{2(n+1)-1}z^{n+1}}{(-2)^{2n-1}z^{n}}=(-2)^2z$$
$$q=4z$$ $$W_0=(-2)^3z^2$$ $$A(z)=10z+\frac{(-2)^3z^2}{1-4z}$$
Is this solved correctly? If so, could anyone explain $W_0$ part? I think its something connected to conditions at the start.
Is this the right answer?
Using the sum of a geometric series, $$ \begin{align} 10x+\sum_{k=2}^\infty(-2)^{2n-1}x^n &=10x-\frac12\sum_{k=2}^\infty(4x)^n\\ &=10x-\frac{8x^2}{1-4x}\tag1 \end{align} $$ which agrees with your answer.
The sum of an infinite geometric series with first term $a$ and common ratio $r$ (with $|r|\lt1$) is $$ \sum_{n=0}^\infty ar^n=\frac{a}{1-r}\tag2 $$ The $W_0$ part of your answer is just the first term: $a$ in $(2)$.