Prove that $g(x)_n = f(x)_{n+1}$

Question

Both $f(x)$ and $g(x)$ are generating functions in closed form. $f(x) = \frac{1-x^2}{1-x-x^2} $ and $ g(x) = \frac{1}{1-x-x^2} $. Prove that $g(x)_n = f(x)_{n+1}$ for $n \ge 0$

How can I prove that? Can you guide me to the right path?

Answer 1

Let \begin{align*} f(x)&=\sum_{n=0}^\infty f_nx^n\\ g(x)&=\sum_{n=0}^\infty g_nx^n \end{align*} notice that \begin{align*} f(x)=\frac{x}{1-x-x^2}+1=xg(x)+1 \end{align*} which implies \begin{align*} \sum_{n=0}^\infty f_nx^n=1+\sum_{n=0}^\infty g_{n}x^{n+1} \end{align*} so the corresponding coefficient for $x^n$ is equal, then we get $g_n=f_{n+1}, \ \forall \ n\ge 0$.