I have a question regarding following exercise. I have to prove by induction. Given is: $$ \begin{cases} g(0) = 0 \\ g(n)= (y-2) g(n-1)+1 \end{cases} $$ I have to prove following : For all $n \in \mathbb N$ : $g(n+1) = (y-2)^n +g(n)$
Induction basis and assumption are clear but i get stuck in induction final steps for n=k+1. Can someone help with the steps?
$g(n)= (y-2)g(n-1)+1$ for all $n$
Suppose for some $k$ that $g(k) = (y-2)^{k-1} +g(k-1)$ (induction hypothesis).
$(y-2)g(k)+1 = (y-2)((y-2)^{k-1} +g(k-1))+1$
$g(k+1)=(y-2)g(k)+1 =(y-2)^{k} +(y-2)g(k-1)+1=(y-2)^k+g(k)$
Then $g(n+1)=(y-2)^n+g(n)$ for all $n$