prove by induction: $x_n=a^nx_0+b(1+a+\cdots+a^{n-1})$ given $f(x)=ax+b$ with initial value $x_0$

Question

prove by induction: $x_n=a^nx_0+b(1+a+\cdots+a^{n-1})$ given $f(x)=ax+b$ with initial value $x_0$

I'm fine with base case and hypothesis, but having some problems showing that it is true for $P(n+1)$ by using $P(n)$, can someone show me how to use $P(n)$?

Answer 1

The formula in question holds for $n=1$. Assume that it holds for some $n\geq1$. Then $$\eqalign{x_{n+1}&:=f(x_n)=ax_n+b\cr &\>=a\bigl(a^nx_0+b(1+\ldots+a^{n-1})\bigr)+b\cr &\>=a^{n+1}x_0+b(1+a+a^2+\ldots+a^n)\ ,\cr}$$ as it should be.