Proof by Induction of Sum of Squares of Fibonacci using Difference Opperators Consider the sequence of Fibonacci numbers $\{F_n\}_{n\geq0}$ where $F_0=0,F_1=1$ and $F_n=F_{n-1}+F_{n-2}$ for $n\geq2$. It is proved that 
\begin{equation}\sum_{i=0}^nF_i^2=F_nF_{n+1}.\end{equation} 
Suppose we generalize the Fibonacci numbers so that we have a sequence $\{A_n\}_{n\geq0}$ such that $A_0=0, A_1=1$ and $A_{n}=aA_{n-1}+bA_{n-2}$ for $n\geq2$.  We will assume that $b \ne 0$.  We can show that 
$$b^nA_0^2+b^{n-1}A_1^2+\cdots+bA_{n-1}^2+A_n^2=\frac{A_nA_{n+1}}{a}.$$ 
This formula can be proved by induction. I am not very good at proofs nor am I good at induction. If I could get some help that would be great.

Let $\{A_n\}_{n\geq 0}$ be s.t. $A_0=0$, $A_1=1$, and $A_{n+2}=a A_{n+1}+b A_n$ for all non-negative integer $n$.  We'll prove by induction on $n$ that
$$(1)\ \ \ \ \ a\sum_{k=0}^n b^{n-k}A_k^2=A_nA_{n+1}.$$
I do not see why we need the restriction $b\neq 0$.  If we treat $0^0$ as $1$, then (1) holds even in this case.
The claim is trivial for $n=0$ (since we have $0=0$).  For $n=1$, we have $$a(bA_0+A_1)=a(0+1)=a=1\cdot a=A_1A_2.$$
Suppose (1) holds for $n$.  Then,
$$a\sum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=ab\sum_{k=0}^nb^{n-k}A_k^2+aA_{n+1}^2.$$
By induction $$ab\sum_{k=0}^nb^{n-k}A_k^2=b\left(a\sum_{k=0}^nb^{n-k}A_k^2\right)=b\left(A_nA_{n+1}\right).$$  That is,
$$a\sum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=b(A_nA_{n+1})+aA_{n+1}^2=A_{n+1}(aA_{n+1}+bA_n).$$
Since $A_{n+2}=aA_{n+1}+bA_n$, we obtain
$$a\sum_{k=0}^{n+1}b^{(n+1)-k}A_k^2=A_{n+1}A_{n+2}$$
and the induction is complete.
