Suppose that the lucas numbers are $l_n=l_{n-1}+l_{n-2}$ for all $n \ge 1$ where $l_0=2$ and $l_1=1$.
Prove that $l_0^2+l_1^2+...+l_n^2=l_n*l_{n+1}+2$ for $n \ge 0$
I think the easiest way to prove this would be induction. For the base case when $n=0$, we get that $l_0^2=2^2=4=2+2=(2*1)+2=l_0*l_1+2$.
I've tried my hand at proving the inductive step but can't solve it.
We have: $$L_0=2,L_1=1$$ $$L_n=L_{n-1}+L_{n-2}$$ We must prove: $$\sum_{k=0}^{n}L_k^2=L_n\cdot L_{n+1}+2$$ for $n\geq 0$. First, notice how: $$\sum_{k=0}^{0}L_k^2=L_0^2=4=2\cdot1+2=L_0\cdot L_1+2$$ Assume that: $$\sum_{k=0}^{n-1}L_k^2=L_{n-1}\cdot L_n+2$$ Now: $$\sum_{k=0}^{n}L_k^2=L_n^2+\sum_{k=0}^{n-1}L_k^2=L_n^2+L_{n-1}\cdot L_n+2=L_n(L_n+L_{n-1})+2=L_n\cdot L_{n+1}+2$$