I am trying to learn Fibonacci tricks and I have one that I can not prove. I know it works because Ive tried it multiple times but I have not a clue how to prove. Here it is:
f(0)^2 + f(1)^2 + f(2)^2 + f(3)^2 = f(3)f(3+1)
0 + 1 + 1 + 4 = 2 * 3
= 6 =6
Is there a way to prove this?
Basis : ok
Inductive step : Assume $∑_{i=0}^{n}f(i)^2=f(n)f(n+1)$. $$∑_{i=0}^{n+1}f(i)^2=f(n)f(n+1)+f(n+1)^2$$$$∑_{i=0}^{n+1}f(i)^2=f(n+1)(f(n)+f(n+1))$$$$∑_{i=0}^{n+1}f(i)^2=f(n+1)f(n+2)$$
There is a geometric interpretation:
You've already proven that it works for one example $k=3$. Now write up the general equation $$ \sum_{k=0}^n F_k^2 = F_n\cdot F_{n+1} $$ and add $F_{n+1}^2$ on both sides. You get $$ F_{n+1}^2+ \sum_{k=0}^n F_k^2 =\sum_{k=0}^{n+1} F_k^2 = F_{n+1}^2+ F_n\cdot F_{n+1}=\left(F_{n+1}+ F_n\right)\cdot F_{n+1} $$ and use $F_{n+2}=F_{n+1}+ F_n$, the definition of Fibonacci numbers.
Since the base case is true, assume that $$ \sum_{i=0}^{n}f(i)^2=f(n)f(n+1)$$ is also true and use this assumption to prove that $$ \sum_{i=0}^{n+1}f(i)^2=f(n+1)f(n+2).$$
