How to prove this Fibonacci Statement?

Question

So here is the problem:

$$F_1F_2 + F_2F_3 + F_3F_4 + ... + F_{2n}F_{2n+1} = F_{2n+1}^2 - 1$$

How could I go about proving this? I tried induction but got lost in the meaning of $F_{2n}$. Any proofs would be appreciated (not just induction).

Answer 1

The mathematical induction way. \begin{align} \sum_{i=1}^{n+1}(F_{2i-1}F_{2i} + F_{2i}F_{2i+1} )&= F_{2n+1}^2-1+F_{2n+1}F_{2n+2}+F_{2n+2}F_{2n+3} \\ &=F_{2n+1}(F_{2n+1}+F_{2n+2})+F_{2n+2}F_{2n+3}-1\\ &=F_{2n+1}F_{2n+3}+F_{2n+2}F_{2n+3}-1\\ &=F_{2n+3}(F_{2n+1}+F_{2n+2})-1\\ &= F_{2n+3}^2-1 \end{align}

Remember to check base case.

Answer 2

$F_{2n+1}^2 - F_{2n}F_{2n+1} = F_{2n+1}(F_{2n+1} - F_{2n}) = F_{2n+1}F_{2n-1} = F_{2n}^2 + (-1)^{2n}$

$F_{2n}^2 - F_{2n}F_{2n-1} = F_{2n}(F_{2n} - F_{2n-1}) = F_{2n}F_{2n-2} = F_{2n-1}^2 + (-1)^{2n-1}$

...

$F_{2}^2 - F_{2}F_{1} = F_{1}^2 + (-1)^{1}$

Adding them all up you will get:

$F_{2n+1}^2 - \sum_{i=1}^{2n} F_{i}F_{i+1} = F_1^2 = 1$