User:Alecscooper/Sandbox

Let $$F_k$$ be the $$k$$'th Fibonacci number.

Then:


 * $$\sum_{i = 0}^{n} F_i F_{i+1} = F_{n+1}^2$$ if $$n$$ is even


 * $$\sum_{i = 0}^{n} F_i F_{i+1} = F_{n+1}^2 - 1$$ if $$n$$ is odd

We will prove the first part by induction and then demonstrate that the second part follows.

Proof of the Even Case
Trivially, $$\sum_{i=0}^0 F_i\cdot F_{i+1} = F_0 \cdot F_1 = 1\cdot 1 = 1 = F_1^2$$, so the base case holds.

We now assume the statement holds for $$n$$, that is: $$\sum_{i = 0}^{n} F_i \cdot F_{i+1} = F_{n+1}^2$$. We must show that it holds for $$n+2$$.

$$ $$ $$ $$ $$ $$

So we have demonstrated our assertion for even $$n$$ by induction.

Proof of the Odd Case
We now examine the case of odd $$n$$.

We start with a lemma: $$F_n \cdot F_{n+2} = F_{n+1}^2 - 1$$ for odd $$n$$, which we will prove by induction.

The base case holds since $$F_1 \cdot F_3 = 1\cdot3 = 3 = 4 - 1 = F_2^2 -1$$.

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Too be honest, the last form is not really any simpler on it's own than that given in the step immediately proceeding it, I just think it is a little more elegant, and it's the form given on proofwiki, from which I derived this proof.