Sum of Odd Sequence of Products of Consecutive Fibonacci Numbers

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

Then:


 * $$\sum_{j = 1}^{2n-1} F_j F_{j+1} = F_{2n}^2$$


 * $$\sum_{j = 1}^{2n} F_j F_{j+1} = F_{2n+1}^2 - 1$$

Proof
Proof by induction:

For all $$n \in \N^*$$, let $$P \left({n}\right)$$ be the proposition $$\sum_{j = 1}^{2n-1} F_j F_{j+1} = F_{2n}^2$$.

Basis for the Induction

 * $$P(1)$$ is true, as this just says $$F_1 F_2 = 1 \times 1 = 1 = F_2^2$$.

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 2$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:
 * $$\sum_{j = 1}^{2k-1} F_j F_{j+1} = F_{2k}^2$$.

Then we need to show:
 * $$\sum_{j = 1}^{2k+1} F_j F_{j+1} = F_{2\left({k + 1}\right)}^2$$.

Induction Step
This is our induction step:

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So $$P \left({k}\right) \implies P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore $$\forall n \ge 1: \sum_{j = 1}^{2n-1} F_j F_{j+1} = F_{2n}^2$$.

For the second result, we have:

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