Sum of Odd Sequence of Products of Consecutive Fibonacci Numbers

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

Then:
 * $(1): \quad \displaystyle \sum_{j \mathop = 1}^{2 n - 1} F_j F_{j + 1} = {F_{2 n} }^2$


 * $(2): \quad \displaystyle \sum_{j \mathop = 1}^{2 n} F_j F_{j + 1} = {F_{2 n + 1} }^2 - 1$

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{j \mathop = 1}^{2 n - 1} F_j F_{j + 1} = {F_{2 n} }^2$

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 $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.

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

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

Induction Step
This is our induction step:

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

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

and proposition $(1)$ has been proved.

For proposition $(2)$, we have: