Sum of Odd Sequence of Products of Consecutive Fibonacci Numbers

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

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

Proof
Proof by induction:

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

Basis for the Induction
$\map P 1$ 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 $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

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

Then we need to show:
 * $\ds \sum_{j \mathop = 1}^{2 k + 1} F_j F_{j + 1} = {F_{2 \paren {k + 1} } }^2$

Induction Step
This is our induction step:

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

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