Fibonacci Number of Index 2n as Sum of Squares of Fibonacci Numbers

Theorem
Let $F_n$ denote the $n$th Fibonacci number.

Then:
 * $F_{2 n} = {F_{n + 1} }^2 - {F_{n - 1} }^2$

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 1}$, let $P \left({n}\right)$ be the proposition:
 * $F_{2 n} = {F_{n + 1} }^2 - {F_{n - 1} }^2$

Basis for the Induction
$P \left({1}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown 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 the induction hypothesis:
 * $F_{2 k} = {F_{k + 1} }^2 - {F_{k - 1} }^2$

from which it is to be shown that:
 * $F_{2 \left({k + 1}\right)} = {F_{k + 2} }^2 - {F_k}^2$

Induction Step
This is the induction step:

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

Therefore:
 * $\forall n \in \Z_{\ge 1}: F_{2 n} = {F_{n + 1} }^2 - {F_{n - 1} }^2$