Sum of Sequence of Squares of Fibonacci Numbers

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

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

That is:
 * ${F_1}^2 + {F_2}^2 + {F_3}^2 + \cdots + {F_n}^2 = F_n F_{n + 1}$

Proof
Proof by induction:

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

Basis for the Induction
$\map P 1$ is the case ${F_1}^2 = 1 = F_3 - 1$, which holds from the definition of Fibonacci numbers.

demonstrating that $\map P 1$ holds.

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:
 * $\ds \sum_{j \mathop = 1}^k {F_j}^2 = F_k F_{k + 1}$

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

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