Summation of Squares of Fibonacci numbers Is the following Proof Correct?
Theorem. Consider the following statement.
 $$\forall n\in\mathbf{N}\left(\sum_{j=0}^{n}(F_i)^2 = F_nF_{n+1}\right)$$
Proof. We prove the statement by recourse to Strong-Induction. Assume for an arbitrary natural number $n$ that 
 $$\forall n<k\left(\sum_{j=0}^{k}(F_i)^2 = F_kF_{k+1}\right)$$
 Now consider the following Cases.
Case-1: $n = 0$,then since $F_0 = 0$ we have $\sum_{i=0}^{0}(F_i)^2 = (F_0)^2 = 0^2 = 0 = 0\cdot F_1 = F_0F_1$.
Case-2: $n = 1$, then since $F_1 = 1$ we have $\sum_{i=0}^{0}(F_i)^2 = (F_0)^2+ (F_1)^2 = 0^2 +1^2 = 1 =  1\cdot(1+0) = F_1\cdot F_2$
Case-3: $n \ge 2$,then by considering the inductive hypothesis in the particular for $k = n-1$ we see that 
 $$\sum_{j=0}^{n}(F_i)^2 = \sum_{j=0}^{k}(F_i)^2 +(F_n)^2 = F_{n-1}F_n+(F_n)^2$$
 from this point onwards we must show that $F_{n-1}F_n+(F_n)^2 = F_nF_{n+1}$, instead we prove the equivalent statement $(F_n)^2 = F_nF_{n+1}-F_nF_{n-1}$, equivalently we have $F_{n}\cdot(F_{n+1}-F_{n-1})$ from definition of the Fibonacci sequence we know that $\forall n\ge 2(F_{n+1} = F_{n}+F_{n-1})$ consequently we have $F_{n} = F_{n+1}-F_{n-1}$ implying that $F_n\cdot(F_{n+1}-F_{n-1}) = F_n\cdot F_n = (F_n)^2$.
 A: Your proof looks correct apart from some typos: $\forall n<k$ should be $\forall k < n$,
and (as @mathlove
noticed) there is a mix-up between $i$ and $j$ in the indices.
But it can be simplified. In particular you
don't need strong induction since only the inductive hypothesis
for $n-1$ is used to prove the statement for $n$, so “simple induction” is sufficient.
Also $F_{n+1} = F_{n}+F_{n-1}$ holds for $n\ge 1$ and not only
for $n \ge 2$, therefore it is sufficient to consider a single
base case ($n = 0$).
For $n \ge 1$ the inductive step then would be: From the
inductive hypothesis we have
$$
\sum_{i=0}^{n-1}(F_i)^2 = F_{n-1}F_n \, .
$$
It follows that
$$
\sum_{i=0}^{n}(F_i)^2 = F_{n-1}F_n+(F_n)^2 = (\underbrace{F_{n-1} + F_n}_{F_{n+1}}) F_n  = F_{n+1}F_n \, .
$$
In the final step, the recurrence relation of the Fibonacci numbers
is used directly, without transformation to an “equivalent statement”.
A: From the figure below the theorem's proof comes. The rectangle's area is $F_{i}F_{i+1}$.

A: This identity can be seen readily in the Fibonacci mosaic below. Clearly, the area of the overall rectangle, $F_{n+1}\times F_{n}$ is the sum of the areas of the individual squares $F_{k}\times F_{k}$ from $k=1:n$.

