A formula for the sum of the squares of the first $n$ Fibonacci numbers.

(Recall that the fibonacci sequence is defined recursively by $F_0=1, F_1=1, \cdots, F_n=F_{n-1}+F_{n-2}, \cdots$.)

There's an interesting tweet from @fermatslibrary https://twitter.com/fermatslibrary/status/1533442276831154178. It shows a picture which proved to me a fascinating formula for the sum of the squares of the Fibonacci numbers up to any particular term. I wanted to make an interactive picture to check this for several more values.

The interactive diagram below gives a pretty convincing visual argument for the theorem that the sum of the squares of the Fibonacci numbers out to $F_n$ is the product of $F_n$ and $F_{n+1}$, that is, for $n=2,3,\cdots$, $\sum_0^n F_i^2 = F_n \times F_{n+1}$.

Press the activate button, and then use to slider to check this out up to $n=25$.

For those who aren't convinced by picture proofs, here's the inductive proof.

For the case $n=2$, $F_0^2+F_1^2+F_2^2=1+1+4=2\times 3=F_2\times F_3$. Now suppose the case $n=k>2$ is true. Then $F_0^2+\cdots +F_k^2=F_k\times F_{k+1}$. Add $F_{k+1}^2$ to both sides and factor the right hand side. Then use the inductive definition of $F_{k+2}=F_{k+1}+F_k$ to get that the theorem holds for the case $n=k+1$. Conclude the formula is true for all $n>2$.