# 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 $F_n\cdot(F_{n+1}-F_{n-1}) = F_n\cdot F_n = (F_n)^2$.

• You need to change $j$ to $i$ in the sums. – mathlove Jan 20 '18 at 16:28
• This is related to math.stackexchange.com/q/442459/147357 (which asked for a proof of the fact and didn't provide it's own attempt). – Teepeemm Jan 20 '18 at 22:06
• There are several Q&A's about this identity, but this one specifically asks "Is the following Proof Correct?" – Martin R Jan 20 '18 at 22:13

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”.

From the figure below the theorem's proof comes. The rectangle's area is $F_{i}F_{i+1}$.

• This doesn't address the question asked, which was if the OP's proof was correct. While I think it is often appropriate to supply additional information such as this, you should always state that you are doing so instead of answering the question. – Paul Sinclair Jan 21 '18 at 4:50

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 $k=1:n$.

• This does not address the question asked and was posted over an hour after Pet123's post, from which it differs only in arranging the squares in a spiralling pattern instead of building from a corner. What was the point? – Paul Sinclair Jan 21 '18 at 4:53
• @PaulSinclair Thanks for your comment. (1) I didn't see Pet123s at the time I typed this up and (2) my diagram is actually the template for the Fibonacci spiral and is more recognizable and well known. As for why I didn't see the post, I had the page open for over an hour while I falsely pursued the identity $\sum_1^n F_kF_{k-1}=F_n^2,~n \text{ even}$ with another mosaic. – Cye Waldman Jan 21 '18 at 16:33