Fibonacci summation proof using matrices?

Question

I have a standard proof for the theorem:

$$\sum_{}^n f_1+f_3+f_5+...+f_{2n-1} = f_{2n}$$

$$f_i$$ refers to the Fibonacci numbers for future reference.

It involves setting p(k) as p(k+1) and proving it through weak induction, it has been graded by my professor and is correct.

However, I have recently came across the fibonacci matrix formulation from here:

How to prove Fibonacci sequence with matrices?

I am curious how I would go about solving this theorem with matrices.

I have tried using the product operator: $$\Pi$$

but I am not experienced enough to correctly formulate it so that they equal each other, for example:

$$\Pi_{i=1}^n \begin{bmatrix}1 & 1\\1 & 0\end{bmatrix}^{2n-1} = \begin{bmatrix}1 & 1\\1 & 0\end{bmatrix}^{2n}$$

Using the product operator may be completely pointless, but I honestly just don't know since I have never really used them before.

Any idea on how the original theorem is shown in matrices?

Thank you for any help in advance.

Yes, $$f_i$$ refers to the fibonacci numbers, i'll edit that in. Thank you. — Antas Santas, Sep 10 '18 at 19:25
@DietrichBurde That is a completely different Fibonacci identity. — Mike Earnest, Sep 10 '18 at 19:31
Ooops, will be corrected soon (this one is the right identity., but not yet with matrices). — Dietrich Burde, Sep 10 '18 at 19:31
One nice setup for the matrix proof is to begin by noting that $$ \pmatrix{f_1\f_0} + \pmatrix{f_3\f_2} + \cdots + \pmatrix{f_{2n-1}\f_{2n-2}} = \ \pmatrix{1\0} + F^2 \pmatrix{1\0} + \cdots + F^{2n-2}\pmatrix{1\0} $$ where $F = \pmatrix{1&1\1&0}$ — Ben Grossmann, Sep 10 '18 at 19:32
I have never seen that put that way before @Omnomnomnom, thank you and it is starting to make more sense on how to go about it. — Antas Santas, Sep 10 '18 at 19:34

Mike Earnest · Answer 1 · 2018-09-10T21:45:12.470

6

Letting $F=\begin{bmatrix}1&1\\1&0\end{bmatrix}$, so $F^2=F+I$, then the finite geometric series formula gives $$ I + F^2 + F^4 + \dots + F^{2n-2} = (F^{2n}-I)(F^2-I)^{-1} = (F^{2n}-I)F^{-1}=F^{2n-1}-F^{-1}. $$ Since $F^{-1}=\begin{bmatrix}0&1\\1&-1\end{bmatrix}$, conclude by examining the upper left entry of the equation $$ I + F^2 + F^4 + \dots + F^{2n-2} = F^{2n-1}-F^{-1}. $$ and recalling that $F^n=\begin{bmatrix}F_{n+1}&F_{n}\\F_{n}&F_{n-1}\end{bmatrix}$.

edited Sep 10 '18 at 21:45

answered Sep 10 '18 at 19:26

Mike Earnest

75,930

Also, $F^{-1}=F-I$. – lhf Oct 18 '18 at 00:37

Fibonacci summation proof using matrices?

1 Answers1