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I'm lost on where to start on this proof:

Using the fact that $A^m A^n = A^{m+n}$ , prove the identity

$F_m F_n + F_{m−1} F_{n−1} = F_{m+n−1}$

I want to use induction starting with n = 1, but would I also have to make m = 1? I haven't done induction with 2 variables before.

or because of $A^m A^n = A^{m+n}$ should I setup the problem as a matrix (in that case what would the columns/rows be)?

I tried doing it mathematically however I think my algebra is wrong so I won't post it here. Am I correct to believe that $F_{m-1} = F_m*-1$ is not the same as $2^{n+1} = 2^n*2$?

Any help would be appreciated, thanks.

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  • $\begingroup$ I can guess what $F_n$ is from the title, but do we have to guess about $A$ as well? What is it and what relation with the Fibonacci number $F_i$ do you know? I cannot imagine they ask you this without saying anything about $A$. Maybe it is a matrix you have to find? $\endgroup$ – Marc van Leeuwen Apr 22 '13 at 4:28
  • $\begingroup$ Unfortunately that is all they say, nothing about A, except for this extra statement which I don't think is related to A: The Fibonacci numbers are given by the formula F1 = F2 = 1 and for n ≥ 3. We also define F0 = 0. Fn = Fn−1 + Fn−2. $\endgroup$ – Goose Apr 22 '13 at 4:29
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You can actually use induction here. We induct on $n$ proving that the relation holds for all $m$ at each step of the way. For $n=2$, $F_1 = F_2 =1$ and the identity $F_m+F_{m-1}=F_{m+1}$ is true for all $m$ by the definition of the Fibonacci sequence. We now have a strong induction hypothesis that the identity holds for values up until $n$, for all $m$. To show that it holds for $n+1$, for all $m$ we note that $$ F_m F_{n+1} + F_{m-1} F_n = F_m(F_{n-1} + F_n) + F_{m-1}(F_{n-2} + F_{n-1}) = $$ $$ (F_mF_n+F_{m-1}F_{n-1}) + (F_mF_{n-1} + F_{m-1}F_{n-2}) = F_{m+n-1} + F_{m+n-2} = F_{m+n}. $$ This completes the induction.

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  • $\begingroup$ I'm wondering why 𝐹𝑚𝐹𝑛+𝐹𝑚−1𝐹𝑛−1 is the same as 𝐹𝑚+𝑛−1 $\endgroup$ – Florian Nov 14 '19 at 17:24
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Fibonacci numbers have a matrix representation:

$$\left( \begin{smallmatrix} F_{n+1} & F_n \\ F_n & F_{n-1}\end{smallmatrix} \right) = \left( \begin{smallmatrix} 1 & 1 \\ 1 & 0 \end{smallmatrix}\right)^n$$

This is probably what you were meant to use for this problem.

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Hint: If $$v_{n}=\left[ \begin{array} {c} F_{n+1} \\ F_{n}\end{array}\right]$$Then:$$F_{m}F_{n}+F_{m-1}F_{n-1}=\langle v_{m-1},v_{n-1}\rangle$$where $\langle \cdot, \cdot \rangle$ is the standard inner product on $\mathbb{R}^2$. This along with Vadim123's hint should get the job done.

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