2

Question: Let $n,m,\in\mathbb{N^*}$ with $n>1$ and let $u_n$ denote the $n$-th term of the Fibonacci sequence, then $$u_{n+m}=u_{n-1}u_m+u_nu_{m+1}$$

I know these theorems:

  1. Two consecutive terms of the Fibonacci sequence are relatively prime.
  2. The Fibonacci sequence, we have $u_n\mid u_m\Longleftrightarrow n\mid m$
  3. In the Fibonacci sequence, we have $(u_m,u_n)=u_{(m,n)}$$$$$I tried:

    I) $n=2$ then $$u_{m+2}=u_1u_m+u_2u_{m+1}\Longrightarrow u_{m+2}=u_m+u_{m+1}\;\;\;OK$$

II) Hypothesis: $$u_{n+m}=u_{n-1}u_m+u_{n}u_{m+1}$$

III) Thesis: $$u_{n+m+1}=u_{n}u_m+u_{n+1}u_{m+1}$$

Show:$$u_{n+m+1}=\underbrace{u_{n+m}}_{\text{hypot.}}+u_{n+m-1}=\\=u_{n-1}u_m+u_{n}u_{m+1}+u_{n+m-1}$$ and now?

Andrés E. Caicedo
  • 81,687
benjamin_ee
  • 3,825
  • I would use induction, you know $u_{1+m} = u_0u_m + u_1u_{m+1}$ and $u_{2+m} = u_1u_m + u_2u_{m+1}$. Then use induction. – Daniel Fischer Aug 15 '13 at 15:18
  • @DanielFischer I have great difficulty making induction with two "letters" in trouble, could a little more detail? – benjamin_ee Aug 15 '13 at 15:22
  • Keep $m$ fixed. Then induct over $n$. "Let $m \in \mathbb{N}$. Then for all $n \in \mathbb{N}$, we have $u_{m+n} = u_{n-1}u_m + u_nu_{m+1}$." – Daniel Fischer Aug 15 '13 at 15:28
  • @marcelolpjunior, http://www.cut-the-knot.org/arithmetic/algebra/FibonacciGCD.shtml and http://www.math.hmc.edu/funfacts/ffiles/20004.5.shtml -- Observe that $1),2)$ are corollary of $3)$ – lab bhattacharjee Aug 15 '13 at 15:29
  • Then @DanielFischer, $$u_{m+2}=u_m+u_{m+1};;?$$It is consistent with the Fibonacci sequence. – benjamin_ee Aug 15 '13 at 15:37
  • @DanielFischer I'm sorry, but I honestly do not understand how to do it, how to solve the problem. – benjamin_ee Aug 15 '13 at 19:12

3 Answers3

5

One slick way -- which actually was supposed to be a motivating example for the linear algebra course I am currently teaching, but I already spent the first two lectures on other motivating examples without getting to it -- is to make use of matrix multiplication. Namely, if

$M = \left[ \begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array} \right]$

then for all positive integers $n$,

$M^n = \left[ \begin{array}{cc} F_{n+1} & F_n \\ F_n & F_{n-1} \end{array} \right]$.

This is very straightforward to prove by induction. And it is amazing how many of the basic identities about Fibonacci numbers drop right out of it. Yours certainly does, simply using $M^{m+n} = M^m M^n$.

Pete L. Clark
  • 100,402
  • Very nice! Also glad to hear that this is used "worldwide" as a motivational example in linear algebra courses. Another possible spot to use this is when you diagonalize $M$ and derive Binet's formula by calculating $M^n$ with the aid of the diagonalization. – Jyrki Lahtonen Aug 15 '13 at 15:55
  • @Jyrki: Yes, I will derive Binet's formula in that way as well. – Pete L. Clark Aug 17 '13 at 05:19
2

Let's prove this using induction:

First we'll prove that it holds for one case and the we'll just add 1 to m and 1 to n:

Step 1:Basis

$n = 2, m = 1$

$$U_3 = U_1*U_1 + U_2*U_2$$ $$U_3 = 1*1 + 1*1$$ $$U_3 = 2\text{, which is true}$$

$n = 2, m = 2$

$$U_4 = U_1*U_2 + U_2*U_3$$ $$U_4 = 1*1 + 1*2$$ $$U_4 = 3\text{, which is true}$$

$n = 3, m = 1$

$$U_4 = U_2*U_1 + U_3*U_2$$ $$U_4 = 1*1 + 2*1$$ $$U_4 = 3\text{, which is true}$$

Step 2:Inductive hypothesis

In the basis we've shown that it works for at least two consecutive numbers, so the statement holds for at least one pair of $U_{n+m}$ and $U_{n+m+1}$

Step 3: Inductive step

Now we'll show that it holds for every 3 consecutive number:

We have:

$$U_{n+m+2} = U_{n+m+1} + U_{n+m}$$ $$U_{n+m+2} = U_{n-1}*U_{m+1} + U_n*U_{m+2} + U_{n-1}*U_m + U_n*U_{m+1}$$ $$U_{n+m+2} = U_{n-1}(U_m + U_{m+1}) + U_n(U_{m+1} + U_{m+2})$$ $$U_{n+m+2} = U_{n-1}*U_{m+2} + U_n*U_{m+3}$$

Q.E.D.

We've proven that if it holds for $U_{k}$ and $U_{k+1}$, then it holds also for $U_{k+2}$. Then if it holds for $U_{k+1}$ and $U_{k+2}$ it also holds for $U_{k+3}$ and so on.

Edit:

I prove it, but I don't think that this answer would satisfy you.

First let me explain the method to you. First we prove that it work for the smallest values of $n$ and $m$, the will prove that it's true for $n$ and $m+1$. Knowing that this is true, we'll show that this is work also for $n$ and $m+2$. So the idea is:

Let $n$ be constant term, you can choose any number $\ge 2$ in the basis we prove that the statement holds for $m = 1$ and $m = 2$. We'll use this fact to prove that it also holds for $m = 3$. Then we'll use the fact that for $m = 2$ and $m = 3$ the statement is true, so we'll simularly prove it holds for $m = 4$ and so on to infinity.

Basis

n = 2,m = 1

$$U_3 = U_1*U_1 + U_2*U_2$$ $$U_3 = 1*1 + 1*1$$ $$U_3 = 2\text{, which is true}$$

$n = 2, m = 2$

$$U_4 = U_1*U_2 + U_2*U_3$$ $$U_4 = 1*1 + 1*2$$ $$U_4 = 3\text{, which is true}$$

Inductive hypothesis

Statement holds for at least one par of consecutive Fibonacci numbers, let those be: $U_{n+m}$ and $U_{n+(m+1)}$.

Inductive step

We need to prove:

$$U_{n+(m+2)} = U_{n-1}*U_{m+2} + U_n*U_{m+3}$$ $$U_{n+(m+2)} = U_{n-1}(U_m + U_{m+1}) + U_n*(U_{m+1} + U_{m+2})$$ $$U_{n+(m+2)} = U_{n-1}*U_m + U_{n-1}*U_{m+1} + U_n*U_{m+1} + U_n*U_{m+2}$$

If we rearrange the terms we'll and make a substitition we'll get:

$$U_{n+(m+2)} = U_{n+(m+1)} + U_{n+m}$$

Which is obviously true.

Hope this make it clearer.

Stefan4024
  • 36,357
0

I tried:

I) $n=2$ then $$u_{m+2}=u_1u_m+u_2u_{m+1}\Longrightarrow u_{m+2}=u_m+u_{m+1}\;\;\;OK$$

II) Hypothesis: $$u_{n+m}=u_{n-1}u_m+u_{n}u_{m+1}$$

III) Thesis: $$u_{n+m+1}=u_{n}u_m+u_{n+1}u_{m+1}$$

Show:$$u_{n+m+1}=\underbrace{u_{n+m}}_{\text{hypot.}}+u_{n+m-1}=\\=u_{n-1}u_m+u_{n}u_{m+1}+u_{n+m-1}$$ and now?

Community
  • 1
benjamin_ee
  • 3,825