Prove for the Fibonacci sequence: $F(m+n) = F(m-1) cdot F(n) + F(m) cdot F(n+1)$












0












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The following formula shall be proved by induction:
$$F(m+n) = F(m-1) cdot F(n) + F(m) cdot F(n+1)$$
Where $F(i), i in mathbb{N}_0$ is the Fibonacci sequence defined as:
$F(0) = 0$, $F(1) = 1$ amd
$$F(n+1) = F(n) + F(n-1) text{ for } n geq 1.$$




How would you go about this task, especially considering that you have two variables which may be changed? Do you have to look at three cases ($m$ increases, $n$ increases, both $m$ and $n$ increase) in order to fully prove it or is there an easier way?










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  • $begingroup$
    You actually only need to prove the case $mmapsto m+1$ with $n$ kept constant, since the equation is, in fact, symmetric (replacing $n$ by $n-1$)
    $endgroup$
    – BAI
    Dec 31 '18 at 17:22










  • $begingroup$
    This identity have been here couple of times, for example Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$ and Product of consecutive Fibonacci numbers divisibility
    $endgroup$
    – Sil
    Dec 31 '18 at 18:02
















0












$begingroup$



The following formula shall be proved by induction:
$$F(m+n) = F(m-1) cdot F(n) + F(m) cdot F(n+1)$$
Where $F(i), i in mathbb{N}_0$ is the Fibonacci sequence defined as:
$F(0) = 0$, $F(1) = 1$ amd
$$F(n+1) = F(n) + F(n-1) text{ for } n geq 1.$$




How would you go about this task, especially considering that you have two variables which may be changed? Do you have to look at three cases ($m$ increases, $n$ increases, both $m$ and $n$ increase) in order to fully prove it or is there an easier way?










share|cite|improve this question











$endgroup$












  • $begingroup$
    You actually only need to prove the case $mmapsto m+1$ with $n$ kept constant, since the equation is, in fact, symmetric (replacing $n$ by $n-1$)
    $endgroup$
    – BAI
    Dec 31 '18 at 17:22










  • $begingroup$
    This identity have been here couple of times, for example Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$ and Product of consecutive Fibonacci numbers divisibility
    $endgroup$
    – Sil
    Dec 31 '18 at 18:02














0












0








0


1



$begingroup$



The following formula shall be proved by induction:
$$F(m+n) = F(m-1) cdot F(n) + F(m) cdot F(n+1)$$
Where $F(i), i in mathbb{N}_0$ is the Fibonacci sequence defined as:
$F(0) = 0$, $F(1) = 1$ amd
$$F(n+1) = F(n) + F(n-1) text{ for } n geq 1.$$




How would you go about this task, especially considering that you have two variables which may be changed? Do you have to look at three cases ($m$ increases, $n$ increases, both $m$ and $n$ increase) in order to fully prove it or is there an easier way?










share|cite|improve this question











$endgroup$





The following formula shall be proved by induction:
$$F(m+n) = F(m-1) cdot F(n) + F(m) cdot F(n+1)$$
Where $F(i), i in mathbb{N}_0$ is the Fibonacci sequence defined as:
$F(0) = 0$, $F(1) = 1$ amd
$$F(n+1) = F(n) + F(n-1) text{ for } n geq 1.$$




How would you go about this task, especially considering that you have two variables which may be changed? Do you have to look at three cases ($m$ increases, $n$ increases, both $m$ and $n$ increase) in order to fully prove it or is there an easier way?







induction fibonacci-numbers






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share|cite|improve this question













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edited Dec 31 '18 at 20:47









Robert Z

98.8k1068139




98.8k1068139










asked Dec 31 '18 at 17:18









StckXchnge-nub12StckXchnge-nub12

355




355












  • $begingroup$
    You actually only need to prove the case $mmapsto m+1$ with $n$ kept constant, since the equation is, in fact, symmetric (replacing $n$ by $n-1$)
    $endgroup$
    – BAI
    Dec 31 '18 at 17:22










  • $begingroup$
    This identity have been here couple of times, for example Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$ and Product of consecutive Fibonacci numbers divisibility
    $endgroup$
    – Sil
    Dec 31 '18 at 18:02


















  • $begingroup$
    You actually only need to prove the case $mmapsto m+1$ with $n$ kept constant, since the equation is, in fact, symmetric (replacing $n$ by $n-1$)
    $endgroup$
    – BAI
    Dec 31 '18 at 17:22










  • $begingroup$
    This identity have been here couple of times, for example Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$ and Product of consecutive Fibonacci numbers divisibility
    $endgroup$
    – Sil
    Dec 31 '18 at 18:02
















$begingroup$
You actually only need to prove the case $mmapsto m+1$ with $n$ kept constant, since the equation is, in fact, symmetric (replacing $n$ by $n-1$)
$endgroup$
– BAI
Dec 31 '18 at 17:22




$begingroup$
You actually only need to prove the case $mmapsto m+1$ with $n$ kept constant, since the equation is, in fact, symmetric (replacing $n$ by $n-1$)
$endgroup$
– BAI
Dec 31 '18 at 17:22












$begingroup$
This identity have been here couple of times, for example Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$ and Product of consecutive Fibonacci numbers divisibility
$endgroup$
– Sil
Dec 31 '18 at 18:02




$begingroup$
This identity have been here couple of times, for example Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$ and Product of consecutive Fibonacci numbers divisibility
$endgroup$
– Sil
Dec 31 '18 at 18:02










4 Answers
4






active

oldest

votes


















5












$begingroup$

Here is a proof in the same spirit as RobertZ's.



First, let's relate the Fibonacci numbers to the following problem:




Suppose that you want to go up some flight of stairs and at every step you can take either one or two stairs: in how many ways can you get up the stairs?




Well, if we say that there are $n$ stairs, then it turns out there are $F_{n+1}$ ways to do it.



A very easy inductive proof shows why:



Base cases: If there is $1$ stairs, you can do it in only $1$ way, and indeed $F_{1+1}=F_2=1$. If there are $2$ stairs, then there are two ways: either take two steps of $1$, or take one step of $2$. And indeed, $F_{2+1}=F_3=2$



Inductive step: Say you have $n >2$ stairs. For your first step you can either go one up or two stairs up. By inductive hypothesis, there are $F_{n}$ ways to finish climbing the $n-1$ stairs after having taken a step of $1$ stairs, and there are $F_{n-1}$ ways to finish climbing the $n-2$ stairs after having taken a step of $2$ stairs. So, there are $F_{n}+F_{n-1}=F_{n+1}$ ways to climb $n$ stairs.



OK, so now that we have made a connection between the Fibonacci numbers and the number of ways to climb stairs in this way, we can prove your desired result very quickly:



Let's climb $m+n-1$ stairs. We now know we can do this in $F_{m+n}$ ways. But note that there are two different possibilities for us climbing those $m+n-1$ stairs:




  1. The first way is that as we climb the stairs, we will at some point have climbed exactly $n$ stairs. If this happens, then there are $m-1$ stairs left to climb, which can be done in $F_m$ ways. And since then there are $F_{n+1}$ ways to climb the first $n$ stairs, that means that there are $F_m cdot F_{n+1}$ ways to climb all the $m+n-1$ stairs this way.


  2. The second way is that at some point we will have climbed exactly $n-1$ stairs, which can be done in $F_{n}$ ways, after which we take a single step of $2$ stairs, and then finish climbing the remaining $m-2$ stairs. This can therefore be done in $F_{m-1} cdot F_{n}$ ways



The total number of ways to climb the $m+n-1$ stairs, then, is the sum of these two ways, and so it must be true that:



$$F_{m+n} = F_{m-1} cdot F_{n} + F_{m} cdot F_{n+1}$$



So note that after I established the connection between climbing stairs and the Fobonacci numbers, the proof was straightforward (and did not use induction). But to establish the connection, I used induction.






share|cite|improve this answer









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  • $begingroup$
    Nice variant (+1). See also math.stackexchange.com/a/11527/299698
    $endgroup$
    – Robert Z
    Dec 31 '18 at 20:53










  • $begingroup$
    @RobertZ :) And here I thought I was original ...
    $endgroup$
    – Bram28
    Jan 1 at 13:25










  • $begingroup$
    Have you any suggestion to improve my answer? Unfortunately it did not have much success :-(
    $endgroup$
    – Robert Z
    Jan 1 at 13:34










  • $begingroup$
    @RobertZ Maybe you could try and show a picture of the domino covering ... both to establish the connection between F numbers and dominoes and then applied to this particular problem
    $endgroup$
    – Bram28
    Jan 1 at 13:54



















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You can get away with inducting on just $n$:



$$ begin{align} F(m + n) & = F(m + (n - 1)) + F(m + (n -2))\
& = F(m-1)F(n-1) + F(m)F(n) + F(m-1)F(n-2) + F(m)F(n-1) \
& = F(m-1)[F(n-1) + F(n-2)] + F(m)[F(n)+F(n-1)] \
& = F(m-1)F(n) + F(m)F(n+1)end{align}$$






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  • $begingroup$
    Can you explain in greater detail the steps you took? You already lost me on the very first one. It seems you have somehow replaced $F(n)$ with the recursive definition of the fibonacci sequence (the last formula I listed), but I didn't even know you can split apart the term partially like that. Isn't the $F(m+n)$ to be seen as static, i.e. if $m=2$ and $n = 3$, you have to solve the entire sum $F(5)$, and can't just partially solve the "3 part" of the fibonacci? Hard to explain my issue, I hope you get what I mean.
    $endgroup$
    – StckXchnge-nub12
    Dec 31 '18 at 17:42










  • $begingroup$
    Hi, using your example: if $m = 2, n = 3$, $n + m = 5$ then $F(5) = F(4) + F(3)$ by the definition of the Fibonacci sequence. Note then that $n + m -1 = 4$ and $n + m - 2 = 3$ which is my first step.
    $endgroup$
    – ODF
    Dec 31 '18 at 17:46










  • $begingroup$
    In the second step I induct on $n$, in the third I just gather like terms and in the final I use the fact that $F(n) = F(n-1) + F(n-2)$ and that $F(n+1) = F(n) + F(n-1)$.
    $endgroup$
    – ODF
    Dec 31 '18 at 17:47










  • $begingroup$
    Alright, I've finally understood your solution; thank you! Just for the record, in order to turn it into an actual proof by induction, I believe you have to proof $F(m+n+1) = F(m-1)cdot F(n+1) + F(m) cdot F(n+2)$ using the very same steps you took.
    $endgroup$
    – StckXchnge-nub12
    Dec 31 '18 at 18:28












  • $begingroup$
    I must revise myself. I think there's an issue here. In the second line, you would apply the induction precondition. If we are inducting on $n$, that means we can only apply it on an $n$, but not e.g. on an $n-1$. But in your calculations you do apply it on just any $n$, so you basically assume that the precondition already applies for every single $n$ (i.e. $n, n+1, n-1, ldots$) when in reality we only defined it to apply to a fix $n$ for the sake of the proof.
    $endgroup$
    – StckXchnge-nub12
    Dec 31 '18 at 19:59





















1












$begingroup$

I won't mention every use of induction. Define $$M:=left(begin{array}{cc}
0 & 1\
1 & 1
end{array}right),,V_{n}:=left(begin{array}{c}
F_{n}\
F_{n+1}
end{array}right)=M^{n}left(begin{array}{c}
0\
1
end{array}right)$$
so $M^{m}=left(begin{array}{cc}
F_{m-1} & F_{m}\
F_{m} & F_{m+1}
end{array}right)$
and $$left(begin{array}{c}
F_{m+n}\
F_{m+n+1}
end{array}right)=M^{m}left(begin{array}{c}
F_{n}\
F_{n+1}
end{array}right)=left(begin{array}{c}
F_{m-1}F_{n}+F_{m}F_{n+1}\
F_{m}F_{n}+F_{m+1}F_{n+1}
end{array}right).$$






share|cite|improve this answer









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    1












    $begingroup$

    A combinatorial proof (no induction).



    Using the representation of the Fibonacci numbers as the numbers of domino coverings the given identity can be shown in a quick and elegant way.



    It is known that the number of domino coverings of a $2times N$ grid is $F_{N+1}$. Consider the number of domino coverings of a $2times (m+n-1)$ grid. Any covering can be of two types.



    1) The covering has a pair of horizontal dominoes at position $m-1$ and $m$.
    enter image description here



    Therefore on the left side we have a covering of a $2times (m-2)$ grid
    and on the right side we have a covering of a $2times (n-1)$ grid. Therefore the number of such coverings is
    $F_{m-1}cdot F_{n}$.



    2) The covering can be split into two coverings, one of a $2times (m-1)$ grid and another of a $2times n$ grid.
    enter image description here



    Therefore the number of such coverings is
    $F_{m}cdot F_{n+1}$.



    Finally we may conclude that the number of domino coverings of a $2times (m+n-1)$ grid is
    $$F_{m+n}=F_{m-1}cdot F_{n}+F_{m}cdot F_{n+1}.$$






    share|cite|improve this answer











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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Here is a proof in the same spirit as RobertZ's.



      First, let's relate the Fibonacci numbers to the following problem:




      Suppose that you want to go up some flight of stairs and at every step you can take either one or two stairs: in how many ways can you get up the stairs?




      Well, if we say that there are $n$ stairs, then it turns out there are $F_{n+1}$ ways to do it.



      A very easy inductive proof shows why:



      Base cases: If there is $1$ stairs, you can do it in only $1$ way, and indeed $F_{1+1}=F_2=1$. If there are $2$ stairs, then there are two ways: either take two steps of $1$, or take one step of $2$. And indeed, $F_{2+1}=F_3=2$



      Inductive step: Say you have $n >2$ stairs. For your first step you can either go one up or two stairs up. By inductive hypothesis, there are $F_{n}$ ways to finish climbing the $n-1$ stairs after having taken a step of $1$ stairs, and there are $F_{n-1}$ ways to finish climbing the $n-2$ stairs after having taken a step of $2$ stairs. So, there are $F_{n}+F_{n-1}=F_{n+1}$ ways to climb $n$ stairs.



      OK, so now that we have made a connection between the Fibonacci numbers and the number of ways to climb stairs in this way, we can prove your desired result very quickly:



      Let's climb $m+n-1$ stairs. We now know we can do this in $F_{m+n}$ ways. But note that there are two different possibilities for us climbing those $m+n-1$ stairs:




      1. The first way is that as we climb the stairs, we will at some point have climbed exactly $n$ stairs. If this happens, then there are $m-1$ stairs left to climb, which can be done in $F_m$ ways. And since then there are $F_{n+1}$ ways to climb the first $n$ stairs, that means that there are $F_m cdot F_{n+1}$ ways to climb all the $m+n-1$ stairs this way.


      2. The second way is that at some point we will have climbed exactly $n-1$ stairs, which can be done in $F_{n}$ ways, after which we take a single step of $2$ stairs, and then finish climbing the remaining $m-2$ stairs. This can therefore be done in $F_{m-1} cdot F_{n}$ ways



      The total number of ways to climb the $m+n-1$ stairs, then, is the sum of these two ways, and so it must be true that:



      $$F_{m+n} = F_{m-1} cdot F_{n} + F_{m} cdot F_{n+1}$$



      So note that after I established the connection between climbing stairs and the Fobonacci numbers, the proof was straightforward (and did not use induction). But to establish the connection, I used induction.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Nice variant (+1). See also math.stackexchange.com/a/11527/299698
        $endgroup$
        – Robert Z
        Dec 31 '18 at 20:53










      • $begingroup$
        @RobertZ :) And here I thought I was original ...
        $endgroup$
        – Bram28
        Jan 1 at 13:25










      • $begingroup$
        Have you any suggestion to improve my answer? Unfortunately it did not have much success :-(
        $endgroup$
        – Robert Z
        Jan 1 at 13:34










      • $begingroup$
        @RobertZ Maybe you could try and show a picture of the domino covering ... both to establish the connection between F numbers and dominoes and then applied to this particular problem
        $endgroup$
        – Bram28
        Jan 1 at 13:54
















      5












      $begingroup$

      Here is a proof in the same spirit as RobertZ's.



      First, let's relate the Fibonacci numbers to the following problem:




      Suppose that you want to go up some flight of stairs and at every step you can take either one or two stairs: in how many ways can you get up the stairs?




      Well, if we say that there are $n$ stairs, then it turns out there are $F_{n+1}$ ways to do it.



      A very easy inductive proof shows why:



      Base cases: If there is $1$ stairs, you can do it in only $1$ way, and indeed $F_{1+1}=F_2=1$. If there are $2$ stairs, then there are two ways: either take two steps of $1$, or take one step of $2$. And indeed, $F_{2+1}=F_3=2$



      Inductive step: Say you have $n >2$ stairs. For your first step you can either go one up or two stairs up. By inductive hypothesis, there are $F_{n}$ ways to finish climbing the $n-1$ stairs after having taken a step of $1$ stairs, and there are $F_{n-1}$ ways to finish climbing the $n-2$ stairs after having taken a step of $2$ stairs. So, there are $F_{n}+F_{n-1}=F_{n+1}$ ways to climb $n$ stairs.



      OK, so now that we have made a connection between the Fibonacci numbers and the number of ways to climb stairs in this way, we can prove your desired result very quickly:



      Let's climb $m+n-1$ stairs. We now know we can do this in $F_{m+n}$ ways. But note that there are two different possibilities for us climbing those $m+n-1$ stairs:




      1. The first way is that as we climb the stairs, we will at some point have climbed exactly $n$ stairs. If this happens, then there are $m-1$ stairs left to climb, which can be done in $F_m$ ways. And since then there are $F_{n+1}$ ways to climb the first $n$ stairs, that means that there are $F_m cdot F_{n+1}$ ways to climb all the $m+n-1$ stairs this way.


      2. The second way is that at some point we will have climbed exactly $n-1$ stairs, which can be done in $F_{n}$ ways, after which we take a single step of $2$ stairs, and then finish climbing the remaining $m-2$ stairs. This can therefore be done in $F_{m-1} cdot F_{n}$ ways



      The total number of ways to climb the $m+n-1$ stairs, then, is the sum of these two ways, and so it must be true that:



      $$F_{m+n} = F_{m-1} cdot F_{n} + F_{m} cdot F_{n+1}$$



      So note that after I established the connection between climbing stairs and the Fobonacci numbers, the proof was straightforward (and did not use induction). But to establish the connection, I used induction.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Nice variant (+1). See also math.stackexchange.com/a/11527/299698
        $endgroup$
        – Robert Z
        Dec 31 '18 at 20:53










      • $begingroup$
        @RobertZ :) And here I thought I was original ...
        $endgroup$
        – Bram28
        Jan 1 at 13:25










      • $begingroup$
        Have you any suggestion to improve my answer? Unfortunately it did not have much success :-(
        $endgroup$
        – Robert Z
        Jan 1 at 13:34










      • $begingroup$
        @RobertZ Maybe you could try and show a picture of the domino covering ... both to establish the connection between F numbers and dominoes and then applied to this particular problem
        $endgroup$
        – Bram28
        Jan 1 at 13:54














      5












      5








      5





      $begingroup$

      Here is a proof in the same spirit as RobertZ's.



      First, let's relate the Fibonacci numbers to the following problem:




      Suppose that you want to go up some flight of stairs and at every step you can take either one or two stairs: in how many ways can you get up the stairs?




      Well, if we say that there are $n$ stairs, then it turns out there are $F_{n+1}$ ways to do it.



      A very easy inductive proof shows why:



      Base cases: If there is $1$ stairs, you can do it in only $1$ way, and indeed $F_{1+1}=F_2=1$. If there are $2$ stairs, then there are two ways: either take two steps of $1$, or take one step of $2$. And indeed, $F_{2+1}=F_3=2$



      Inductive step: Say you have $n >2$ stairs. For your first step you can either go one up or two stairs up. By inductive hypothesis, there are $F_{n}$ ways to finish climbing the $n-1$ stairs after having taken a step of $1$ stairs, and there are $F_{n-1}$ ways to finish climbing the $n-2$ stairs after having taken a step of $2$ stairs. So, there are $F_{n}+F_{n-1}=F_{n+1}$ ways to climb $n$ stairs.



      OK, so now that we have made a connection between the Fibonacci numbers and the number of ways to climb stairs in this way, we can prove your desired result very quickly:



      Let's climb $m+n-1$ stairs. We now know we can do this in $F_{m+n}$ ways. But note that there are two different possibilities for us climbing those $m+n-1$ stairs:




      1. The first way is that as we climb the stairs, we will at some point have climbed exactly $n$ stairs. If this happens, then there are $m-1$ stairs left to climb, which can be done in $F_m$ ways. And since then there are $F_{n+1}$ ways to climb the first $n$ stairs, that means that there are $F_m cdot F_{n+1}$ ways to climb all the $m+n-1$ stairs this way.


      2. The second way is that at some point we will have climbed exactly $n-1$ stairs, which can be done in $F_{n}$ ways, after which we take a single step of $2$ stairs, and then finish climbing the remaining $m-2$ stairs. This can therefore be done in $F_{m-1} cdot F_{n}$ ways



      The total number of ways to climb the $m+n-1$ stairs, then, is the sum of these two ways, and so it must be true that:



      $$F_{m+n} = F_{m-1} cdot F_{n} + F_{m} cdot F_{n+1}$$



      So note that after I established the connection between climbing stairs and the Fobonacci numbers, the proof was straightforward (and did not use induction). But to establish the connection, I used induction.






      share|cite|improve this answer









      $endgroup$



      Here is a proof in the same spirit as RobertZ's.



      First, let's relate the Fibonacci numbers to the following problem:




      Suppose that you want to go up some flight of stairs and at every step you can take either one or two stairs: in how many ways can you get up the stairs?




      Well, if we say that there are $n$ stairs, then it turns out there are $F_{n+1}$ ways to do it.



      A very easy inductive proof shows why:



      Base cases: If there is $1$ stairs, you can do it in only $1$ way, and indeed $F_{1+1}=F_2=1$. If there are $2$ stairs, then there are two ways: either take two steps of $1$, or take one step of $2$. And indeed, $F_{2+1}=F_3=2$



      Inductive step: Say you have $n >2$ stairs. For your first step you can either go one up or two stairs up. By inductive hypothesis, there are $F_{n}$ ways to finish climbing the $n-1$ stairs after having taken a step of $1$ stairs, and there are $F_{n-1}$ ways to finish climbing the $n-2$ stairs after having taken a step of $2$ stairs. So, there are $F_{n}+F_{n-1}=F_{n+1}$ ways to climb $n$ stairs.



      OK, so now that we have made a connection between the Fibonacci numbers and the number of ways to climb stairs in this way, we can prove your desired result very quickly:



      Let's climb $m+n-1$ stairs. We now know we can do this in $F_{m+n}$ ways. But note that there are two different possibilities for us climbing those $m+n-1$ stairs:




      1. The first way is that as we climb the stairs, we will at some point have climbed exactly $n$ stairs. If this happens, then there are $m-1$ stairs left to climb, which can be done in $F_m$ ways. And since then there are $F_{n+1}$ ways to climb the first $n$ stairs, that means that there are $F_m cdot F_{n+1}$ ways to climb all the $m+n-1$ stairs this way.


      2. The second way is that at some point we will have climbed exactly $n-1$ stairs, which can be done in $F_{n}$ ways, after which we take a single step of $2$ stairs, and then finish climbing the remaining $m-2$ stairs. This can therefore be done in $F_{m-1} cdot F_{n}$ ways



      The total number of ways to climb the $m+n-1$ stairs, then, is the sum of these two ways, and so it must be true that:



      $$F_{m+n} = F_{m-1} cdot F_{n} + F_{m} cdot F_{n+1}$$



      So note that after I established the connection between climbing stairs and the Fobonacci numbers, the proof was straightforward (and did not use induction). But to establish the connection, I used induction.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 31 '18 at 17:44









      Bram28Bram28

      63.2k44793




      63.2k44793












      • $begingroup$
        Nice variant (+1). See also math.stackexchange.com/a/11527/299698
        $endgroup$
        – Robert Z
        Dec 31 '18 at 20:53










      • $begingroup$
        @RobertZ :) And here I thought I was original ...
        $endgroup$
        – Bram28
        Jan 1 at 13:25










      • $begingroup$
        Have you any suggestion to improve my answer? Unfortunately it did not have much success :-(
        $endgroup$
        – Robert Z
        Jan 1 at 13:34










      • $begingroup$
        @RobertZ Maybe you could try and show a picture of the domino covering ... both to establish the connection between F numbers and dominoes and then applied to this particular problem
        $endgroup$
        – Bram28
        Jan 1 at 13:54


















      • $begingroup$
        Nice variant (+1). See also math.stackexchange.com/a/11527/299698
        $endgroup$
        – Robert Z
        Dec 31 '18 at 20:53










      • $begingroup$
        @RobertZ :) And here I thought I was original ...
        $endgroup$
        – Bram28
        Jan 1 at 13:25










      • $begingroup$
        Have you any suggestion to improve my answer? Unfortunately it did not have much success :-(
        $endgroup$
        – Robert Z
        Jan 1 at 13:34










      • $begingroup$
        @RobertZ Maybe you could try and show a picture of the domino covering ... both to establish the connection between F numbers and dominoes and then applied to this particular problem
        $endgroup$
        – Bram28
        Jan 1 at 13:54
















      $begingroup$
      Nice variant (+1). See also math.stackexchange.com/a/11527/299698
      $endgroup$
      – Robert Z
      Dec 31 '18 at 20:53




      $begingroup$
      Nice variant (+1). See also math.stackexchange.com/a/11527/299698
      $endgroup$
      – Robert Z
      Dec 31 '18 at 20:53












      $begingroup$
      @RobertZ :) And here I thought I was original ...
      $endgroup$
      – Bram28
      Jan 1 at 13:25




      $begingroup$
      @RobertZ :) And here I thought I was original ...
      $endgroup$
      – Bram28
      Jan 1 at 13:25












      $begingroup$
      Have you any suggestion to improve my answer? Unfortunately it did not have much success :-(
      $endgroup$
      – Robert Z
      Jan 1 at 13:34




      $begingroup$
      Have you any suggestion to improve my answer? Unfortunately it did not have much success :-(
      $endgroup$
      – Robert Z
      Jan 1 at 13:34












      $begingroup$
      @RobertZ Maybe you could try and show a picture of the domino covering ... both to establish the connection between F numbers and dominoes and then applied to this particular problem
      $endgroup$
      – Bram28
      Jan 1 at 13:54




      $begingroup$
      @RobertZ Maybe you could try and show a picture of the domino covering ... both to establish the connection between F numbers and dominoes and then applied to this particular problem
      $endgroup$
      – Bram28
      Jan 1 at 13:54











      3












      $begingroup$

      You can get away with inducting on just $n$:



      $$ begin{align} F(m + n) & = F(m + (n - 1)) + F(m + (n -2))\
      & = F(m-1)F(n-1) + F(m)F(n) + F(m-1)F(n-2) + F(m)F(n-1) \
      & = F(m-1)[F(n-1) + F(n-2)] + F(m)[F(n)+F(n-1)] \
      & = F(m-1)F(n) + F(m)F(n+1)end{align}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Can you explain in greater detail the steps you took? You already lost me on the very first one. It seems you have somehow replaced $F(n)$ with the recursive definition of the fibonacci sequence (the last formula I listed), but I didn't even know you can split apart the term partially like that. Isn't the $F(m+n)$ to be seen as static, i.e. if $m=2$ and $n = 3$, you have to solve the entire sum $F(5)$, and can't just partially solve the "3 part" of the fibonacci? Hard to explain my issue, I hope you get what I mean.
        $endgroup$
        – StckXchnge-nub12
        Dec 31 '18 at 17:42










      • $begingroup$
        Hi, using your example: if $m = 2, n = 3$, $n + m = 5$ then $F(5) = F(4) + F(3)$ by the definition of the Fibonacci sequence. Note then that $n + m -1 = 4$ and $n + m - 2 = 3$ which is my first step.
        $endgroup$
        – ODF
        Dec 31 '18 at 17:46










      • $begingroup$
        In the second step I induct on $n$, in the third I just gather like terms and in the final I use the fact that $F(n) = F(n-1) + F(n-2)$ and that $F(n+1) = F(n) + F(n-1)$.
        $endgroup$
        – ODF
        Dec 31 '18 at 17:47










      • $begingroup$
        Alright, I've finally understood your solution; thank you! Just for the record, in order to turn it into an actual proof by induction, I believe you have to proof $F(m+n+1) = F(m-1)cdot F(n+1) + F(m) cdot F(n+2)$ using the very same steps you took.
        $endgroup$
        – StckXchnge-nub12
        Dec 31 '18 at 18:28












      • $begingroup$
        I must revise myself. I think there's an issue here. In the second line, you would apply the induction precondition. If we are inducting on $n$, that means we can only apply it on an $n$, but not e.g. on an $n-1$. But in your calculations you do apply it on just any $n$, so you basically assume that the precondition already applies for every single $n$ (i.e. $n, n+1, n-1, ldots$) when in reality we only defined it to apply to a fix $n$ for the sake of the proof.
        $endgroup$
        – StckXchnge-nub12
        Dec 31 '18 at 19:59


















      3












      $begingroup$

      You can get away with inducting on just $n$:



      $$ begin{align} F(m + n) & = F(m + (n - 1)) + F(m + (n -2))\
      & = F(m-1)F(n-1) + F(m)F(n) + F(m-1)F(n-2) + F(m)F(n-1) \
      & = F(m-1)[F(n-1) + F(n-2)] + F(m)[F(n)+F(n-1)] \
      & = F(m-1)F(n) + F(m)F(n+1)end{align}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Can you explain in greater detail the steps you took? You already lost me on the very first one. It seems you have somehow replaced $F(n)$ with the recursive definition of the fibonacci sequence (the last formula I listed), but I didn't even know you can split apart the term partially like that. Isn't the $F(m+n)$ to be seen as static, i.e. if $m=2$ and $n = 3$, you have to solve the entire sum $F(5)$, and can't just partially solve the "3 part" of the fibonacci? Hard to explain my issue, I hope you get what I mean.
        $endgroup$
        – StckXchnge-nub12
        Dec 31 '18 at 17:42










      • $begingroup$
        Hi, using your example: if $m = 2, n = 3$, $n + m = 5$ then $F(5) = F(4) + F(3)$ by the definition of the Fibonacci sequence. Note then that $n + m -1 = 4$ and $n + m - 2 = 3$ which is my first step.
        $endgroup$
        – ODF
        Dec 31 '18 at 17:46










      • $begingroup$
        In the second step I induct on $n$, in the third I just gather like terms and in the final I use the fact that $F(n) = F(n-1) + F(n-2)$ and that $F(n+1) = F(n) + F(n-1)$.
        $endgroup$
        – ODF
        Dec 31 '18 at 17:47










      • $begingroup$
        Alright, I've finally understood your solution; thank you! Just for the record, in order to turn it into an actual proof by induction, I believe you have to proof $F(m+n+1) = F(m-1)cdot F(n+1) + F(m) cdot F(n+2)$ using the very same steps you took.
        $endgroup$
        – StckXchnge-nub12
        Dec 31 '18 at 18:28












      • $begingroup$
        I must revise myself. I think there's an issue here. In the second line, you would apply the induction precondition. If we are inducting on $n$, that means we can only apply it on an $n$, but not e.g. on an $n-1$. But in your calculations you do apply it on just any $n$, so you basically assume that the precondition already applies for every single $n$ (i.e. $n, n+1, n-1, ldots$) when in reality we only defined it to apply to a fix $n$ for the sake of the proof.
        $endgroup$
        – StckXchnge-nub12
        Dec 31 '18 at 19:59
















      3












      3








      3





      $begingroup$

      You can get away with inducting on just $n$:



      $$ begin{align} F(m + n) & = F(m + (n - 1)) + F(m + (n -2))\
      & = F(m-1)F(n-1) + F(m)F(n) + F(m-1)F(n-2) + F(m)F(n-1) \
      & = F(m-1)[F(n-1) + F(n-2)] + F(m)[F(n)+F(n-1)] \
      & = F(m-1)F(n) + F(m)F(n+1)end{align}$$






      share|cite|improve this answer









      $endgroup$



      You can get away with inducting on just $n$:



      $$ begin{align} F(m + n) & = F(m + (n - 1)) + F(m + (n -2))\
      & = F(m-1)F(n-1) + F(m)F(n) + F(m-1)F(n-2) + F(m)F(n-1) \
      & = F(m-1)[F(n-1) + F(n-2)] + F(m)[F(n)+F(n-1)] \
      & = F(m-1)F(n) + F(m)F(n+1)end{align}$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 31 '18 at 17:24









      ODFODF

      1,486510




      1,486510












      • $begingroup$
        Can you explain in greater detail the steps you took? You already lost me on the very first one. It seems you have somehow replaced $F(n)$ with the recursive definition of the fibonacci sequence (the last formula I listed), but I didn't even know you can split apart the term partially like that. Isn't the $F(m+n)$ to be seen as static, i.e. if $m=2$ and $n = 3$, you have to solve the entire sum $F(5)$, and can't just partially solve the "3 part" of the fibonacci? Hard to explain my issue, I hope you get what I mean.
        $endgroup$
        – StckXchnge-nub12
        Dec 31 '18 at 17:42










      • $begingroup$
        Hi, using your example: if $m = 2, n = 3$, $n + m = 5$ then $F(5) = F(4) + F(3)$ by the definition of the Fibonacci sequence. Note then that $n + m -1 = 4$ and $n + m - 2 = 3$ which is my first step.
        $endgroup$
        – ODF
        Dec 31 '18 at 17:46










      • $begingroup$
        In the second step I induct on $n$, in the third I just gather like terms and in the final I use the fact that $F(n) = F(n-1) + F(n-2)$ and that $F(n+1) = F(n) + F(n-1)$.
        $endgroup$
        – ODF
        Dec 31 '18 at 17:47










      • $begingroup$
        Alright, I've finally understood your solution; thank you! Just for the record, in order to turn it into an actual proof by induction, I believe you have to proof $F(m+n+1) = F(m-1)cdot F(n+1) + F(m) cdot F(n+2)$ using the very same steps you took.
        $endgroup$
        – StckXchnge-nub12
        Dec 31 '18 at 18:28












      • $begingroup$
        I must revise myself. I think there's an issue here. In the second line, you would apply the induction precondition. If we are inducting on $n$, that means we can only apply it on an $n$, but not e.g. on an $n-1$. But in your calculations you do apply it on just any $n$, so you basically assume that the precondition already applies for every single $n$ (i.e. $n, n+1, n-1, ldots$) when in reality we only defined it to apply to a fix $n$ for the sake of the proof.
        $endgroup$
        – StckXchnge-nub12
        Dec 31 '18 at 19:59




















      • $begingroup$
        Can you explain in greater detail the steps you took? You already lost me on the very first one. It seems you have somehow replaced $F(n)$ with the recursive definition of the fibonacci sequence (the last formula I listed), but I didn't even know you can split apart the term partially like that. Isn't the $F(m+n)$ to be seen as static, i.e. if $m=2$ and $n = 3$, you have to solve the entire sum $F(5)$, and can't just partially solve the "3 part" of the fibonacci? Hard to explain my issue, I hope you get what I mean.
        $endgroup$
        – StckXchnge-nub12
        Dec 31 '18 at 17:42










      • $begingroup$
        Hi, using your example: if $m = 2, n = 3$, $n + m = 5$ then $F(5) = F(4) + F(3)$ by the definition of the Fibonacci sequence. Note then that $n + m -1 = 4$ and $n + m - 2 = 3$ which is my first step.
        $endgroup$
        – ODF
        Dec 31 '18 at 17:46










      • $begingroup$
        In the second step I induct on $n$, in the third I just gather like terms and in the final I use the fact that $F(n) = F(n-1) + F(n-2)$ and that $F(n+1) = F(n) + F(n-1)$.
        $endgroup$
        – ODF
        Dec 31 '18 at 17:47










      • $begingroup$
        Alright, I've finally understood your solution; thank you! Just for the record, in order to turn it into an actual proof by induction, I believe you have to proof $F(m+n+1) = F(m-1)cdot F(n+1) + F(m) cdot F(n+2)$ using the very same steps you took.
        $endgroup$
        – StckXchnge-nub12
        Dec 31 '18 at 18:28












      • $begingroup$
        I must revise myself. I think there's an issue here. In the second line, you would apply the induction precondition. If we are inducting on $n$, that means we can only apply it on an $n$, but not e.g. on an $n-1$. But in your calculations you do apply it on just any $n$, so you basically assume that the precondition already applies for every single $n$ (i.e. $n, n+1, n-1, ldots$) when in reality we only defined it to apply to a fix $n$ for the sake of the proof.
        $endgroup$
        – StckXchnge-nub12
        Dec 31 '18 at 19:59


















      $begingroup$
      Can you explain in greater detail the steps you took? You already lost me on the very first one. It seems you have somehow replaced $F(n)$ with the recursive definition of the fibonacci sequence (the last formula I listed), but I didn't even know you can split apart the term partially like that. Isn't the $F(m+n)$ to be seen as static, i.e. if $m=2$ and $n = 3$, you have to solve the entire sum $F(5)$, and can't just partially solve the "3 part" of the fibonacci? Hard to explain my issue, I hope you get what I mean.
      $endgroup$
      – StckXchnge-nub12
      Dec 31 '18 at 17:42




      $begingroup$
      Can you explain in greater detail the steps you took? You already lost me on the very first one. It seems you have somehow replaced $F(n)$ with the recursive definition of the fibonacci sequence (the last formula I listed), but I didn't even know you can split apart the term partially like that. Isn't the $F(m+n)$ to be seen as static, i.e. if $m=2$ and $n = 3$, you have to solve the entire sum $F(5)$, and can't just partially solve the "3 part" of the fibonacci? Hard to explain my issue, I hope you get what I mean.
      $endgroup$
      – StckXchnge-nub12
      Dec 31 '18 at 17:42












      $begingroup$
      Hi, using your example: if $m = 2, n = 3$, $n + m = 5$ then $F(5) = F(4) + F(3)$ by the definition of the Fibonacci sequence. Note then that $n + m -1 = 4$ and $n + m - 2 = 3$ which is my first step.
      $endgroup$
      – ODF
      Dec 31 '18 at 17:46




      $begingroup$
      Hi, using your example: if $m = 2, n = 3$, $n + m = 5$ then $F(5) = F(4) + F(3)$ by the definition of the Fibonacci sequence. Note then that $n + m -1 = 4$ and $n + m - 2 = 3$ which is my first step.
      $endgroup$
      – ODF
      Dec 31 '18 at 17:46












      $begingroup$
      In the second step I induct on $n$, in the third I just gather like terms and in the final I use the fact that $F(n) = F(n-1) + F(n-2)$ and that $F(n+1) = F(n) + F(n-1)$.
      $endgroup$
      – ODF
      Dec 31 '18 at 17:47




      $begingroup$
      In the second step I induct on $n$, in the third I just gather like terms and in the final I use the fact that $F(n) = F(n-1) + F(n-2)$ and that $F(n+1) = F(n) + F(n-1)$.
      $endgroup$
      – ODF
      Dec 31 '18 at 17:47












      $begingroup$
      Alright, I've finally understood your solution; thank you! Just for the record, in order to turn it into an actual proof by induction, I believe you have to proof $F(m+n+1) = F(m-1)cdot F(n+1) + F(m) cdot F(n+2)$ using the very same steps you took.
      $endgroup$
      – StckXchnge-nub12
      Dec 31 '18 at 18:28






      $begingroup$
      Alright, I've finally understood your solution; thank you! Just for the record, in order to turn it into an actual proof by induction, I believe you have to proof $F(m+n+1) = F(m-1)cdot F(n+1) + F(m) cdot F(n+2)$ using the very same steps you took.
      $endgroup$
      – StckXchnge-nub12
      Dec 31 '18 at 18:28














      $begingroup$
      I must revise myself. I think there's an issue here. In the second line, you would apply the induction precondition. If we are inducting on $n$, that means we can only apply it on an $n$, but not e.g. on an $n-1$. But in your calculations you do apply it on just any $n$, so you basically assume that the precondition already applies for every single $n$ (i.e. $n, n+1, n-1, ldots$) when in reality we only defined it to apply to a fix $n$ for the sake of the proof.
      $endgroup$
      – StckXchnge-nub12
      Dec 31 '18 at 19:59






      $begingroup$
      I must revise myself. I think there's an issue here. In the second line, you would apply the induction precondition. If we are inducting on $n$, that means we can only apply it on an $n$, but not e.g. on an $n-1$. But in your calculations you do apply it on just any $n$, so you basically assume that the precondition already applies for every single $n$ (i.e. $n, n+1, n-1, ldots$) when in reality we only defined it to apply to a fix $n$ for the sake of the proof.
      $endgroup$
      – StckXchnge-nub12
      Dec 31 '18 at 19:59













      1












      $begingroup$

      I won't mention every use of induction. Define $$M:=left(begin{array}{cc}
      0 & 1\
      1 & 1
      end{array}right),,V_{n}:=left(begin{array}{c}
      F_{n}\
      F_{n+1}
      end{array}right)=M^{n}left(begin{array}{c}
      0\
      1
      end{array}right)$$
      so $M^{m}=left(begin{array}{cc}
      F_{m-1} & F_{m}\
      F_{m} & F_{m+1}
      end{array}right)$
      and $$left(begin{array}{c}
      F_{m+n}\
      F_{m+n+1}
      end{array}right)=M^{m}left(begin{array}{c}
      F_{n}\
      F_{n+1}
      end{array}right)=left(begin{array}{c}
      F_{m-1}F_{n}+F_{m}F_{n+1}\
      F_{m}F_{n}+F_{m+1}F_{n+1}
      end{array}right).$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        I won't mention every use of induction. Define $$M:=left(begin{array}{cc}
        0 & 1\
        1 & 1
        end{array}right),,V_{n}:=left(begin{array}{c}
        F_{n}\
        F_{n+1}
        end{array}right)=M^{n}left(begin{array}{c}
        0\
        1
        end{array}right)$$
        so $M^{m}=left(begin{array}{cc}
        F_{m-1} & F_{m}\
        F_{m} & F_{m+1}
        end{array}right)$
        and $$left(begin{array}{c}
        F_{m+n}\
        F_{m+n+1}
        end{array}right)=M^{m}left(begin{array}{c}
        F_{n}\
        F_{n+1}
        end{array}right)=left(begin{array}{c}
        F_{m-1}F_{n}+F_{m}F_{n+1}\
        F_{m}F_{n}+F_{m+1}F_{n+1}
        end{array}right).$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          I won't mention every use of induction. Define $$M:=left(begin{array}{cc}
          0 & 1\
          1 & 1
          end{array}right),,V_{n}:=left(begin{array}{c}
          F_{n}\
          F_{n+1}
          end{array}right)=M^{n}left(begin{array}{c}
          0\
          1
          end{array}right)$$
          so $M^{m}=left(begin{array}{cc}
          F_{m-1} & F_{m}\
          F_{m} & F_{m+1}
          end{array}right)$
          and $$left(begin{array}{c}
          F_{m+n}\
          F_{m+n+1}
          end{array}right)=M^{m}left(begin{array}{c}
          F_{n}\
          F_{n+1}
          end{array}right)=left(begin{array}{c}
          F_{m-1}F_{n}+F_{m}F_{n+1}\
          F_{m}F_{n}+F_{m+1}F_{n+1}
          end{array}right).$$






          share|cite|improve this answer









          $endgroup$



          I won't mention every use of induction. Define $$M:=left(begin{array}{cc}
          0 & 1\
          1 & 1
          end{array}right),,V_{n}:=left(begin{array}{c}
          F_{n}\
          F_{n+1}
          end{array}right)=M^{n}left(begin{array}{c}
          0\
          1
          end{array}right)$$
          so $M^{m}=left(begin{array}{cc}
          F_{m-1} & F_{m}\
          F_{m} & F_{m+1}
          end{array}right)$
          and $$left(begin{array}{c}
          F_{m+n}\
          F_{m+n+1}
          end{array}right)=M^{m}left(begin{array}{c}
          F_{n}\
          F_{n+1}
          end{array}right)=left(begin{array}{c}
          F_{m-1}F_{n}+F_{m}F_{n+1}\
          F_{m}F_{n}+F_{m+1}F_{n+1}
          end{array}right).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 31 '18 at 17:53









          J.G.J.G.

          28.1k22844




          28.1k22844























              1












              $begingroup$

              A combinatorial proof (no induction).



              Using the representation of the Fibonacci numbers as the numbers of domino coverings the given identity can be shown in a quick and elegant way.



              It is known that the number of domino coverings of a $2times N$ grid is $F_{N+1}$. Consider the number of domino coverings of a $2times (m+n-1)$ grid. Any covering can be of two types.



              1) The covering has a pair of horizontal dominoes at position $m-1$ and $m$.
              enter image description here



              Therefore on the left side we have a covering of a $2times (m-2)$ grid
              and on the right side we have a covering of a $2times (n-1)$ grid. Therefore the number of such coverings is
              $F_{m-1}cdot F_{n}$.



              2) The covering can be split into two coverings, one of a $2times (m-1)$ grid and another of a $2times n$ grid.
              enter image description here



              Therefore the number of such coverings is
              $F_{m}cdot F_{n+1}$.



              Finally we may conclude that the number of domino coverings of a $2times (m+n-1)$ grid is
              $$F_{m+n}=F_{m-1}cdot F_{n}+F_{m}cdot F_{n+1}.$$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                A combinatorial proof (no induction).



                Using the representation of the Fibonacci numbers as the numbers of domino coverings the given identity can be shown in a quick and elegant way.



                It is known that the number of domino coverings of a $2times N$ grid is $F_{N+1}$. Consider the number of domino coverings of a $2times (m+n-1)$ grid. Any covering can be of two types.



                1) The covering has a pair of horizontal dominoes at position $m-1$ and $m$.
                enter image description here



                Therefore on the left side we have a covering of a $2times (m-2)$ grid
                and on the right side we have a covering of a $2times (n-1)$ grid. Therefore the number of such coverings is
                $F_{m-1}cdot F_{n}$.



                2) The covering can be split into two coverings, one of a $2times (m-1)$ grid and another of a $2times n$ grid.
                enter image description here



                Therefore the number of such coverings is
                $F_{m}cdot F_{n+1}$.



                Finally we may conclude that the number of domino coverings of a $2times (m+n-1)$ grid is
                $$F_{m+n}=F_{m-1}cdot F_{n}+F_{m}cdot F_{n+1}.$$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  A combinatorial proof (no induction).



                  Using the representation of the Fibonacci numbers as the numbers of domino coverings the given identity can be shown in a quick and elegant way.



                  It is known that the number of domino coverings of a $2times N$ grid is $F_{N+1}$. Consider the number of domino coverings of a $2times (m+n-1)$ grid. Any covering can be of two types.



                  1) The covering has a pair of horizontal dominoes at position $m-1$ and $m$.
                  enter image description here



                  Therefore on the left side we have a covering of a $2times (m-2)$ grid
                  and on the right side we have a covering of a $2times (n-1)$ grid. Therefore the number of such coverings is
                  $F_{m-1}cdot F_{n}$.



                  2) The covering can be split into two coverings, one of a $2times (m-1)$ grid and another of a $2times n$ grid.
                  enter image description here



                  Therefore the number of such coverings is
                  $F_{m}cdot F_{n+1}$.



                  Finally we may conclude that the number of domino coverings of a $2times (m+n-1)$ grid is
                  $$F_{m+n}=F_{m-1}cdot F_{n}+F_{m}cdot F_{n+1}.$$






                  share|cite|improve this answer











                  $endgroup$



                  A combinatorial proof (no induction).



                  Using the representation of the Fibonacci numbers as the numbers of domino coverings the given identity can be shown in a quick and elegant way.



                  It is known that the number of domino coverings of a $2times N$ grid is $F_{N+1}$. Consider the number of domino coverings of a $2times (m+n-1)$ grid. Any covering can be of two types.



                  1) The covering has a pair of horizontal dominoes at position $m-1$ and $m$.
                  enter image description here



                  Therefore on the left side we have a covering of a $2times (m-2)$ grid
                  and on the right side we have a covering of a $2times (n-1)$ grid. Therefore the number of such coverings is
                  $F_{m-1}cdot F_{n}$.



                  2) The covering can be split into two coverings, one of a $2times (m-1)$ grid and another of a $2times n$ grid.
                  enter image description here



                  Therefore the number of such coverings is
                  $F_{m}cdot F_{n+1}$.



                  Finally we may conclude that the number of domino coverings of a $2times (m+n-1)$ grid is
                  $$F_{m+n}=F_{m-1}cdot F_{n}+F_{m}cdot F_{n+1}.$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 1 at 14:13

























                  answered Dec 31 '18 at 17:33









                  Robert ZRobert Z

                  98.8k1068139




                  98.8k1068139






























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