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Prove the following combinatorially:

$$F_{n} = \sum_{i=0}^{\left\lfloor n/2 \right\rfloor}\binom{n-i}{i}$$

So, I know that the Fibonacci number counts the number of ways to cover a $1 \times n$ rectangle with checkers (covering a single square) and dominoes (covering two squares). I believe that will come up somewhere in the explanation, but I am having difficulty seeing how to show that the RHS counts the same as the Fibonacci number.

EDIT: I think I have got it figured out. If we say that the LHS is the way to cover a nx1 rectangle with checkers and dominoes, then the floor of n/2 is the max amount of dominoes you could have. In this case, i represents the number of dominoes in that particular solution, hence why you have n-i positions. Then you figure out how many unique solutions there are when you have i dominoes, adding up all the solutions with the different number of dominoes.

Blue
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  • This is not $F_n$, it should be $F_{n+1}$. Put $n=2$ and see – Sayan Dutta Mar 09 '23 at 21:32
  • Depends on what you mean by Combinatorial proof. Personally, I take the (arguable) position that asserting that in Pascal's Triangle, each number is the sum of the two numbers above it is a Combinatorial argument. My first try, which might not succeed, would be to attempt to use this property of Pascal's triangle to prove the assertion by induction. I would split into two cases, depending on whether $~n~$ is odd or even. In each case, I would use that each Fibonacci # is the sum of the two previous Fibonacci #'s, & simply explore where this leads, against the backdrop of Pascal's tri. – user2661923 Mar 09 '23 at 21:58
  • @SayanDutta Different people take different indexing for the Fibonacci sequence -- this question is internally consistent (correspodning to the indexing $F_0 = F_1 = 1$) – JBL Mar 10 '23 at 13:19
  • @user2661923 The proof you suggest is pure algebraic manipulation, there is no combinatorics at all in it. – JBL Mar 10 '23 at 13:22
  • @JBL I disagree. You can construe $~\displaystyle \binom{n}{k}~$ as the $~a^k b^{n-k}~$ coefficient of $~(a+b)^n.~$ Then, consider the product $~(a + b)^{n+1} = (a+b)^n \times (a+b).~$ This consideration leads to the combinatoric (rather than algebraic) argument that the coefficient of $~a^{n+1}b^n~$ in the product $~(a+b)^{n+1}~$ is the sum of the $~a^nb^n~$ and $~a^{n+1}b^{n-1}~$ coefficients from $~(a + b)^n.~$ This provides a combinatoric argument that $$\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}.$$ This combinatoric argument is then the foundation of my approach. – user2661923 Mar 10 '23 at 19:04
  • @user2661923: Yes, Pascal's recurrence can be proved combinatorially. No, it does not follow that a proof using induction and Pascal's recurrence is a combinatorial proof. Obviously the term "combinatorial proof" is not sufficiently well defined for this to pass beyond being a question of opinion -- but by any reasonable definition a proof that uses only algebraic properties of the objects involved should not be called combinatorial. (FWIW I have a PhD in combinatorics, so you don't need to explain elementary things to me.) – JBL Mar 10 '23 at 19:24

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Define $$G_n = \sum_{i=0}^{\left\lfloor n/2 \right\rfloor}\binom{n-i}{i}$$ so that we have $$G_1=1,G_2=2$$ and \begin{align*} G_{n}+G_{n+1}&=\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}\cdots +\binom{n+1}{0}+\binom{n}{1}+\binom{n-1}{2}\cdots\\ &=\binom{n+1}{0}+\left(\binom n0+\binom{n}{1}\right)+\left(\binom{n-1}{1}+\binom{n-1}{2}\right)+\left(\binom{n-2}{2}+\binom{n-2}{3}\right)+\cdots\\ &=\binom{n+2}{0}+\binom{n+1}{1}+\binom{n}{2}+\cdots\\ &=G_{n+2} \end{align*} And so, $G_{n}$ is $n+1$-th Fibonacci number.

Edit: Just realized that you wanted a combinatorial proof- that one is actually easier.

Try proving that the number of $(n+1)$ length binary strings with no consecutive $1$'s relate to the Fibonacci numbers.

Now count the number of binary $(n+1)$ length strings that contain exactly $k$ $1$'s.

Sayan Dutta
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