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I've been struggled for this identity for a while, how can I use combinatorial proof to prove the Fibonacci identity $$F_2+F_5+\dots+F_{3n-1}=\frac{F_{3n+1}-1}{2}$$ I know that $F_n$ is number of tilings for the board of length $n-1$, so if I rewrite the identity and let $f_n$ be the number of tilings for the board of length $n$, then I got $$f_1+f_4+\dots+f_{3n-2}=\frac{f_{3n}-1}{2}$$ the only thing that I know so far is the Right hand side, $f_{3n}-1$ is the number of tilings for the $3n$ board with at least one $(1\times 2)$ tile (or maybe I am wrong), but I have no idea of what the fraction $\frac{1}{2}$ is doing here. Can anyone help?

(P.S.: In general, when it comes to this kind of combinatorial proof question, is it ok to rewrite the question in a different way? Or is it ok to rewrite this question as $2(f_1+f_4+\dots+f_{3n-1})=f_{3n}-1$, then process the proof?

Thank you for all your useful proofs, but this is an identity from a course that I am taking recently, and it is all about combinatorial proof, so some hint about how to find the number of tilings for the board of length $3n$ would be really helpful.

Thanks for dtldarek's help, I finally came up with:

Rewrite the identity as $2F_2+2F_5+\dots+2F_{3n-1}=\frac{F_{3n+1}-1}{2}$, then the Left hand side becomes $F_2+F_2+F_5+F_5+\dots+F_{3n-1}+F_{3n-1}=F_0+F_1+F_2+F_3+\dots+F_{3n-3}+F_{3n-2}+F_{3n-1}=\sum^{3n-1}_{i=0}F_{i}\implies \sum^{3n-1}_{i=0} F_i=F_{3n+1}-1$, and recall that $f_n$ is the number of tilings for the board of length $n$, so we have $\sum^{3n-2}_{i=0}f_i=f_{3n}-1$.

For the Right hand side $f_{3n}$ is the number of tilings for the length of $3n$ board, then $f_{3n}-1$ is the number of tilings for a $3n$ board use at least one $1\times 2$ tile. Now, for the Left hand side, conditioning on the last domino in the $k^{th}$ cell, for any cells before the $k^{th}$ cell, there are only one way can be done, and all cells after the $k+1$ cell can be done in $f_{3n-k-1}$, finally sum up $k$ from 0 to $3n-1$, which is the Left hand side.

Is it ok? did I change the meaning of the original identity?

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3  
Why are you wanting to prove it in a combinatorial way? Induction seems like a much easier way to proceed with a proof. –  Andrew D Mar 8 '13 at 17:59
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A quick answer to the P.S.: Yes, it’s perfectly fine to do so. –  Brian M. Scott Mar 8 '13 at 17:59
    
I recently proved in another problem that $F_0+F_3+\dots F_{3n} = \frac{F_{3n+2}-1}{2}$. This proof works here, too. (Skip to the section with matrices.) –  Thomas Andrews Mar 8 '13 at 18:06
    
thank you for the proof, but I still don't see that how I can apply a combinatorial proof here. This identity is from a course that I am taking recently, it is all about combinatorial proof, so basically algebraic proof is not really useful here, but thanks for the proof anyways. –  user62453 Mar 8 '13 at 18:17

4 Answers 4

This is rather easy using algebra:

$$F_2+F_5+\dots+F_{3n-1}=\frac{F_{3n+1}-1}{2}$$ $$ 2F_2+2F_5+\dots+2F_{3n-1}=F_{3n+1}-1 $$ $$ (F_0+F_1)+F_2+(F_3+F_4) + F_5+\dots+(F_{3n-3} + F_{3n-2}) + F_{3n-1}=F_{3n+1}-1 $$ $$ \sum_{k=0}^{m} F_k = F_{m+2} -1 \quad\text{ for } m = 3n-1$$

where the last one is a well know identity for Fibonacci numbers. To prove it by combinatorial interpretation you might want to follow the same way. Let's start with the following from Wikipedia:

The number of binary strings of length $n$ without consecutive $1$s is the Fibonacci number $F_{n+2}$.

Let the dot $\cdot$ denote concatenation and $$F_{n+2} = \big\{ w \in \{\mathtt{0},\mathtt{1}\}^n \mid \text{ there is no two consecutive }\mathtt{1}\text{s in }w\big\}, $$

then the last identity reads: $$F_{m+2} - \{\mathtt{0}^{m+2}\} = \mathtt{10}\cdot F_m \cup \mathtt{010}\cdot F_{m-1} \cup \ldots \cup \mathtt{0}^{m-1}\mathtt{10} \cdot F_1 \cup \mathtt{0}^{m}\mathtt{10}\cdot F_0.$$

However, we could group $F_k \cup F_{k+1}$ into $F_{k+2}$, because $F_{k+2} = \mathtt{0}\cdot F_{k+1} \cup \mathtt{10} \cdot F_{k}$. Going up through all the equations in a similar manner you can find the combinatorial interpretation for the algebraic proof above.

Good luck ;-)

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I see what you are doing, for the first part you went from $F_2+F_5+\dots+F_{3n-1}=\frac{F_{3n+1}-1}{2}$ to $\sum_{i=0}^{3n-1}F_i=F_{3n+1} -1$, this is very useful, Can I just start the proof from this point instead of going into to the proof by using the binary strings? does it change the meaning of the original identity? –  user62453 Mar 8 '13 at 18:56
    
@user62453 The pure-algebraic proof is alright by itself. In the begging there was "how can I use combinatorial proof to prove the Fibonacci identity" so I thought you wanted use interpretation. –  dtldarek Mar 8 '13 at 21:51

The inductive proof works: $F_2=2=\frac {5-1}2=\frac {F_4-1}2$ Assume we have $\sum_{i=0}^n F_{3i+2}=\frac {F_{3i+4}-1}2$ then $$\begin {align}\sum_{i=0}^{n+1} F_{3i+2}&=\sum_{i=0}^n F_{3i+2}+F_{3n+5}\\&=\frac {F_{3i+4}-1}2+F_{3n+5}\\&=-\frac 12+\frac {F_{3n+5}+F_{3n+6}}2\\&=\frac {F_{3n+7}-1}2 \\&=\frac {F_{3(n+1)+4}-1}2\end {align}$$

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It is instructive to look at this matrix formula: $$\left(\begin{matrix}1 & 1\\1&0\end{matrix}\right)^n = \left(\begin{matrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{matrix}\right)$$

So let $A=\left(\begin{matrix}1 & 1\\1&0\end{matrix}\right)$.

Then $\sum_{i=1}^{n} A^{3n-1} = A^{-1}\sum_{i=1}^n A^{3n}$. This matrix has in the non-diagonals the sum you are looking for. But the sum follows the normal geometric series, so we are looking for a non-diagonal member of $$A^{-1}A^{3}(A^{3n}-I)(A^3-I)^{-1}$$ (Also, all these matrices commute.)

Now $A^2 = I + A$, so $A^3 = A + A^2 = I + 2A$, so $A^3-I = 2A$, so this formula becomes:

$$\frac{1}{2} A(A^{3n}-I)=A^{3n+1}-A$$

But we know that $A^{3n+1}$ has $F_{3n+1}$ in the lower left corner, and we know that $A$ has $1$ in the lower left corner, so the lower left corner of the matrix we are computing is $\frac{F_{3n+1}-1}{2}$, which is the result we want.

While the inductive proof works pretty directly, this shows you a more general approach to finding closed formulas for $\sum_{i=0}^{n-1} F_{di+r}$ for any $d,r$.

In this more general case:

$$\sum_{i=0}^{n-1} A^{di+r} = (A^{dn+r}-A^r)(A^d-I)^{-1}$$ we can compute

$$(A^d-I)^{-1} = \frac{1}{F_{d+1}+F_{d-1} - 1 -(-1)^d} \left(\begin{matrix}1-F_{d-1}&F_d\\F_d&1-F_{d+1}\end{matrix}\right)$$

And $$A^{dn+r}-A^r = \left(\begin{matrix}F_{dn+r+1}-F_{r+1}&F_{dn+r}-F_{r}\\F_{dn+r}-F_{r}&F_{dn+r-1}-F_{r-1}\end{matrix}\right)$$

Multiplying to get the lower left corner gives you:

$$\sum_{i=0}^{n-1} F_{di+r} = \frac{F_d(F_{dn+r+1}-F_{r+1})-(F_{d+1}-1)(F_{dn+r}-F_r)}{F_{d+1}+F_{d-1} - 1 -(-1)^d}$$

When $d=3$, then $F_3=2$, $F_4=3$ and $F_2=1$, and, after some simplification, we get:

$$\sum_{i=0}^{n-1} F_{3i+r} = \frac{F_{3n+r-1} - F_{r-1}}{2}$$

When $r=0$ and $d$ is odd, this simplifies to:

$$\sum_{i=0}^{n-1} F_{di} = \frac{F_dF_{dn+1} - (F_{d+1}-1)F_{dn} - F_d}{F_{d+1}+F_{d-1}}$$

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This may not be the quickest approach, but it seems fairly simple, using only the recursion equation $F_i+F_{i+1}=F_{i+2}$ and initial conditions, which I take to be $F_0=0$ and $F_1=1$. Notice first that you can apply the recursion equation to replace, on the left side of your formula, each term by the sum of the two preceding Fibonacci numbers, so this left side is equal to $F_0+F_1+F_3+F_4+F_6+F_7+\dots+F_{3n-3}+F_{3n-2}$, which skips exactly those terms that are present in your formula's left side. So, adding this form of your left side to the original form, you find that twice the left side is the sum of all the Fibonacci numbers up to and including $F_{3n-1}$. So what needs to be proved is that $\sum_{i=0}^{3n-1}F_i=F_{3n+1}-1$. That can be done by induction on $n$. The base case is trivial ($0+1+1=3-1$), and the induction step is three applications of the recursion equation. In $\sum_{i=0}^{3(n+1)-1}F_i$, apply the induction hypothesis to replace all but the last three terms with $F_{3n+1}-1$. To combine that with the last three terms, $F_{3n}+F_{3n+1}+F_{3n+2}$, use the recursion equation to replace $F_{3n}+F_{3n+1}$ and $F_{3n+1}+F_{3n+2}$ with $F_{3n+2}$ and $F_{3n+3}$, respectively, and then again to replace these last two results with $F_{3n+4}=F_{3(n+1)+1}$.

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