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Let's define the Fibonacci numbers by the sequence $F_{0}=0$, $F_{1}=1$, $F_{2}=1$, and $F_{n}$ = $F_{n-1}$ + $F_{n-2}$. Suppose we take a staircase of $n$ stairs, break it in half, and show that $F_{2n}=F_{n}^{2}+F_{n-1}^{2}$. I don't have any clue to get start, anyone help me with this question.

Leucippus
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Student
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  • How do you solve the staircase problem? EDIT: That should say, "a staircase of $2n$ stairs", by the way. – Kevin Long Feb 06 '18 at 16:16
  • What it is the staircase problem never hear about that – Guy Fsone Feb 06 '18 at 16:20
  • @GuyFsone I'm assuming that OP is referring to the proof that the number of ways of climbing a staircase with $n$ steps only by stepping either one step or two steps at a time is equal to the $n$-th Fibonacci number. You prove this by showing they have the same initial conditions and satisfy the same recurrence relation. OP needs to apply the reasoning used to show that those numbers achieve the same recurrence to prove this claim. – Kevin Long Feb 06 '18 at 16:23
  • It should be $F_{2n-1}=F_n^2+F_{n-1}^2$ – robjohn Feb 06 '18 at 17:28

3 Answers3

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It is simple to show that $$ F_nF_k+F_{n-1}F_{k-1}=F_{n+k-1}\tag1 $$ holds for $k=1$ and $k=2$. Suppose that $(1)$ holds for $k-1$ and $k-2$, then $$ \begin{align} F_nF_k+F_{n-1}F_{k-1} &=F_n(\color{#C00}{F_{k-1}}+\color{#090}{F_{k-2}})+F_{n-1}(\color{#C00}{F_{k-2}}+\color{#090}{F_{k-3}})\\ &=\color{#C00}{F_{n+k-2}}+\color{#090}{F_{n+k-3}}\\ &=F_{n+k-1}\tag2 \end{align} $$ Thus, $(1)$ holds for all $k\ge1$.

Setting $n=k$ in $(2)$, we get $$ F_n^2+F_{n-1}^2=F_{2n-1}\tag3 $$

robjohn
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The staircase problem is explained here: How many distinct ways to climb stairs in 1 or 2 steps at a time? where they show that the number of ways to climb $n$ stairs taking only $1$ or $2$ steps is $F_{n+1}$.

The identity to be proven should be $$F_{2n+1}=F_{n+1}^{2}+F_{n}^{2}$$ (note that $3=F_{4}\not=F_{2}^{2}+F_{1}^{2}=1+1$).

Hint. Climbing a staircase with $2n$ steps, we have two cases:

1) we arrive at step $n$ (step $n$ is not broken).

2) we don't arrive at step $n$ (it is broken!) which implies that we arrive at step $n-1$ and then we climb 2 steps.

Robert Z
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We also have the following combinatorial interpretation: $F_{n+2}$ is the number of ways for picking some non-adjacent books from a shelf with $n$ books on it. Let us assume to have $2n-2$ books on a shelf: the number of ways for selecting a subset of non-adjacent books is $F_{2n}$. We either pick the $(n-1)$-th book from the left or we don't: in the former case we are selecting a set of non-adjacent books from a shelf with $n-3$ books on it, then from a shelf with $n-2 $ books on it; in the latter case we are selecting a set of non-adjacent books from a shelf with $n-2$ books, then from a shelf with $n-1$ books. It follows that $$ F_{2n} = F_{n-1} F_n + F_n F_{n+1} = F_n (F_{n-1}+F_{n+1}) = F_nL_n$$ where $L_n$ is the $n$-th Lucas number.

Jack D'Aurizio
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