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Is it true:

$$\sum_{k=\frac{n}{2}}^{n}\binom{k}{n-k}=F_n? $$ where $F_n$ is a Fibonacci number.

Wolfram alpha says that the answer must be $F_{n+1}$.

I've seen many posts with this question but none of them was helpful to me. And this time the sum is different. Can anybody give me a breath explanation how should I solve this equation?

Edit: $n \ge 0, \quad F_1=F_2=1$

Karagum
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1 Answers1

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It depends. If you assume that the Fibonacci numbers start with $F_1=1$ and $F_2=1$, then Wolfram alpha is correct, i.e. $$ \sum_{k=\frac{n}{2}}^n \binom{k}{n-k}=F_{n+1}. $$ If you assume that the Fibonacci numbers start with $F_1=0$ and $F_2=1$, then we have $$ \sum_{k=\frac{n}{2}}^n \binom{k}{n-k}=F_{n}. $$ Actually, with $F_1=F_2=1$, we have $$ \sum_{k=\lceil\frac{n}{2}\rceil}^n \binom{k}{n-k}=F_{n+1}. $$

Let $$ y_n = \sum_{k=\lceil\frac{n}{2}\rceil}^n \binom{k}{n-k}. $$ What I want to proof is that $y_n = y_{n-2}+y_{n-1}$. I will use the fact that $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$. First, I proof this for $n$ is even, such that I can use the fact that $\lceil \frac{n+1}{2} \rceil = \lceil \frac{n}{2} \rceil + 1$: \begin{align} y_n &= 1 + \sum_{k=\lceil\frac{n}{2}\rceil+1}^{n-1} \binom{k}{n-k} + 1 \\ &=\sum_{k=\lceil\frac{n}{2}\rceil+1}^{n-1} \binom{k-1}{n-k-1}+1 + \sum_{k=\lceil\frac{n}{2}\rceil+1}^{n-1} \binom{k-1}{n-k}+1 \\ &=\sum_{k=\lceil\frac{n}{2}\rceil+1}^{n-1} \binom{k-1}{n-2-(k-1)}+1 + \sum_{k=\lceil\frac{n+1}{2}\rceil}^{n-1} \binom{k-1}{n-1-(k-1)}+1 \\ &= \sum_{k=\lceil\frac{n-2}{2}\rceil+1}^{n-2} \binom{k}{n-2-k}+1 + \sum_{k=\lceil\frac{n-1}{2}\rceil}^{n-2} \binom{k}{n-1-k} + 1 \\ &= \sum_{k=\lceil\frac{n-2}{2}\rceil}^{n-2} \binom{k}{n-2-k} + \sum_{k=\lceil\frac{n-1}{2}\rceil}^{n-1} \binom{k}{n-1-k} \\ &=y_{n-2} + y_{n-1}. \end{align}

Now we can do the same for $n$ being odd. I will assume that $\lceil\frac{n}{2}\rceil-1 = \lceil\frac{n-1}{2}\rceil$: \begin{align} y_n &= \sum_{k=\lceil\frac{n}{2}\rceil}^{n-1} \binom{k}{n-k} + 1 \\ &= \sum_{k=\lceil\frac{n}{2}\rceil}^{n-1} \binom{k-1}{n-k-1} + \sum_{k=\lceil\frac{n}{2}\rceil}^{n-1} \binom{k-1}{n-k} + 1 \\ &= \sum_{k=\lceil\frac{n}{2}\rceil}^{n-1} \binom{k-1}{n-2-(k-1)} + \sum_{k=\lceil\frac{n}{2}\rceil}^{n-1} \binom{k-1}{n-1-(k-1)} + 1 \\ &= \sum_{k=\lceil\frac{n-2}{2}\rceil}^{n-2} \binom{k}{n-2-k} + \sum_{k=\lceil\frac{n-1}{2}\rceil}^{n-2} \binom{k}{n-1-k}+1 \\ &= y_{n-2} + \sum_{k=\lceil\frac{n-1}{2}\rceil}^{n-1} \binom{k}{n-1-k} \\ &= y_{n-2} + y_{n-1}. \end{align}

EdG
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