Pascal's Triangle and Fibonacci numbers

Let fnf_{n} denote the nthnth Fibonacci number. Prove that fn=(n−10)+(n−21)+(n−32)+...+(n−kk−1),f_{n} = \dbinom{n-1}{0}+\dbinom{n-2}{1}+\dbinom{n-3}{2}+...+\dbinom{n-k}{k-1}, where k=⌊n+12⌋k= \lfloor\frac{n+1}{2}\rfloor Would appreciate if anyone posts a simple and not that long proof for this.

#NumberTheory

Note by Marc Vince Casimiro
6 years, 6 months ago

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Comments

You can prove this by induction. I'll demonstrate how to show that f10=f9+f8f_{10}=f_9+f_8:

f10=(90)+(81)+(72)+(63)+(54)f_{10}=\left(\begin{matrix}9\\0\end{matrix}\right)+\left(\begin{matrix}8\\1\end{matrix}\right)+\left(\begin{matrix}7\\2\end{matrix}\right)+\left(\begin{matrix}6\\3\end{matrix}\right)+\left(\begin{matrix}5\\4\end{matrix}\right) =(90)+[(70)+(71)]+[(61)+(62)]+[(52)+(53)]+[(43)+(44)]=\left(\begin{matrix}9\\0\end{matrix}\right)+\left[\left(\begin{matrix}7\\0\end{matrix}\right)+\left(\begin{matrix}7\\1\end{matrix}\right)\right]+\left[\left(\begin{matrix}6\\1\end{matrix}\right)+\left(\begin{matrix}6\\2\end{matrix}\right)\right]+\left[\left(\begin{matrix}5\\2\end{matrix}\right)+\left(\begin{matrix}5\\3\end{matrix}\right)\right]+\left[\left(\begin{matrix}4\\3\end{matrix}\right)+\left(\begin{matrix}4\\4\end{matrix}\right)\right] =(90)+[(71)+(62)+(53)+(44)]+[(70)+(61)+(52)+(43)]=\left(\begin{matrix}9\\0\end{matrix}\right)+\left[\left(\begin{matrix}7\\1\end{matrix}\right)+\left(\begin{matrix}6\\2\end{matrix}\right)+\left(\begin{matrix}5\\3\end{matrix}\right)+\left(\begin{matrix}4\\4\end{matrix}\right)\right]+\left[\left(\begin{matrix}7\\0\end{matrix}\right)+\left(\begin{matrix}6\\1\end{matrix}\right)+\left(\begin{matrix}5\\2\end{matrix}\right)+\left(\begin{matrix}4\\3\end{matrix}\right)\right] =[(80)+(71)+(62)+(53)+(44)]+[(70)+(61)+(52)+(43)]=\left[\left(\begin{matrix}8\\0\end{matrix}\right)+\left(\begin{matrix}7\\1\end{matrix}\right)+\left(\begin{matrix}6\\2\end{matrix}\right)+\left(\begin{matrix}5\\3\end{matrix}\right)+\left(\begin{matrix}4\\4\end{matrix}\right)\right]+\left[\left(\begin{matrix}7\\0\end{matrix}\right)+\left(\begin{matrix}6\\1\end{matrix}\right)+\left(\begin{matrix}5\\2\end{matrix}\right)+\left(\begin{matrix}4\\3\end{matrix}\right)\right] =f9+f8=f_9+f_8

Be careful of odd-even parity when proving it! I used the Recursive Formula of binomial coefficient to prove it.

If I have time I may do the whole proof.

Kenny Lau - 6 years, 6 months ago

Long Proof: fn=∑k=0n−1(n−1−kk)f_{n}=\displaystyle\sum_{k=0}^{n-1} \binom{n-1-k}{k} Using Pascal's formula, for each 2≤n2 \leq n gn−1+gn−2=∑k=0n−2(n−2−kk)+∑j=0n−3(n−3−jj)g_{n-1} + g_{n-2} = \displaystyle\sum_{k=0}^{n-2}\binom{n-2-k}{k}+ \displaystyle\sum_{j=0}^{n-3} \binom{n-3-j}{j} =(n−20)+∑k=1n−2(n−2−kk)+∑k=1n−2(n−2−kk−1)=\binom{n-2}{0} + \displaystyle\sum_{k=1}^{n-2} \binom{n-2-k}{k} + \displaystyle\sum_{k=1}^{n-2} \binom{n-2-k}{k-1} =(n−20)+∑k=1n−2((n−2−kk)+(n−2−kk−1))=\binom{n-2}{0}+\displaystyle\sum_{k=1}^{n-2}\left(\binom{n-2-k}{k}+\binom{n-2-k}{k-1}\right) =(n−20)+∑k=1n−2(n−1−kk)=\binom{n-2}{0} + \displaystyle\sum_{k=1}^{n-2}\binom{n-1-k}{k} =(n−20)+∑k=1n−2(n−1−kk)+(0n−1)=\binom{n-2}{0} + \displaystyle\sum_{k=1}^{n-2}\binom{n-1-k}{k} + \binom{0}{n-1} =∑k=0n−1(n−1−kk)=fn=\displaystyle\sum_{k=0}^{n-1}\binom{n-1-k}{k}=f_{n} Pretty long and complex. Looking forward to the simple induction @kenny lau :)

Marc Vince Casimiro - 6 years, 6 months ago
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