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Pascal's Triangle and Fibonacci numbers

Let \(f_{n}\) denote the \(nth\) Fibonacci number. Prove that \[f_{n} = \dbinom{n-1}{0}+\dbinom{n-2}{1}+\dbinom{n-3}{2}+...+\dbinom{n-k}{k-1}, \] where \(k= \lfloor\frac{n+1}{2}\rfloor\) \[\]Would appreciate if anyone posts a simple and not that long proof for this.

Note by Marc Vince Casimiro
2 years, 1 month ago

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You can prove this by induction. I'll demonstrate how to show that \(f_{10}=f_9+f_8\):

\[f_{10}=\left(\begin{matrix}9\\0\end{matrix}\right)+\left(\begin{matrix}8\\1\end{matrix}\right)+\left(\begin{matrix}7\\2\end{matrix}\right)+\left(\begin{matrix}6\\3\end{matrix}\right)+\left(\begin{matrix}5\\4\end{matrix}\right)\] \[=\left(\begin{matrix}9\\0\end{matrix}\right)+\left[\left(\begin{matrix}7\\0\end{matrix}\right)+\left(\begin{matrix}7\\1\end{matrix}\right)\right]+\left[\left(\begin{matrix}6\\1\end{matrix}\right)+\left(\begin{matrix}6\\2\end{matrix}\right)\right]+\left[\left(\begin{matrix}5\\2\end{matrix}\right)+\left(\begin{matrix}5\\3\end{matrix}\right)\right]+\left[\left(\begin{matrix}4\\3\end{matrix}\right)+\left(\begin{matrix}4\\4\end{matrix}\right)\right]\] \[=\left(\begin{matrix}9\\0\end{matrix}\right)+\left[\left(\begin{matrix}7\\1\end{matrix}\right)+\left(\begin{matrix}6\\2\end{matrix}\right)+\left(\begin{matrix}5\\3\end{matrix}\right)+\left(\begin{matrix}4\\4\end{matrix}\right)\right]+\left[\left(\begin{matrix}7\\0\end{matrix}\right)+\left(\begin{matrix}6\\1\end{matrix}\right)+\left(\begin{matrix}5\\2\end{matrix}\right)+\left(\begin{matrix}4\\3\end{matrix}\right)\right]\] \[=\left[\left(\begin{matrix}8\\0\end{matrix}\right)+\left(\begin{matrix}7\\1\end{matrix}\right)+\left(\begin{matrix}6\\2\end{matrix}\right)+\left(\begin{matrix}5\\3\end{matrix}\right)+\left(\begin{matrix}4\\4\end{matrix}\right)\right]+\left[\left(\begin{matrix}7\\0\end{matrix}\right)+\left(\begin{matrix}6\\1\end{matrix}\right)+\left(\begin{matrix}5\\2\end{matrix}\right)+\left(\begin{matrix}4\\3\end{matrix}\right)\right]\] \[=f_9+f_8\]

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 · 2 years, 1 month ago

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Long Proof: \[f_{n}=\displaystyle\sum_{k=0}^{n-1} \binom{n-1-k}{k} \] Using Pascal's formula, for each \(2 \leq n\) \[g_{n-1} + g_{n-2} = \displaystyle\sum_{k=0}^{n-2}\binom{n-2-k}{k}+ \displaystyle\sum_{j=0}^{n-3} \binom{n-3-j}{j}\] \[=\binom{n-2}{0} + \displaystyle\sum_{k=1}^{n-2} \binom{n-2-k}{k} + \displaystyle\sum_{k=1}^{n-2} \binom{n-2-k}{k-1}\] \[=\binom{n-2}{0}+\displaystyle\sum_{k=1}^{n-2}\left(\binom{n-2-k}{k}+\binom{n-2-k}{k-1}\right)\] \[=\binom{n-2}{0} + \displaystyle\sum_{k=1}^{n-2}\binom{n-1-k}{k}\] \[=\binom{n-2}{0} + \displaystyle\sum_{k=1}^{n-2}\binom{n-1-k}{k} + \binom{0}{n-1}\] \[=\displaystyle\sum_{k=0}^{n-1}\binom{n-1-k}{k}=f_{n}\] Pretty long and complex. Looking forward to the simple induction @kenny lau :) Marc Vince Casimiro · 2 years, 1 month ago

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