×

# 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
3 years, 3 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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.

- 3 years, 3 months ago

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 :)

- 3 years, 3 months ago