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Edwin Kofler

Fibonacci Equation for Zth (Nth) Term Using Pascal's Triangle (Part 2 of 2)

If you’re not familiar with Pascal’s Triangle, see part 1.

Finding a Formula

Pascal's Triangle How do we leverage the relationship (in the image above) to obtain an equation that obtains the $z$th term ($Fib(z)$, or $F(z)$) in the Fibonacci sequence? First, we need to figure out what our equation may look like. We know we’re adding up terms of the Fibonacci sequence, so a summation symbol will be used. Additionally, we are adding up terms from Pascal’s triangle, where each term individually can be written as $_nC_r$.

So our final equation will look akin to this:

$$\sum_{\varphi=1}^{z_{end}}{_n}C_k $$

But this is not exactly right, since $n$ and $k$ are different for each $\varphi$th term that’s being added up (dependent on the summation index). Therefore, we can more accurately write:

$$ Fib(z) = f(z) = \sum_{\varphi=1}^{z_{end}}{_{n(\varphi)}}C _{k(\varphi)} $$

Note that the summation index, $\varphi$ is starting from $1$. Also, I’m starting Pascal’s triangle from a row of 1. So since $z$ starts at 1, $z_1 = 1$, $z_2 = 1$, and $z_3 = 2$.

I started by reorganizing all the $_nC_r$ terms from the triangle above into rows. I tried to color the table similarly to the triangle.

Pascal's Triange Reordered

One of the first and more obvious patterns is found in the $z_{end}$ column. Rather than increasing by an increment of $1$ for every row as the $z$ colum does, it increases by an increment of $1$ for every other row.

This is an important pattern because it determines the number of terms being summed.

$$ z_{end} = round(\frac{z}{2}) $$

Now, we know we can better describe what the summation might look like

$$ Fib(z) = f(z) =\sum_{\varphi=1}^{round(\frac{z}{2})}{_{n(\varphi)}}C _{k(\varphi)} $$

So to find the $4$th term in the Fibonacci Sequence, or $z = 4$, we know $z_{end} = 2$. We’re summing $2$ terms.

$$ Fib(4) =\sum_{\varphi=1}^{2}{_{n(\varphi)}}C _{k(\varphi)} =\ _2C_1 +\ _3C_0 $$

We only know that $_2C_1$ and $_3C_0$ are summed due to the table I wrote above. We don’t know why $n(1) = 2$ or why $k(2) = 0$ yet.

There are a few other patterns held within the grid. I found it easier to find the pattern by changing which way the terms were summed. As you can see, I rearrange the order of the ${{n(\varphi)}}C{k(\varphi)}$ terms, making the $\varphi_1, \varphi_2, \ldots {\varphi} _{th}$

$\varphi_1, \varphi_2, … \varphi _{th} $ terms somewhat arbitrary (depending on the grid structure rather than concrete values).

Pascal's Triange Reordered

Now, you can see a clearer pattern for the $ k(\varphi) $ more easily, now that they’re more aligned. It’s value is one less than the current summation index value (the $\varphi$th term, up to $z_{end}$).

$$ k(\varphi) = \varphi - 1 $$

Now, we just have to determine $n(\varphi)$. It’s slightly harder because it depends on the $z$th term and the $\varphi$th term. In other words, it depends on the $z$th term we’re adding to in the Fibonacci sequence and the value of the summation index, $\varphi$.

$$ n(\varphi) = z - \varphi $$

Or, more accurately

$$ n(\varphi, z) = z - \varphi $$

Now we have all of the missing parts! Putting it all together, we obtain

$$ F(z) = Fib(z) = \sum_{\varphi=1}^{z_{end}\ =\ round(\frac{z}{2})} {_{n(\varphi, z)\ =\ z - \varphi}}C _{k(\varphi)\ =\ \varphi - 1} $$

Or, more succinctly

$$ F(z) = \sum_{\varphi=1}^{round(\frac{z}{2})}{_{z - \varphi}}C _{\varphi - 1} $$

Does this equation give us the $z$th term of the Fibonacci sequence by adding up terms of Pascal’s triangle? Yes! It conforms to the original goal.

Let $z = 9$. We obtain

$$ F(9) = \sum_{\varphi=1}^{5}{_{9 - \varphi}}C _{\varphi - 1} $$

$$ =\ _{9 - 1}C _{1 - 1} + \ _{9 - 2}C _{2 - 1} + \ _{9 - 3}C _{3 - 1} + \ _{9 - 4}C _{4 - 1} + \ _{9 - 5}C _{5 - 1} $$

$$ =\ _{8}C _{0} + \ _{7}C _{1} + \ _{6}C _{2} + \ _{5}C _{3} + \ _{4}C _{4} $$

$$=\ 1 + 7 + 15 + 10 + 1$$

$$=\ 34$$

Did make a proof? No. Although I can’t conclusively say so, I want to believe that this works for any $n\ |\ n \in \mathbb{N}$.