# Fibonacci Number by Power of 2

## Theorem

 $\displaystyle \forall n \in \Z_{\ge 0}: \ \$ $\displaystyle 2^{n - 1} F_n$ $=$ $\displaystyle \sum_k 5^k \dbinom n {2 k + 1}$ $\displaystyle$ $=$ $\displaystyle \dbinom n 1 + 5 \dbinom n 3 + 5^2 \dbinom n 5 + \cdots$

where:

$F_n$ denotes the $n$th Fibonacci number
$\dbinom n {2 k + 1} \$ denotes a binomial coefficient.

## Proof 1

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle 2^{n - 1} F_n = \sum_k 5^k \dbinom n {2 k + 1}$

First note the bounds of the summation.

By definition, $\dbinom n k = 0$ where $k < 0$ or $k > n$.

Thus in all cases in the following, terms outside the range $0 \le k \le n$ can be discarded.

$P \left({0}\right)$ is the case:

 $\displaystyle 2^{-1} F_0$ $=$ $\displaystyle 0$ Definition of Fibonacci Numbers: $F_0 = 0$ $\, \displaystyle \forall k \in \Z: \,$ $\displaystyle$ $=$ $\displaystyle 5^k \dbinom 0 {2 k + 1}$ Zero Choose n

Thus $P \left({0}\right)$ is seen to hold.

### Basis for the Induction

$P \left({1}\right)$ is the case:

 $\displaystyle 2^0 F_1$ $=$ $\displaystyle 1$ Definition of Fibonacci Numbers: $F_1 = 1$ $\displaystyle$ $=$ $\displaystyle 5^0 \dbinom 1 {2 \times 0 + 1}$ One Choose n

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true.

So this is the induction hypothesis:

$\displaystyle 2^{r - 1} F_r = \sum_k 5^k \dbinom r {2 k + 1}$

from which it is to be shown that:

$\displaystyle 2^r F_{r + 1} = \sum_k 5^k \dbinom {r + 1} {2 k + 1}$

### Induction Step

This is the induction step:

 $\displaystyle 2^r F_{r + 1}$ $=$ $\displaystyle 2^r \left({F_{r - 1} + F_r}\right)$ Definition of Fibonacci Numbers $\displaystyle$ $=$ $\displaystyle 4 \times 2^{r - 2} F_{r - 1} + 2 \times 2^{r - 1} F_r$ $\displaystyle$ $=$ $\displaystyle 4 \times \sum_k 5^k \dbinom {r - 1} {2 k + 1} + 2 \sum_k 5^k \dbinom r {2 k + 1}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle 2 \times \sum_k 5^k \dbinom {r - 1} {2 k + 1} + 2 \sum_k 5^k \left({\dbinom {r - 1} {2 k + 1} + \dbinom r {2 k + 1} }\right)$

So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\forall n \in \Z_{\ge 0}: \displaystyle 2^{n - 1} F_n = \sum_k 5^k \dbinom n {2 k + 1}$

$\blacksquare$

## Proof 2

 $\displaystyle 2^{n - 1} F_n$ $=$ $\displaystyle \dfrac {2^n} {2 \sqrt 5} \left({\phi^n - \hat \phi^n}\right)$ Euler-Binet Formula $\displaystyle$ $=$ $\displaystyle \dfrac {\left({1 + \sqrt 5}\right)^n - \left({1 - \sqrt 5}\right)^n} {2 \sqrt 5}$ Definition 2 of Golden Mean $\displaystyle$ $=$ $\displaystyle \dfrac 1 {2 \sqrt 5} \sum_{j \mathop = 0}^n \binom n j \sqrt 5^j - \dfrac 1 {2 \sqrt 5} \sum_{j \mathop = 0}^n \left({-1}\right)^j \binom n j \sqrt 5^j$ Binomial Theorem $\displaystyle$ $=$ $\displaystyle \dfrac 1 {\sqrt 5} \sum_{\substack {0 \mathop \le j \mathop \le n \\ j \text { odd} } } \binom n j \sqrt 5^j$ even terms vanish, odd terms double up $\displaystyle$ $=$ $\displaystyle \dfrac 1 {\sqrt 5} \sum_{j \text { odd} } \binom n j 5^{j / 2}$ Definition of Binomial Coefficient: $\dbinom n j = 0$ for $j < 0$ and $j > n$ $\displaystyle$ $=$ $\displaystyle \sum_{j \text { odd} } \binom n j 5^{\left({j - 1}\right) / 2}$ gathering the spare $\sqrt 5$ into the index

Setting $j = 2 k + 1$ for $0 \le k \le \left({j - 1}\right) / 2$ gives:

$\displaystyle \sum_{k \mathop \ge 0} \binom n {2 k + 1} 5^k$

and the limits of the index of the summation are irrelevant, as $\dbinom n {2 k + 1} = 0$ for $j < 0$ and $j > n$.

Hence the result.

## Also presented as

This result can also be seen presented as:

$\displaystyle 2^n F_n = 2 \sum_{k \text { odd} } \dbinom n k 5^{\left({k - 1}\right) / 2}$

## Historical Note

This result was discovered by Eugène Charles Catalan.