Fibonacci Number by Power of 2

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

Proof
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:

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

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

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 $r \ge 2$, 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:

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

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