Fibonacci Number by Power of 2

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Theorem

\(\ds \forall n \in \Z_{\ge 0}: \, \) \(\ds 2^{n - 1} F_n\) \(=\) \(\ds \sum_k 5^k \dbinom n {2 k + 1}\)
\(\ds \) \(=\) \(\ds \dbinom n 1 + 5 \dbinom n 3 + 5^2 \dbinom n 5 + \cdots\)

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 $\map P n$ be the proposition:

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


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.


$\map P 0$ is the case:

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

Thus $\map P 0$ is seen to hold.


Basis for the Induction

$\map P 1$ is the case:

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

Thus $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

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


So this is the induction hypothesis:

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


from which it is to be shown that:

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


Induction Step

This is the induction step:

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


This needs considerable tedious hard slog to complete it.
In particular: To handle the awkward boundary cases, consider splitting the cases up into odd and even $r$
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
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So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

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

$\blacksquare$


Proof 2

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


Setting $j = 2 k + 1$ for $0 \le k \le \paren {j - 1} / 2$ gives:

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

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:

$\ds 2^n F_n = 2 \sum_{k \text { odd} } \dbinom n k 5^{\paren {k - 1} / 2}$


Historical Note

This result was discovered by Eugène Charles Catalan.


Sources

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