Fibonacci Number as Sum of Binomial Coefficients
Theorem
Let $F_n$ denote the $n$th Fibonacci number.
Then:
\(\ds \forall n \in \Z_{>0}: \, \) | \(\ds F_n\) | \(=\) | \(\ds \sum_{k \mathop = 0}^{\floor {\frac {n - 1} 2} } \dbinom {n - k - 1} k\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \binom {n - 1} 0 + \binom {n - 2} 1 + \binom {n - 3} 2 + \dotsb + \binom {n - j} {j - 1} + \binom {n - j - 1} j\) | where $j = \floor {\dfrac {n - 1} 2}$ |
where:
- $\dbinom a b$ denotes a binomial coefficient
- $\floor x$ denotes the floor function, which is the greatest integer less than or equal to $x$.
Proof
By definition of Fibonacci numbers:
- $F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, \ldots$
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
- $\ds F_n = \sum_{k \mathop = 0}^{\floor {\frac {n - 1} 2} } \dbinom {n - k - 1} k$
Basis for the Induction
$\map P 1$ is the case:
\(\ds F_1\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom 0 0\) | Zero Choose Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom {1 - 0 - 1} 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^{\floor {\frac {1 - 1} 2} } \dbinom {1 - k - 1} k\) |
So $\map P 1$ is seen to hold.
$\map P 2$ is the case:
\(\ds F_2\) | \(=\) | \(\ds 1 + 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom 1 0\) | Binomial Coefficient with Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom {2 - 0 - 1} 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^{\floor {\frac {2 - 1} 2} } \dbinom {2 - k - 1} k\) |
So $\map P 2$ is also seen to hold.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P {k - 1}$ and $\map P k$ are true, where $k > 2$ is an even integer, then it logically follows that $\map P {k + 1}$ and $\map P {k + 2}$ are both true.
So this is our induction hypothesis:
- $\ds F_{k - 1} = \sum_{i \mathop = 0}^{\frac k 2 - 1} \dbinom {k - i - 2} i$
- $\ds F_k = \sum_{i \mathop = 0}^{\frac k 2 - 1} \dbinom {k - i - 1} i$
Then we need to show:
- $\ds F_{k + 1} = \sum_{i \mathop = 0}^{\frac k 2} \dbinom {k - i} i$
- $\ds F_{k + 2} = \sum_{i \mathop = 0}^{\frac k 2} \dbinom {k - i + 1} i$
Induction Step
This is our induction step:
For the first part:
\(\ds \sum_{i \mathop = 0}^{\frac k 2} \dbinom {k - i} i\) | \(=\) | \(\ds \dbinom k 0 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i} i + \dbinom {k - \frac k 2} {\frac k 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i} i + \dbinom {\frac k 2} {\frac k 2}\) | Binomial Coefficient with Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i} i + 1\) | Binomial Coefficient with Self | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \paren { \dbinom {k - i - 1} i + \dbinom {k - i - 1} {i - 1} } + 1\) | Pascal's Rule | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i - 1} i + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i - 1} {i - 1} + 1\) | Summation is Linear | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i - 1} i + \sum_{i \mathop = 0}^{\frac k 2 - 2} \dbinom {k - i - 2} i + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom {k - 2} 0 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i - 1} i + \sum_{i \mathop = 0}^{\frac k 2 - 2} \dbinom {k - i - 2} i + 1\) | Binomial Coefficient with Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom {k - 2} 0 + \sum_{i \mathop = 1}^{\frac k 2 - 1} \dbinom {k - i - 1} i + \sum_{i \mathop = 0}^{\frac k 2 - 2} \dbinom {k - i - 2} i + \dbinom {k - \paren {\frac k 2 - 1} - 2} {\frac k 2 - 1}\) | Binomial Coefficient with Self | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 0}^{\frac k 2 - 1} \dbinom {k - i - 1} i + \sum_{i \mathop = 0}^{\frac k 2 - 1} \dbinom {k - i - 2} i\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_k + F_{k - 1}\) | Induction hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds F_{k + 1}\) | Definition of Fibonacci Number |
For the second part:
\(\ds \sum_{i \mathop = 0}^{\frac k 2} \dbinom {k - i + 1} i\) | \(=\) | \(\ds \dbinom k 0 + \sum_{i \mathop = 1}^{\frac k 2} \dbinom {k - i + 1} i\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \sum_{i \mathop = 1}^{\frac k 2} \dbinom {k - i + 1} i\) | Binomial Coefficient with Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \sum_{i \mathop = 1}^{\frac k 2} \paren {\dbinom {k - i} i + \dbinom {k - i} {i - 1} }\) | Pascal's Rule | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \sum_{i \mathop = 1}^{\frac k 2} \dbinom {k - i} i + \sum_{i \mathop = 1}^{\frac k 2} \dbinom {k - i} {i - 1}\) | Summation is Linear | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \sum_{i \mathop = 1}^{\frac k 2} \dbinom {k - i} i + \sum_{i \mathop = 0}^{\frac k 2 - 1} \dbinom {k - i - 1} i\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom {k - 2} 0 + \sum_{i \mathop = 1}^{\frac k 2} \dbinom {k - i} i + \sum_{i \mathop = 0}^{\frac k 2 - 1} \dbinom {k - i - 1} i\) | Binomial Coefficient with Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 0}^{\frac k 2} \dbinom {k - i} i + \sum_{i \mathop = 0}^{\frac k 2 - 1} \dbinom {k - i - 1} i\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_{k + 1} + F_k\) | Induction hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds F_{k + 2}\) | Definition of Fibonacci Number |
So $\map P {k - 1} \land \map P k \implies \map P {k + 1} \land \map P {k + 2}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \in \Z_{>0}: F_n = \sum_{k \mathop = 0}^{\floor {\frac {n - 1} 2} } \dbinom {n - k - 1} k$
$\blacksquare$
Examples
Fibonacci Number $F_{12}$
\(\ds F_{12}\) | \(=\) | \(\ds \binom {11} 0 + \binom {10} 1 + \binom 9 2 + \binom 8 3 + \binom 7 4 + \binom 6 5\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + 10 + 36 + 56 + 35 + 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 144\) |
Historical Note
This result was discovered by François Édouard Anatole Lucas.
Sources
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {3-1}$ Permutations and Combinations: Exercise $13$
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $5$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $5$
- but note the misprint in the statement of the result.