Sum of Sequence of Product of Fibonacci Number with Binomial Coefficient
Jump to navigation
Jump to search
Theorem
Let $F_n$ denote the $n$th Fibonacci number.
Then:
\(\ds \forall n \in \Z_{>0}: \, \) | \(\ds F_{2 n}\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \dbinom n k F_k\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom n 1 F_1 + \dbinom n 2 F_2 + \dbinom n 3 F_3 + \dotsb + \dbinom n {n - 1} F_{n - 1} + \dbinom n n F_n\) |
where $\dbinom n k$ 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:
- $F_{2 n} = \ds \sum_{k \mathop = 1}^n \dbinom n k F_k$
$\map P 0$ is the case:
\(\ds F_0\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^0 \dbinom 0 k F_k\) | vacuously |
Thus $\map P 0$ is seen to hold.
Basis for the Induction
$\map P 1$ is the case:
\(\ds F_2\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom 1 1 F_1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^1 \dbinom 1 k F_k\) |
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 $\map P m$ is true, where $m \ge 1$, then it logically follows that $\map P {m + 1}$ is true.
So this is the induction hypothesis:
- $F_{2 m} = \ds \sum_{k \mathop = 1}^m \dbinom m k F_k$
from which it is to be shown that:
- $F_{2 \paren {m + 1} } = \ds \sum_{k \mathop = 1}^{m + 1} \dbinom {m + 1} k F_k$
Induction Step
This is the induction step:
\(\ds \sum_{k \mathop = 1}^{m + 1} \dbinom {m + 1} k F_k\) | \(=\) | \(\ds \sum_{k \mathop = 1}^m \dbinom {m + 1} k F_k + \dbinom {m + 1} {m + 1} F_{m + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^m \paren {\dbinom m {k - 1} + \dbinom m k} F_k + F_{m + 1}\) | Pascal's Rule | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^m \dbinom m {k - 1} F_k + F_{2 m} + F_{m + 1}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^m \dbinom m {k - 1} \paren {F_{k - 1} + F_{k - 2} } + F_{2 m} + F_{m + 1}\) | Definition of Fibonacci Numbers |
This needs considerable tedious hard slog to complete it. In particular: Can't see where this is going -- anyone care to have a go at this? 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.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \Z_{\ge 0}: F_{2 n} = \ds \sum_{k \mathop = 1}^n \dbinom n k F_k$
Proof 2
\(\ds \sum_{k \mathop = 1}^n \dbinom n k F_k\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \dbinom n k \paren {\frac {\phi^k - \hat \phi^k} {\sqrt 5} }\) | Euler-Binet Formula | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\sqrt 5} \paren {\sum_{k \mathop = 1}^n \dbinom n k \phi^k - \sum_{k \mathop = 1}^n \dbinom n k \hat \phi^k}\) | Summation is Linear | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\sqrt 5} \paren {\sum_{k \mathop = 0}^n \dbinom n k \phi^k 1^{n - k} - \sum_{k \mathop = 0}^n \dbinom n k \hat \phi^k 1^{n - k} }\) | Binomial Coefficient with Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\sqrt 5} \paren {\paren {1 + \phi}^n - \paren {1 + \hat \phi}^n}\) | Binomial Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\sqrt 5} \paren {\paren {\phi^2}^n - \paren {\hat \phi^2}^n}\) | Golden Mean as Root of Quadratic $x^2 = x + 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds F_{2 n}\) | Euler-Binet Formula |
$\blacksquare$
Sources
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {3-1}$ Permutations and Combinations: Exercise $14$