Sum of Sequence of Product of Fibonacci Number with Binomial Coefficient/Proof 2
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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
\(\ds \sum_{k \mathop = 1}^n \dbinom n k F_k\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \dbinom n k \paren {\frac {\phi^k - \hat \phi^k} {\sqrt 5} }\) | 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 |
$\blacksquare$