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

$\blacksquare$