Sum of Sequence of Product of Fibonacci Number with Binomial Coefficient

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Theorem

Let $F_n$ denote the $n$th Fibonacci number.

Then:

\(\, \displaystyle \forall n \in \Z_{>0}: \, \) \(\displaystyle F_{2 n}\) \(=\) \(\displaystyle \sum_{k \mathop = 1}^n \dbinom n k F_k\)
\(\displaystyle \) \(=\) \(\displaystyle \dbinom n 1 F_1 + \dbinom n 2 F_2 + \dbinom n 3 F_3 + \dotsb + \dbinom n {n - 1} F_{n - 1} + \dbinom n n F_n\)

where $\dbinom n k$ denotes a binomial coefficient.


Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$F_{2 n} = \displaystyle \sum_{k \mathop = 1}^n \dbinom n k F_k$


$\map P 0$ is the case:

\(\displaystyle F_0\) \(=\) \(\displaystyle 0\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^0 \dbinom 0 k F_k\) vacuously

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


Basis for the Induction

$\map P 1$ is the case:

\(\displaystyle F_2\) \(=\) \(\displaystyle 1\)
\(\displaystyle \) \(=\) \(\displaystyle \dbinom 1 1 F_1\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^1 \dbinom 1 k F_k\)

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:

$F_{2 m} = \displaystyle \sum_{k \mathop = 1}^m \dbinom m k F_k$


from which it is to be shown that:

$F_{2 \paren {m + 1} } = \displaystyle \sum_{k \mathop = 1}^{m + 1} \dbinom {m + 1} k F_k$


Induction Step

This is the induction step:

\(\displaystyle \sum_{k \mathop = 1}^{m + 1} \dbinom {m + 1} k F_k\) \(=\) \(\displaystyle \sum_{k \mathop = 1}^m \dbinom {m + 1} k F_k + \dbinom {m + 1} {m + 1} F_{m + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^m \paren {\dbinom m {k - 1} + \dbinom m k} F_k + F_{m + 1}\) Pascal's Rule
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^m \dbinom m {k - 1} F_k + F_{2 m} + F_{m + 1}\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^m \dbinom m {k - 1} \paren {F_{k - 1} + F_{k - 2} } + F_{2 m} + F_{m + 1}\) Definition of Fibonacci Numbers



So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{\ge 0}: F_{2 n} = \displaystyle \sum_{k \mathop = 1}^n \dbinom n k F_k$


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