# 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:

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