# Sum of Sequence of Product of Fibonacci Number with Binomial Coefficient

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## Contents

## 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:

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

## Sources

- 1971: George E. Andrews:
*Number Theory*... (previous) ... (next): $\text {3-1}$ Permutations and Combinations: Exercise $14$