# Sum of Sequence of Even Index Fibonacci Numbers

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## Contents

## Theorem

Let $F_k$ be the $k$th Fibonacci number.

Then:

\(\, \displaystyle \forall n \ge 1: \, \) | \(\displaystyle \sum_{j \mathop = 1}^n F_{2 j}\) | \(=\) | \(\displaystyle F_2 + F_4 + F_6 + \cdots + F_{2 n}\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle F_{2 n + 1} - 1\) |

## Proof

Proof by induction:

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

- $\displaystyle \sum_{j \mathop = 1}^n F_{2 j} = F_{2 n + 1} - 1$

### Basis for the Induction

$\map P 1$ is the case $F_2 = 1 = F_3 - 1$, which holds from the definition of Fibonacci numbers.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

- $\displaystyle \sum_{j \mathop = 1}^k F_{2 j} = F_{2 k + 1} - 1$

Then we need to show:

- $\displaystyle \sum_{j \mathop = 1}^{k + 1} F_{2 j} = F_{2 k + 3} - 1$

### Induction Step

This is our induction step:

\(\displaystyle \sum_{j \mathop = 1}^{k + 1} F_{2 j}\) | \(=\) | \(\displaystyle \sum_{j \mathop = 1}^k F_{2 j} + F_{2 \paren {k + 1} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle F_{2 k + 1} - 1 + F_{2 k + 2}\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle F_{2 k + 3} - 1\) | Definition of Fibonacci Number |

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

Therefore:

- $\displaystyle \forall n \ge 1: \sum_{j \mathop = 1}^n F_{2 j} = F_{2 n + 1} - 1$

$\blacksquare$

## Sources

- 1971: George E. Andrews:
*Number Theory*... (previous) ... (next): $\text {1-1}$ Principle of Mathematical Induction: Exercise $9$ - 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $5$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $5$