# Sum of Sequence of Even Index Fibonacci Numbers

## Theorem

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

Then:

 $\ds \forall n \ge 1: \,$ $\ds \sum_{j \mathop = 1}^n F_{2 j}$ $=$ $\ds F_2 + F_4 + F_6 + \cdots + F_{2 n}$ $\ds$ $=$ $\ds F_{2 n + 1} - 1$

## Proof

Proof by induction:

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

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

### 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:

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

Then we need to show:

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

### Induction Step

This is our induction step:

 $\ds \sum_{j \mathop = 1}^{k + 1} F_{2 j}$ $=$ $\ds \sum_{j \mathop = 1}^k F_{2 j} + F_{2 \paren {k + 1} }$ $\ds$ $=$ $\ds F_{2 k + 1} - 1 + F_{2 k + 2}$ Induction Hypothesis $\ds$ $=$ $\ds F_{2 k + 3} - 1$ Definition of Fibonacci Number

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

Therefore:

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

$\blacksquare$