Sum of Sequence of Even Index Fibonacci Numbers

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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:

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

$\blacksquare$


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