Sum of Sequence of Even Index Fibonacci Numbers

Theorem

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

Then:

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

That is:

$F_2 + F_4 + F_6 + \cdots + F_{2n} = F_{2n+1} - 1$

Proof

Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:

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

Basis for the Induction

• $P(1)$ 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 $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

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

Then we need to show:

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

Induction Step

This is our induction step:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{j=1}^{k+1} F_{2j}\)	\(=\)	\(\displaystyle \sum_{j=1}^k F_{2j} + F_{2\left({k + 1}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle F_{2k + 1} - 1 + F_{2k + 2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by induction hypothesis
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle F_{2k + 3} - 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of Fibonacci numbers

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

Therefore:

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

$\blacksquare$

Sources

• George E. Andrews: Number Theory (1971): $\S 1.1$: Exercise $9$