Sum of Sequence of Even Index Fibonacci Numbers

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

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

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

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 $$k \ge 2$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:
 * $$\sum_{j=1}^k F_{2j} = F_{2k + 1} - 1$$.

Then we need to show:
 * $$\sum_{j=1}^{k+1} F_{2j} = F_{2k + 3} - 1$$.

Induction Step
This is our induction step:

$$ $$ $$

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

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