Sum of Fibonacci Terms

The even numbers in the Fibonacci number sequence are the numbers $F_{3n}$ for $n \ge 1$ (technically, the sequence should start with $F_0=0$ , but including this will not affect the sum). It is a simple induction to prove that $\sum_{n=1}^N F_{3n} \; = \; \tfrac12\big[F_{3N+2} - 1\big]$ Since $F_{33} = 3524578$ and $F_{36} = 14930352$ , we want to evaluate $\sum_{n=1}^{11} F_{3n} \; = \; \tfrac12(F_{35} - 1) \; = \; \tfrac12(9227465 - 1) \; = \; \boxed{4613732}$

Excel: Cells A1:A33 contain the first 33 Fibonacci numbers. (Use Joe's formula or more common recursive relation.)

The sum we are seeking can be found using the formula: {=SUM(IF(MOD(($A$1:$A$33),2)=0,$A$1:$A$33,0))}

Formula adds the even numbers in the range.

Alternatively, pick out rows with indices that are multiples of 3: {=SUM(IF(MOD(ROW($A$1:$A$33),3)=0,$A$1:$A$33,0)) }

First we note that every third Fibonacci number or $F_{3n}$ , where $n \in \mathbb N$ , is even. We can find the largest $F_{3n} < 4,000,000$ using $F_k = \left[\dfrac {\varphi^k}{\sqrt 5}\right]$ , where $\varphi = \dfrac {1+\sqrt 5}2$ is the golden ratio and $[ \ \cdot\ ]$ denotes the closest integer function. Putting $\left[\dfrac {\varphi^k}{\sqrt 5}\right] < 4000000$ , we get the largest $F_{3n} < 4000000$ is $F_{33} = 3524578$ . Therefore the sum we are looking for is:

$\begin{aligned} S & = \sum_{n=1}^{11} F_{3n} = \sum_{n=1}^{11} \frac {\varphi^{3n}-(-\varphi)^{-3n}}{\sqrt 5} = \frac {\varphi^3\left(\varphi^{33}-1\right)}{\sqrt 5\left(\varphi^3-1\right)} + \frac {\varphi^{-3}\left(1+\varphi^{-33}\right)}{\sqrt 5\left(1+\varphi^{-3}\right)} = \boxed{4613732} \end{aligned}$