The second problem is: By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

Lets again start with a python script solving this:

from itertools import takewhile, filterfalse
def fib():
    lastfib = 1
    fib = 0
    tmp = 0
    while True:
        yield fib
        tmp = fib
        fib = lastfib + fib
        lastfib = tmp
print(sum(filterfalse(lambda x: x % 2, takewhile(lambda x: x<4000000, fib()))))

Analytically, I have not gotten far. First thing I notice is that $\textrm{fib}\left(n\right)$ is only even if $n \% 3 = 0$, else it is odd.

Think in binary or follow my proof if you do not believe me:

First step, $\textrm{fib}\left(3 \cdot n\right)$ is always even:

Basis:

$$ n = 0 $$

$$ \textrm{fib}\left(0\right) = 0 $$

Assumption: Holds to $n$, therefore $\textrm{fib}\left(n\right)$ is even

Induction: $n \rightarrow n + 1$

$$ \begin{align} \textrm{fib}\left(3\cdot \left(n+1\right) \right) &= \textrm{fib}\left(3\cdot n + 3\right) \\ &= \textrm{fib}\left(3 \cdot n + 2\right) + \textrm{fib}\left(3 \cdot n + 1\right) \\ &= 2 \cdot \textrm{fib}\left(2 \cdot n + 1 \right) + \textrm{fib}\left(3 \cdot n \right) \end{align} $$

$2 \cdot x, x \in N$ is always even, $\textrm{fib}\left(3 \cdot n\right)$ is even and the sum of even numbers is even as well. Ergo $\textrm{fib}\left(3 \cdot n\right)$ is always even!

But we still have to proof that $\textrm{fib}\left(3 \cdot n\right)$ are the ONLY even numbers!

Lets take a look at $\textrm{fib}\left(3 \cdot n + 2\right)$ which should be always odd:

Basis:

$$ n = 0 \\ \textrm{fib}\left(2\right) = 1 $$

Assumption: Holds to $n$, therefore $\textrm{fib}\left(n + 2\right)$ is odd

Induction: $n \rightarrow n + 1$

$$ \begin{align} \textrm{fib}\left(3 \cdot \left(n+1\right) + 2\right) &= \textrm{fib}\left(3 \cdot n + 5\right) \\ &= \textrm{fib}\left(3n + 4\right) + \textrm{fib}\left(3 \cdot \left(n + 1\right)\right) \\ &= 2 \cdot \textrm{fib}\left(3 \cdot \left(n + 1\right)\right) + \textrm{fib}\left(3 \cdot n + 2\right) \end{align} $$

A sum of a odd and an even number is always odd hence $\textrm{fib}\left(3 \cdot n+2\right)$ is odd

What is left? $\textrm{fib}\left(3 \cdot n + 1\right)$!

Basis:

$$ n = 0 \\ \textrm{fib}\left(1\right) = 1 $$

Assumption: Holds to $n$, therefore $\textrm{fib}\left(n + 1\right)$ is odd

Induction: $n \rightarrow n + 1$

$$ \begin{align} \textrm{fib}\left(3 \cdot \left(n+1\right) + 1\right) &= \textrm{fib}\left(3 \cdot n + 4\right) \\ &= \textrm{fib}\left(3 \cdot n + 2 \right) + \textrm{fib}\left(3 \cdot \left(n + 1\right)\right) \\ \end{align} $$

And again, odd + even = odd!

So the only even numbers in the Fibonacci sequence are the ones with $n \% 3 = 0$!

But that's it, no idea how to battle this problem further.

Update

Well, after some thinking, I guess I found an analytic solution.

Lets have a look which Fibonacci numbers are in the sum of all even Fibonacci numbers: (remember, every third is even)

$$ \begin{array}{c|ccccccccccccc} N & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} $$

But these could also be written like this:

$$ \begin{array}{c|ccccccccccccc} N & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ 3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ \end{array} $$

This means that $$ \sum_{i=0}^N f_{\textrm{even}} = \sum_{i=0}^N f_{\textrm{odd}} $$ and therefore we can also write it as: $$ \sum_{i=0}^N f_{\textrm{even}} = \frac{1}{2} \sum_{i=0}^N f_i $$

We now only need to know what the sum of all Fibonacci numbers to some n is: $$ \sum_{i=0}^N f_i = f_{N+2} - 1 $$

Wherefore, the solution is: $$ \sum_{i=0}^N f_{\textrm{even}} = \frac{f_{n+2} - 1}{2} $$

with $n$ chosen to be the biggest value fulfilling: $f_n \leq 4000000$