Project Euler Q2 - Sum of even Fibonacci numbers
Daniel's solution to another question applies here equally.

On another question, I showed that the sum is equal to

$$
\frac{F_{3n+2} - 1}{2}
$$

where \$3n\$ is the largest multiple of \$3\$ not to exceed the maximum.

If you wish to use this for asymptotic improvements, note that since the Fibonacci sequence has a closed-form solution, you can use it to find an approximate bound for \$3n\$, and then use fast exact mechanisms to find the exact value of \$F_{3n+2}\$.

$$
F_{3n} \approx \phi^{3n}/\sqrt 5 \\
3n \approx \log_\phi \left(F_{3n} \sqrt 5 \right)
$$

Finding a lower bound for \$n\$ thus takes very little time.

Personally I'd stick with the simple, slower solution, but it's interesting to note how far you can get with a bit of mathematical analysis.