<s>[INST] Closed form for the sum of even fibonacci numbers? I recently took a look a project euler, and I am trying to think of a smart way to do number 2. I looked at the sequence, and I saw that the question is basically asking for
$$
\sum_{i=1}^n F_{3i}
$$
For whatever n is gives me the nth even under a million. So, I did some fiddling around with this, and I got this to be equivalent to
$$
\sum_{i=0}^{n-1} F_{3i-1}(3^{n-i}-1)
$$
I was wondering, is there even a closed form for something like this? Or am I wasting my time? I couldn't find a closed form online anywhere.
 [/INST] $$F_k=\frac{\left(1+\sqrt{5}\right)^k}{2^k\sqrt{5}}-\frac{\left(1-\sqrt{5}\right)^k}{2^k\sqrt{5}}$$
$$\sum_{k=1}^nF_{3k}=\sum_{k=1}^n\frac{\left(1+\sqrt{5}\right)^{3k}}{2^{3k}\sqrt{5}}-\sum_{k=1}^n\frac{\left(1-\sqrt{5}\right)^{3k}}{2^{3k}\sqrt{5}}$$
$$=\frac{1}{\sqrt5}\sum_{k=1}^n\left(\frac{1+\sqrt{5}}{2}\right)^{3k}-\frac{1}{\sqrt5}\sum_{k=1}^n\left(\frac{1-\sqrt{5}}{2}\right)^{3k}$$
$$\text{ but we have , } x^3+x^6+x^9...x^{3n}=x^3\frac{x^{3n}-1}{x^3-1}$$
$$\text{ so then, }$$
$$=\frac{1}{\sqrt5}\sum_{k=1}^n\left(\frac{1+\sqrt{5}}{2}\right)^{3k}-\frac{1}{\sqrt5}\sum_{k=1}^n\left(\frac{1-\sqrt{5}}{2}\right)^{3k}$$
$$=\frac{1}{\sqrt5}\left(\left(\frac{1+\sqrt{5}}{2}\right)^3\frac{\left(\frac{1+\sqrt{5}}{2}\right)^{3n}-1}{\left(\frac{1+\sqrt{5}}{2}\right)^3-1}-\left(\frac{1-\sqrt{5}}{2}\right)^3\frac{\left(\frac{1-\sqrt{5}}{2}\right)^{3n}-1}{\left(\frac{1-\sqrt{5}}{2}\right)^3-1}\right)=\frac{F_{3n+2}-1}{2}$$
$$=\sum_{k=1}^{n}F_{3k}$$
</s>