Prove the sum of the even Fibonacci numbers Let $f_n$ denote the $nth$ Fibonacci number. Prove that 
$f_2\:+\:f_4\:+...+f_{2n}=f_{2n+1}-1$
I am having trouble proving this. I thought to use induction as well as Binet's formula where,
$f_n=\frac{\tau^2-\sigma^2}{\sqrt5}$ where $\tau=\frac{1+\sqrt5}{2}$ and $\sigma=\frac{-1}{\tau}$.
Can someone give me a hand?
Thanks!

This can be proven with induction. If we're saying the sequence starts $1,1,2,\ldots$ then clearly $f_2 = f_3 - 1$.
Now assume $f_2\:+\:f_4\:+\ldots+f_{2n}=f_{2n+1}-1$. Add $f_{2n+2}$ to both sides. Then 
$$f_2\:+\:f_4\:+...+f_{2n} + f_{2n+2}= f_{2n+2}+f_{2n+1}-1$$
And since $f_{2n+2}+f_{2n+1} = f_{2n+3}$, the result follows.
