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!

Using the recurrence relation for $f_n$ we find
\begin{align}f_2 + f_4 + \cdots + f_{2n} = (f_3 - f_1) + (f_5 - f_3) + \cdots + (f_{2n+1} - f_{2n-1}),
\end{align}
which telescopes to $f_{2n+1} - f_1 = f_{2n+1} - 1$.
