Sum of even-indexed Fibonacci numbers

Statement

The identity $F_2+F_4+F_6+\dots+F_{2n}=F_{2n+1}-1$ holds. Verify it for $n=4$.

Solution

With $n=4$ the terms are $F_2,F_4,F_6,F_8=1,3,8,21$. The sum: $1+3+8+21=\boxed{33}=F_9-1=34-1.$