Sum of odd-indexed Fibonacci numbers

Statement

The identity $F_1+F_3+F_5+\dots+F_{2n-1}=F_{2n}$ holds. Verify it for $n=4$.

Solution

With $n=4$ the terms are $F_1,F_3,F_5,F_7=1,2,5,13$. The sum: $1+2+5+13=\boxed{21}=F_8.$