Sum of even indexed Fibonacci numbers
From specialfunctionswiki
Theorem
The following formula holds: $$\displaystyle\sum_{k=1}^n F_{2k} = F_{2n+1}-1,$$ where $F_{2k}$ denotes the $2k$th Fibonacci number.
The following formula holds: $$\displaystyle\sum_{k=1}^n F_{2k} = F_{2n+1}-1,$$ where $F_{2k}$ denotes the $2k$th Fibonacci number.
