Here are the sums of the first few even Fibonacci numbers:
$$\begin{align} &f_2=2 \\ &f_2 + f_4 = 4 \\ &f_2 + f_4 + f_6 = 12 \\ &f_2 + f_4 +f_6 + f_8 = 33 \end{align}$$
And here are the first few odd Fibonacci numbers:
$$\begin{align} f_1 &=1 \\ f_3 &= 2 \\ f_5 &= 5 \\ f_7 &= 13 \\ f_9 &= 34 \end{align}$$
This seems to suggest that:
$$ \sum^n_{k=1} f_{2k} = f_{2n+1} - 1 $$
The sum of the first odd \(n\) Fibonacci numbers (derived here) is:
$$\sum^n_{k=1} f_{2k-1} = f_{2n}$$
The sum of the first \(n\) Fibonacci numbers (derived here) is:
$$\sum^n_{k=1} f_k = f_{n+2} - 1$$
The sum of the first \(n\) even and the first \(n\) odd Fibonacci numbers can be written as:
$$\begin{align}\sum^{n}_{k=1} f_{2k-1} + \sum^{n}_{k=1} f_{2k} &= \sum^{2n}_{k=1} f_k \\ f_{2n} + \sum^{n}_{k=1} f_{2k} &= f_{2n+2} - 1 \end{align}$$
Rearranging:
$$\begin{align} \sum^{n}_{k=1} f_{2k} &= f_{2n+2} - f_{2n} - 1 \\ \sum^{n}_{k=1} f_{2k} &= f_{2n+1} - 1\end{align}$$