Here are the sums of the first few odd Fibonacci numbers:
$$\begin{align} &f_1=1 \\ &f_1 + f_3 = 3 \\ &f_1 + f_3 + f_5 = 8 \\ &f_1 + f_3 +f_5 + f_7 = 21 \end{align}$$
And here are the first few even Fibonacci numbers:
$$\begin{align} f_2 &=1 \\ f_4 &= 3 \\ f_6 &= 8 \\ f_8 &= 21 \end{align}$$
This seems to suggest that:
$$ \sum^n_{k=1} f_{2k-1} = f_{2n} $$
To prove the above statment for all cases, we can use proof by induction. We already have some base cases. Now we need the induction step:
$$\begin{align} \sum^{n+1}_{k=1} f_{2k-1} &= \sum^n_{k=1} f_{2k-1} + f_{2(n+1)-1} \\ &= f_{2n} + f_{2n+1} \end{align}$$
By definition \(f_{2n-1} = f_{2n} + f_{2n+1}\). This means:
$$\begin{align} \sum^{n+1}_{k=1} f_{2k-1} &= f_{2n-1} \\ &= f_{2(n+1)-1} \end{align}$$
This shows that our statement works for \(n+1\) as well.