Most questions related to partial sums of binomials I have come across was in MO. After your comment I checked for differences between MO and MSE. The question here, is the problem from my research, so it fits (though the solution appeared simple).

Thanks to your answer, I managed to find the proof. Please feel free, to include it in your post. Using $\binom{4g+1}{2k-1}=\binom{4g}{2k-1}+\binom{4g}{2k-2}$: $\sum_{k=1}^g \binom{4g+1}{2k-1}=\sum_{k=1}^g \binom{4g}{2k-1}+\sum_{k=0}^{g-1} \binom{4g}{2k}=\sum_{k=0}^{2g-1} \binom{4g}{k}=\dfrac{1}{2}(2^{4g}-\binom{4g}{2g})$