You have to use the concept of odd-even here. Let's test some cases.
1) Odd case ( 2 odd cases)
- G+2: odd + even is always odd
- 3G + 1: (odd * odd) + odd is even
- 4G - 1 : (even * odd) - odd is odd
- 2(2G - 1): even * anything is even
2) Even case (2 cases)
- G+2: even + even is always even
- 3G + 1: (odd * even) + odd is odd
- 4G - 1 : (even * even) - odd is odd
- 2(2G - 1): even * anything is even

We see that when we choose odd or even numbers, we always get 2 cases out of 4 which are odd. So, C is the answer.