You are correct about the evaluation; so choice (c) does produce an odd number at least for a=2.
Can you convince yourself that its value will be odd for any integer a? Why?
And can you see why the others can all produce an even number, so that they can't be the answer? All you need there is to give an example for each.
1)Sum of 2 odds numbers are even
2)Sum of 2 even numbers are even
**) sum of odd and even is odd
3)If you square different integers you will get some even results and some odd results
4)Every even number can be written as 2 times an integer (so an even*even=even and odd*even=even)
5) odd*odd is odd
Using these 6 rules above we can look at each choice and either reject it or accept it.
a
2: we reject by rule 3
a
2+1: we do not know if a
2 is even or odd, so when we add 1 to a
2 (to get a
2+1)we still don't know if it is even or odd, so we reject it.
2a
2+1: Since 2a
2 is even by rule 4, when we add 1 we get odd. So 2a
2+1 is odd.
3a
2+2: By rule 3, we know that a
2 can be either even or odd. If a
2 is even and you multiply by 3 (by rule 4) you get an even and if you add 1, it becomes odd. So we reject 3a
2+2
4a
2+4: 4a
2 is even (since even *
odd or even = even). 4 is even. The sum of these two even number by rule 2 is even. THIS IS THE ANSWER