F(n)=2+4+6+...+2n is the same as F(n)=2(1+2+3+...+n)
The sum of the numbers from 1 to n for positive n = n*(n+1)/2.
So by substitution, F(n) = 2*n*(n+1) / 2 = n*(n+1), since the 2's cancel. This appears to be the first option in your list.
- Steve J