We have a formula for finding the nth term, and we have an actual value for n=1. We can find f(3) by substituting n=3 in the general function: f(3)=f(1)+3=7+3=10, and we can find f(5)=f(3)+3=10+3=13, so f(7)=16 and f(9)=19. The first 5 "terms" are 7, 10, 13, 16, 19, i.e., f(1), f(3), f(5), f(7), f(9). We have no formula for f(2), f(4), etc. can we create one? Let's assume a linear function f(x)=ax+b. When x=1, f(1)=7, so a+b=7. When x=3, 3a+b=10. f(3)-f(1)=3, so 2a=3 and a=3/2. Therefore b=7-3/2=11/2. Therefore f(x)=3x/2+11/2 or 1/2(3x+11). To prove we have the right equation, let's put in x=5: f(x)=13. Checks out! Therefore the first five terms, starting at zero are: 11/2, 7, 17/2, 10, 23/2.