Short answer: I don't know, but that doesn't mean I can't give you something that you can decide for yourself. y = 4*2^(2n - 2) is the pattern. Go for broke. Try n = 4. You should get 256. Let's try it. y = 4 * 2^(2*4 - 2) y = 4 * 2^(8 - 2) y = 4 * 2^6 y = 4 * 64 y = 256 yup it works.
The other end is just as important. Suppose n = 1 Then y = 4 * 2^(2*1 - 2) = 4 * 2^0 = 4*1 = 4 Both work.
If this formula is correct, we can abbreviate it to make your task easier. y = 4 * 2^(2n - 2) y = 2^2 * 2^(2n - 2) y = 2^(2n - 2 + 2) y = 2^(2n) Now try the two end points again.
n = 4 y = 2^(2*4) y = 2^8 y = 256
n = 1 y = 2^(2*1) y = 2^2 y = 4 which again checks.
so y = 2^(2n) I think is an exponential function. Sorry my explanation is so long.
