find a sequence for f(n)=4(2)n−1

Answer 1

I suspect a typo, did you really mean
f(n) = 4(2)n -1 or f(n) = 4(2)^(n-1) ?

I suspect the latter.
we could write
f(n) = (2^2)(2^(n-1)
= 2^(2+n-1) = 2^(n+1)

easy to find the terms to be 4, 8, 16, ...