Anonymous

The nth term is 2^(n - 1). Look: 2^(n - 1) yields the list of numbers you provided. Let n = 1 2^(1 - 1) 2^(0) = 1 Let n = 2 2^(2 - 1) 2^(1) = 2 Let n = 3 2^(3 - 1) 2^(2) = 4 Let n = 4 2^(4 - 1) 2^(3) = 8

Anonymous

This is a series. A(n)= 2^(n-1). Khanacademy.org has a great tutorial on series equations.

Anonymous

Note that there are infinity possible answers, for example: nth term = 2^(n - 1) nth term = (1/6)n^3 - (1/2)n^2 + (4/3)n There are infinity other possibilities as well.

Anonymous

For the nth term this sequence is a(n) = 2^(n - 1).

Anonymous

1, 2, 4, 8, … an = 2^(n − 1)