Question

Jerry has thought of a pattern that shows powers of two. Here are the first six numbers of Jerry’s sequence:

1, 2, 4, 8, 16, 32, ….
Write an expression for the nth number of Jerry’s sequence.

Answer 1

nth number of Jerry’s sequence = 2ⁿ⁻¹

Explanation:

This is an example of geometric progression.

First term, a = 1

Common ratio, r = 2

We have expression for n th term of GP as

`t_(n)=ar^(n-1)`

Substituting

`t_(n)=1* 2^(n-1)\\\\t_n=2^(n-1)`

nth number of Jerry’s sequence = 2ⁿ⁻¹