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.

Respuesta :

Answer:

nth number of Jerry’s sequence = 2ⁿ⁻¹

Step-by-step 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

             [tex]t_{n}=ar^{n-1}[/tex]

Substituting

             [tex]t_{n}=1\times 2^{n-1}\\\\t_n=2^{n-1}[/tex]

nth number of Jerry’s sequence = 2ⁿ⁻¹