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# geometric sequence

Select he first five terms of the geometric sequence an=(2)^n^-1, starting with n=1

I believe the answer is 2,4,8,16,64

I get arithmetic sequence and geometric sequence mixed up. Is there a way to remember how to keep them separate so I do not mix up these sequences.

an=(2)n-1

n  | n-1 | 2n-1
--------------
1  |  0  |  1
2  |  1  |  2
3  |  2  |  4
4  |  3  |  8
5  |  4  | 16

Note that you did not write the sequence as (2)n-1. You wrote it as (2)^n^-1, where n-1 = 1/n1 = 1/n. (The superscript feature of this editor does not express a power raised to a power.) However, if this was the intent, then instead of the above we would actually have:

n | n-1 | 2^n-1
-----------------
1 | 1    | 2
2 | 1/2 | 21/2 = the square root of 2
3 | 1/3 | 21/3 = the cube root of 2
4 | 1/4 | 21/4 = the fourth root of 2
5 | 1/5 | 21/5 = the fifth root of 2

Thank you Kevin!
Hi again Katie;
an=(2)^n^-1
This is 2 to the exponential of n to the exponential of -1.
Let's multiply the exponentials..(n)(-1)=-1n=-n
an=2-n
n=1
a(1)=2-1
a=1/2

n=2
a(2)=2-2
2a=1/4
a=1/8

n=3
a(3)=2-3
3a=1/8
a=1/24

n=4
a(4)=2-4
4a=1/16
a=1/64

n=5
a(5)=2-5
5a=1/32
a=1/160