geometric sequence

Question

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.

Answer 1

an=(2)^n-1

 

n  | n-1 | 2^n-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 | 2^1/2 = the square root of 2

3 | 1/3 | 2^1/3 = the cube root of 2

4 | 1/4 | 2^1/4 = the fourth root of 2

5 | 1/5 | 2^1/5 = the fifth root of 2

 

 

 

Answer 2

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

Answer 3

It should be 1,2,4,8,16 because first term is 2⁰ = 1.