Question

How do you write the first five terms of the sequence `a_n=2-1/3^n`?

Answer 1

`5/3, 17/9, 53/27, 161/81, 485/243`

Explanation:

First term would be for n=1, thus `a_n= 2- 1/3^1 = 5/3`

2nd term would be for n=2.thus `a_2= 2-1/3^2= 2-1/9= 17/9`

3rd term would be for n=3, thus `a_3= 2- 1/3^3= 2-1/27= 53/27`

4th term would be for n=4, thus `a_4= 2-1/3^4= 2-1/81= 161/81`

5th term would be for n=5, thus `a_5= 2- 1/3^5= 2- 1/243=485/243`

Answer 2

`5/3, 17/9, 53/27, 161/81, 485/243`

Explanation:

Let `c_n = 3^n`

Then `c_1, c_2,...` starts:

`3, 9, 27, 81, 243`

Let `b_n = 2*3^n-1 = 2c_n-1`

Then `b_1, b_2,...` starts:

`5, 17, 53, 161, 485`

Then:

`a_n = 2-1/3^n = (2*3^n-1)/3^n = b_n/c_n`

So `a_1, a_2,...` starts:

`5/3, 17/9, 53/27, 161/81, 485/243`