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

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Answer 1 · 2017-02-18T09:22:57

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

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 · 2017-02-18T09:24:44

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

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#