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

Feb 18, 2017

$\frac{5}{3} , \frac{17}{9} , \frac{53}{27} , \frac{161}{81} , \frac{485}{243}$

Explanation:

First term would be for n=1, thus ${a}_{n} = 2 - \frac{1}{3} ^ 1 = \frac{5}{3}$

2nd term would be for n=2.thus ${a}_{2} = 2 - \frac{1}{3} ^ 2 = 2 - \frac{1}{9} = \frac{17}{9}$

3rd term would be for n=3, thus ${a}_{3} = 2 - \frac{1}{3} ^ 3 = 2 - \frac{1}{27} = \frac{53}{27}$

4th term would be for n=4, thus ${a}_{4} = 2 - \frac{1}{3} ^ 4 = 2 - \frac{1}{81} = \frac{161}{81}$

5th term would be for n=5, thus ${a}_{5} = 2 - \frac{1}{3} ^ 5 = 2 - \frac{1}{243} = \frac{485}{243}$

Feb 18, 2017

$\frac{5}{3} , \frac{17}{9} , \frac{53}{27} , \frac{161}{81} , \frac{485}{243}$

Explanation:

Let ${c}_{n} = {3}^{n}$

Then ${c}_{1} , {c}_{2} , \ldots$ starts:

$3 , 9 , 27 , 81 , 243$

Let ${b}_{n} = 2 \cdot {3}^{n} - 1 = 2 {c}_{n} - 1$

Then ${b}_{1} , {b}_{2} , \ldots$ starts:

$5 , 17 , 53 , 161 , 485$

Then:

${a}_{n} = 2 - \frac{1}{3} ^ n = \frac{2 \cdot {3}^{n} - 1}{3} ^ n = {b}_{n} / {c}_{n}$

So ${a}_{1} , {a}_{2} , \ldots$ starts:

$\frac{5}{3} , \frac{17}{9} , \frac{53}{27} , \frac{161}{81} , \frac{485}{243}$