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$$Express $0.\overline{21}_3$ as a base 10 fraction in reduced form.$$

Mellie  Jun 7, 2015

Best Answer 

 #3
avatar+92806 
+15

You got me thinking Chris :)

 

$$Express $0.\overline{21}_3$ as a base 10 fraction in reduced form.$$

 

$$21_3=2*3+1=7$$

 

So we have here a sum

 

    $$\\0.\overline{21}_3=\frac{7}{3^2}+\frac{7}{3^4}+\frac{7}{3^6}+\frac{7}{3^8}+...\\\\
$This is the infinite sum of a GP$\\\\
a=\frac{7}{9}\qquad r=\frac{1}{9}\\\\
S_{\infty}=\frac{a}{1-r}}\\\\
S_{\infty}=\frac{\frac{7}{9}}{1-\frac{1}{9}}}\\\\
S_{\infty}=\frac{7}{9}\div \frac{8}{9}}\\\\
S_{\infty}=\frac{7}{9}\times \frac{9}{8}}\\\\
S_{\infty}=\frac{7}{8}\\\\$$

Melody  Jun 8, 2015
 #1
avatar+87323 
+15

We have two sums to consider.....

 

2*3^(-1) + 2*3^(-3) + 2*3^(-5)+ ....+2*3^-(2n-1)  =  (2/3) /(1 - 3^(-2)) = (2/3) / (1 - 1/9)  =

(2/3)/(8/9)  = (2/3)*(9/8)  = 18/24 = 3/4   ...... and...... 

 

3^(-2) + 3^(-4) + 3^(-6) + ....+ 3^-(2n)  =  (1/9) / (1 - 3^(-2))  = (1/9)/ ( 1 - 1/9)  = (1/9)/(8/9) =

(1/9) * (9/8)  =  9/72  = 1/8

 

So .......  3/4 + 1/8  =  6/8 + 1/8 =  7/8

 

 

CPhill  Jun 8, 2015
 #2
avatar+92806 
0

Thanks Chris, I had not thought about doing it like that :)

Melody  Jun 8, 2015
 #3
avatar+92806 
+15
Best Answer

You got me thinking Chris :)

 

$$Express $0.\overline{21}_3$ as a base 10 fraction in reduced form.$$

 

$$21_3=2*3+1=7$$

 

So we have here a sum

 

    $$\\0.\overline{21}_3=\frac{7}{3^2}+\frac{7}{3^4}+\frac{7}{3^6}+\frac{7}{3^8}+...\\\\
$This is the infinite sum of a GP$\\\\
a=\frac{7}{9}\qquad r=\frac{1}{9}\\\\
S_{\infty}=\frac{a}{1-r}}\\\\
S_{\infty}=\frac{\frac{7}{9}}{1-\frac{1}{9}}}\\\\
S_{\infty}=\frac{7}{9}\div \frac{8}{9}}\\\\
S_{\infty}=\frac{7}{9}\times \frac{9}{8}}\\\\
S_{\infty}=\frac{7}{8}\\\\$$

Melody  Jun 8, 2015
 #4
avatar+87323 
0

Yeah, Melody......I like yours better........we don't have to split the sums that way.......

 

 

CPhill  Jun 8, 2015

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