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Find \(1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{36} + \dotsb,\) where we alternately multiply by 1/2 and 1/3 to get the next term.

 Feb 18, 2019
 #1
avatar+363 
-3

Are you saying that you multiply this finally number in the set of digits by 1/2 or 1/3 ? or multiply the previous fraction by those to fractions.

 Feb 18, 2019
edited by HiylinLink  Feb 18, 2019
edited by HiylinLink  Feb 18, 2019
 #2
avatar+101871 
+3

We have

 

1 + 1/2 + 1/6 + 1/12 + 1/36 + 1/72 +  ......

 

Notice that we can split this up like so

 

1 + 1/6 + 1/36 + ......+

 

1/2 + 1/12 +  1/72 +  ......+

 

The sum of the first series is            1 / (1 - 1/6)  =   1 / (5/6) =   6/5

 

The sum of the second series is  (1/2) / ( 1 - 1/6)  =  (1/2) /(5./6) = 6/10 =  3/5

 

So....the sum of the series is   6/5 + 3/5  =   9/5

 

 

cool cool cool

 Feb 18, 2019
 #3
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+1

Here is another way:

∑ [ 2^(-n) * 3^(2 - n), n, 1,∞] = 9/5

 Feb 18, 2019
 #4
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+1

You can also use infinite series to sum them up:

Sum =F / [1 - R], where F =First term = 1.5, R =Common ratio = 1/6

Sum =1.5/ [1 - 1/6]

Sum =1.5 / (5/6)

Sum =1.5 x 6/5

Sum = 9 / 5.

 Feb 18, 2019

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