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Find the sum \(\frac{1}{4}+\frac{3}{16}+\frac{5}{64}+\frac{7}{256}+...+\frac{(2n-1)}{4^n}+...\)

mathtoo  Sep 14, 2018

Best Answer 

 #2
avatar+27128 
+2

As follows:

 

Alan  Sep 15, 2018
 #1
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a(n) =( 2^(-2 n) (2 n - 1))

Let n = 100

∑[( 2^(-2 n) (2 n - 1)), n, 1, 100]  = 5/9

Guest Sep 14, 2018
 #2
avatar+27128 
+2
Best Answer

As follows:

 

Alan  Sep 15, 2018
 #3
avatar+90968 
+1

Nice, Alan  !!!

 

 

cool cool cool

CPhill  Sep 15, 2018
 #4
avatar+27128 
+3

Thanks Chris!

Alan  Sep 16, 2018

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