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Let S = \(\frac{1}{2^3}\) + \(\frac{1}{4^3}\) + \(\frac{1}{6^3}\)\( \dotsb\) and T = \(\frac{1}{1^3}\) + \(\frac{1}{3^3}\) + \(\frac{1}{5^3}\) + \(\dotsb\). Find S / T. Hint(s): Consider the sum S + T. Can you relate this sum to S or T?

 Jun 24, 2019
edited by Guest  Jun 24, 2019

Best Answer 

 #1
avatar+28130 
+4

Multiply S by 23 so 8S = (2/2)3 + (2/4)3 + (2/6)3 + ... = 1/13 + 1/23 + 1/33 + ... = S + T

 

So:  8S = S + T

 

Divide both sides by T and rearrange. I’ll leave you to do this.

 Jun 24, 2019
 #1
avatar+28130 
+4
Best Answer

Multiply S by 23 so 8S = (2/2)3 + (2/4)3 + (2/6)3 + ... = 1/13 + 1/23 + 1/33 + ... = S + T

 

So:  8S = S + T

 

Divide both sides by T and rearrange. I’ll leave you to do this.

Alan Jun 24, 2019
 #2
avatar+1013 
+3

Nice one 

Nickolas  Jun 24, 2019
 #3
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We have that

 

 \(S + T = \frac{1}{1^3} + \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{4^3} + \dotsb.\)

 

Dividing both sides by \(8 = 2^3,\) we get

 

\(\frac{S + T}{8} = \frac{1}{2^3} + \frac{1}{4^3} + \frac{1}{6^3} + \frac{1}{8^3} + \dots = S.\)

 

Rearranging this equation, we get \(S/T = \boxed{1/7}.\)

 

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 Jun 24, 2019

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