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# HELP HELP HELP!

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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

#1
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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
+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.

Alan Jun 24, 2019
#2
+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 $$S/T = \boxed{1/7}.$$            Jun 24, 2019