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Find the sum (2^1)/(4^1 - 1) + 2^2/(4^2 - 1) + (2^4)/(4^4 - 1) + 2^8/(4^8 - 1) + .....

 Feb 9, 2020
 #1
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∑[(2^2^n) /(4^2^n - 1), n, 0, ∞] = 1

 Feb 9, 2020
 #2
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Find the sum \(\dfrac{2^1}{4^1 - 1} + \dfrac{2^2}{4^2 - 1} + \dfrac{2^4}{4^4 - 1} + \dfrac{2^8}{4^8 - 1} + \dots\)

 

 

\(\begin{array}{|rcll|} \hline && \dfrac{2^1}{4^1 - 1} + \dfrac{2^2}{4^2 - 1} + \dfrac{2^4}{4^4 - 1} + \dfrac{2^8}{4^8 - 1} + \dots \\\\ &=& \dfrac{2^1}{2^2 - 1} + \dfrac{2^2}{2^4 - 1} + \dfrac{2^4}{2^8 - 1} + \dfrac{2^8}{2^{16} - 1} + \dots \\\\ && \boxed{ \text{We substitute }y =2 } \\ &=& \dfrac{y^1}{y^2 - 1} + \dfrac{y^2}{y^4 - 1} + \dfrac{y^4}{y^8 - 1} + \dfrac{y^8}{y^{16} - 1} + \dots \\\\ && \boxed{\dfrac{y^1}{y^2 - 1}\\ =\dfrac{1+y-1}{y^2-1}\\ = \dfrac{1+y}{y^2-1} -\dfrac{1}{y^2-1}\\ = \dfrac{1+y}{(y-1)(y+1)} -\dfrac{1}{y^2-1} \\ = \dfrac{1}{y-1} -\dfrac{1}{y^2-1}\\ \mathbf{\dfrac{y^1}{y^2 - 1}} = \mathbf{\dfrac{1}{y-1} -\dfrac{1}{y^2-1}}} \\ &=& \left(\dfrac{1}{y-1} -\dfrac{1}{y^2-1}\right) \\ & +& \left(\dfrac{1}{y^2-1} -\dfrac{1}{y^4-1}\right)\\ & +& \left(\dfrac{1}{y^4-1} -\dfrac{1}{y^8-1}\right) \\ &+& \left(\dfrac{1}{y^8-1} -\dfrac{1}{y^{16}-1}\right) + \dots \quad | \quad \text{The sum is a telescoping one} \\\\ &=& \dfrac{1}{y-1} \quad | \quad y=2 \\ &=& \dfrac{1}{2-1} \\ &=& \dfrac{1}{1} \\ &=& \mathbf{1} \\ \hline \end{array} \)

 

\(\dfrac{2^1}{4^1 - 1} + \dfrac{2^2}{4^2 - 1} + \dfrac{2^4}{4^4 - 1} + \dfrac{2^8}{4^8 - 1} + \dots = \mathbf{1}\)

 

laugh

 Feb 10, 2020

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