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The product 

\(4^{1/4} \cdot 16^{1/16} \cdot 64^{1/64} \cdot 256^{1/256} \dotsm\)
can be expressed in the form \(\sqrt[a]{b},\) where a and b are positive integers. Find the smallest possible value of a + b.

 Mar 27, 2021
 #1
avatar+420 
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\(4^{\frac{1}{4}} \cdot 16^{\frac{1}{16}} \cdot 64^{\frac{1}{64}} \cdot 256^{\frac{1}{256}} \dotsm \\ =4^{1\cdot\frac{1}{4}} \cdot 4^{2\cdot\frac{1}{16}} \cdot 4^{3\cdot\frac{1}{64}} \cdot 4^{4\cdot\frac{1}{256}} \dotsm\\ =4^{\frac{1}{4}+\frac{2}{16}+\frac{3}{64}+\frac{4}{256}+\dotsm}\)

First, let's calculate the infinite series in the exponent; let it equal x. Therefore,

\(x =\frac{1}{4}+\frac{2}{16}+\frac{3}{64}+\frac{4}{256}+\dotsm\\ \frac{1}{4}x=\frac{1}{16}+\frac{2}{64}+\frac{3}{256}+\dotsm\\ \text{subtract } \frac{1}{4}x \text{ from }x\text{ to get:}\\ \frac{3}{4}x=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\dotsm \)

That infinite geometric series on the right-hand side is just equal to \(\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}\) using the infinite geometric sequence formula. Then,

\(\frac{3}{4}x=\frac{1}{3}\\ x=\frac{4}{9}\)

Therefore, the original product is equal to:

\(4^{\frac{4}{9}}=\sqrt[9]{4^4}=\sqrt[9]{256}\)

\(a+b = 9+256=\boxed{265}\)

This is the smallest possible value because the expression \(\sqrt[9]{256}\) cannot be reduced.

 Mar 27, 2021
 #2
avatar+336 
0

Yep got the same answer as you

wolfiechan  Mar 27, 2021

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