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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
+283
+2

$$4^{\frac{1}{4}} \cdot 16^{\frac{1}{16}} \cdot 64^{\frac{1}{64}} \cdot 256^{\frac{1}{256}} \dotsm \\ =4^{1\cdot\frac{1}{4}} \cdot 4^{2\cdot\frac{1}{16}} \cdot 4^{3\cdot\frac{1}{64}} \cdot 4^{4\cdot\frac{1}{256}} \dotsm\\ =4^{\frac{1}{4}+\frac{2}{16}+\frac{3}{64}+\frac{4}{256}+\dotsm}$$

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
+315
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Yep got the same answer as you

wolfiechan  Mar 27, 2021