AMC12 2022 A

AMC12 2022 A · Q8

AMC12 2022 A · Q8. It mainly tests Exponents & radicals, Sequences & recursion (algebra).

The infinite product \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots\] evaluates to a real number. What is that number?

无限积 \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots\] 收敛到一个实数。该实数是多少？

(A) \sqrt{10} \sqrt{10}

(B) \sqrt[3]{100} \sqrt[3]{100}

(D) 10 10

(E) 10\sqrt[3]{10} 10\sqrt[3]{10}

Answer

Correct choice: (A)

正确答案：（A）

Solution

We can write $\sqrt[3]{10}$ as $10 ^ \frac{1}{3}$. Similarly, $\sqrt[3]{\sqrt[3]{10}} = (10 ^ \frac{1}{3}) ^ \frac{1}{3} = 10 ^ \frac{1}{3^2}$. By continuing this, we get the form \[10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} \cdots,\] which is \[10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots}.\] Using the formula for an infinite geometric series $S = \frac{a}{1-r}$, we get \[\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}.\] Thus, our answer is $10 ^ \frac{1}{2} = \boxed{\textbf{(A) }\sqrt{10}}$.

我们可以将 $\sqrt[3]{10}$ 写成 $10 ^ \frac{1}{3}$。类似地，$\sqrt[3]{\sqrt[3]{10}} = (10 ^ \frac{1}{3}) ^ \frac{1}{3} = 10 ^ \frac{1}{3^2}$。继续下去，我们得到形式 \[10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} \cdots,\] 即 \[10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots}。\] 使用无限等比级数公式 $S = \frac{a}{1-r}$，有 \[\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}。\] 因此，答案是 $10 ^ \frac{1}{2} = \boxed{\textbf{(A) }\sqrt{10}}$。

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