AMC12 2004 A

AMC12 2004 A · Q17

AMC12 2004 A · Q17. It mainly tests Functions basics, Sequences & recursion (algebra).

Let $f$ be a function with the following properties: (i) $f(1) = 1$, and (ii) $f(2n) = n \cdot f(n)$ for any positive integer $n$. What is the value of $f(2^{100})$?

设 $f$ 是一个具有以下性质的函数： (i) $f(1) = 1$，且 (ii) 对任意正整数 $n$，$f(2n) = n \cdot f(n)$。 $ f(2^{100})$ 的值是多少？

(A) 1 1

(B) $2^{99}$ $2^{99}$

(D) $2^{4950}$ $2^{4950}$

(E) $2^{9999}$ $2^{9999}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

From (ii), note that \begin{alignat*}{8} f(2) &= 1\cdot f(1) &&= 1, \\ f\left(2^2\right) &= 2\cdot f(2) &&= 2, \\ f\left(2^3\right) &= 2^2\cdot f\left(2^2\right) &&= 2^{2+1}, \\ f\left(2^4\right) &= 2^3\cdot f\left(2^3\right) &&= 2^{3+2+1}, \end{alignat*} and so on. In general, we have \[f\left(2^n\right)=2^{(n-1)+(n-2)+(n-3)+\cdots+3+2+1}\] for any positive integer $n.$ Therefore, the answer is \begin{align*} f\left(2^{100}\right)&=2^{99+98+97+\cdots+3+2+1} \\ &=2^{99\cdot100/2} \\ &= \boxed{\textbf {(D)}\ 2^{4950}}. \end{align*}

由 (ii) 可得 \begin{alignat*}{8} f(2) &= 1\cdot f(1) &&= 1, \\ f\left(2^2\right) &= 2\cdot f(2) &&= 2, \\ f\left(2^3\right) &= 2^2\cdot f\left(2^2\right) &&= 2^{2+1}, \\ f\left(2^4\right) &= 2^3\cdot f\left(2^3\right) &&= 2^{3+2+1}, \end{alignat*} 依此类推。一般地，对任意正整数 $n$，有 \[f\left(2^n\right)=2^{(n-1)+(n-2)+(n-3)+\cdots+3+2+1}\] 因此答案为 \begin{align*} f\left(2^{100}\right)&=2^{99+98+97+\cdots+3+2+1} \\ &=2^{99\cdot100/2} \\ &= \boxed{\textbf {(D)}\ 2^{4950}}. \end{align*}

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