2004 AMC 12A 第 17 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

17.

ff 是满足以下性质的函数:

(i) f(1)=1f(1) = 1, 且

(ii) 对任意正整数 nn,都有 f(2n)=nf(n)f(2n) = n \cdot f(n)

f(2100)f(2^{100}) 的值是多少?

Let ff be a function with the following properties:

(i) f(1)=1,f(1) = 1, and

(ii) f(2n)=nf(n)f(2n) = n \cdot f(n) for any positive integer n.n.

What is the value of f(2100)?f(2^{100})?

11

2992^{99}

21002^{100}

249502^{4950}

299992^{9999}

答案:D
知识点:函数方程递推求和
难度评级:1720
解答:

f(2n)=nf(n)f(2n) = n \cdot f(n) 中令 n=2kn = 2^{k} 得到 f(2k+1)=2kf(2k)f(2^{k+1}) = 2^{k} \cdot f(2^{k})

f(21)=f(2)=1f(1)=20f(2^1) = f(2) = 1 \cdot f(1) = 2^0, 展开,指数累加为 f(2n)=20+1+2++(n1)=2n(n1)/2. \begin{aligned} f(2^n) &= 2^{0 + 1 + 2 + \cdots + (n-1)} \\ &= 2^{n(n-1)/2}. \end{aligned}

因此 f(2100)=210099/2=24950f(2^{100}) = 2^{100 \cdot 99 / 2} = 2^{4950}

所以正确答案是 D

Applying f(2n)=nf(n)f(2n) = n \cdot f(n) with n=2k,n = 2^{k}, we get f(2k+1)=2kf(2k).f(2^{k+1}) = 2^{k} \cdot f(2^{k}).

Unwinding from f(21)=f(2)=1f(1)=20,f(2^1) = f(2) = 1 \cdot f(1) = 2^0, the exponents accumulate: f(2n)=20+1+2++(n1)=2n(n1)/2. \begin{aligned} f(2^n) &= 2^{0 + 1 + 2 + \cdots + (n-1)} \\ &= 2^{n(n-1)/2}. \end{aligned}

Therefore f(2100)=210099/2=24950.f(2^{100}) = 2^{100 \cdot 99 / 2} = 2^{4950}.

Thus, the correct answer is D.

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