AMC12 2017 A

AMC12 2017 A · Q7

AMC12 2017 A · Q7. It mainly tests Sequences & recursion (algebra), Patterns & sequences (misc).

Define a function on the positive integers recursively by $f(1) = 2$, $f(n) = f(n-1) + 1$ if $n$ is even, and $f(n) = f(n-2) + 2$ if $n$ is odd and greater than 1. What is $f(2017)$?

在正整数上定义一个递归函数：$f(1) = 2$，如果 $n$ 是偶数则 $f(n) = f(n-1) + 1$，如果 $n$ 是奇数且大于 1 则 $f(n) = f(n-2) + 2$。$f(2017)$ 是多少？

(A) 2017 2017

(B) 2018 2018

(D) 4035 4035

(E) 4036 4036

Answer

Correct choice: (B)

正确答案：（B）

Solution

It is clear after listing the first few values, $f(1) = 2$, $f(2) = f(1) + 1 = 3$, $f(3) = f(1) + 2 = 4$, $f(4) = f(3) + 1 = 5$, and so on, that $f(n) = n + 1$ for all positive integers $n$. Indeed, the function is uniquely determined by the recursive description, and the function defined by $f(n) = n + 1$ fits the description. Therefore $f(2017) = 2018$.

列出前几个值显然，$f(1) = 2$，$f(2) = f(1) + 1 = 3$，$f(3) = f(1) + 2 = 4$，$f(4) = f(3) + 1 = 5$，依此类推，对所有正整数 $n$ 有 $f(n) = n + 1$。的确，这个函数由递归定义唯一确定，而 $f(n) = n + 1$ 符合该描述。因此 $f(2017) = 2018$。

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