AMC12 2020 B

AMC12 2020 B · Q6

AMC12 2020 B · Q6. It mainly tests Polynomials, Perfect squares & cubes.

For all integers $n \geq 9$, the value of $\frac{(n + 2)! - (n + 1)!}{n!}$ is always which of the following?

对于所有整数 $n \geq 9$，表达式 $\frac{(n + 2)! - (n + 1)!}{n!}$ 的值总是下列哪一项？

(A) a multiple of 4 4的倍数

(B) a multiple of 10 10的倍数

(D) a perfect square 完全平方数

(E) a perfect cube 完全立方数

Answer

Correct choice: (D)

正确答案：（D）

Solution

For all nonnegative integers $n$, $$\frac{(n+2)! - (n+1)!}{n!} = (n+2)(n+1) - (n+1) = (n+1)(n+2-1) = (n+1)^2,$$ a perfect square. When $n = 10$, the value is 121, which is not a multiple of 4, a multiple of 10, a prime number, or a perfect cube.

对于所有非负整数 $n$， $$\frac{(n+2)! - (n+1)!}{n!} = (n+2)(n+1) - (n+1) = (n+1)(n+2-1) = (n+1)^2,$$ 是一个完全平方数。当 $n = 10$ 时，该值为 121，它不是 4 的倍数、10 的倍数、质数或完全立方数。

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