2009 AMC 12A 第 11 题

Question

图中的

F_1

，

F_2

，

F_3

，和

F_4

是一个图形序列的前几项。对于

n \ge 3

，

F_n

由

F_{n-1}

构造而来：在它周围加一个正方形，并且在新正方形的每一边上，比

F_{n-1}

外部正方形的每一边多放一个菱形。例如，图形

F_3

有

13

个菱形。图形

F_{20}

中有多少个菱形？

The figures

F_1,

F_2,

F_3,

and

F_4

shown are the first in a sequence of figures. For

n \ge 3,

F_n

is constructed from

F_{n-1}

by surrounding it with a square and placing one more diamond on each side of the new square than

F_{n-1}

had on each side of its outside square. For example, figure

F_3

has

13

diamonds. How many diamonds are there in figure

F_{20}?

401

485

585

626

761

答案：E

知识点：等差数列求和

难度评级：1630

小提示：

F_n

的外部正方形上有

4(n - 1)

个菱形。

The outside square of

F_n

has

4(n - 1)

diamonds

大提示：

把所有层相加，

F_n

有

1 + 4\big(1 + 2 + \cdots + (n - 1)\big)

个菱形。

Summing over all rings,

F_n

has

1 + 4\big(1 + 2 + \cdots + (n - 1)\big)

diamonds

解答：

F_n

的外部正方形比

F_{n-1}

的外部正方形多

4

个菱形，而

F_2

的外部正方形有

4

个菱形，所以

F_n

的外部正方形有

4(n - 1)

个菱形。

把所有层相加，

\begin{gathered} 1 + 4\big(1 + 2 + \cdots + (n - 1)\big) \\ = 1 + 4\cdot\frac{(n - 1)n}{2} \\ = 1 + 2(n - 1)n \end{gathered}\text{。}

当

n = 20

时，它等于

1 + 2\cdot 19\cdot 20 = 761

。

因此，正确答案是 E。

The outside square of

F_n

has

4

more diamonds than that of

F_{n-1},

and the outside square of

F_2

has

4,

so the outside square of

F_n

has

4(n - 1)

diamonds.

Adding all the rings,

\begin{gathered} 1 + 4\big(1 + 2 + \cdots + (n - 1)\big) \\ = 1 + 4\cdot\frac{(n - 1)n}{2} \\ = 1 + 2(n - 1)n. \end{gathered}

For

n = 20,

this is

1 + 2\cdot 19\cdot 20 = 761.

Thus, the correct answer is E.

2009 AMC 12A 第 11 题

答案：E

难度评级：1630

小提示：

大提示：

解答：

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