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

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

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

解答：

其他年份的第 11 题