Question

图中的 $F_1$ 、 $F_2$ 、 $F_3$ 和 $F_4$ 是一个图形序列的前几项。当 $n \ge 3$ 时， $F_n$ 由 $F_{n-1}$ 构造而成：在它外面围上一个正方形，并且新正方形每条边上的菱形数量比 $F_{n-1}$ 外部正方形每条边上的菱形数量多一个。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：递推等差数列求和

难度评级：1400

解答：

从 $F_{n-1}$ 到 $F_n$ ，新的外部正方形带来 $4(n-1)$ 个菱形。从 $F_1$ 的单个菱形开始， $\begin{aligned} F_n &= 1 \\ &\quad {}+ 4\big(1 + 2 + \cdots + (n-1)\big) \\ &= 1 + 4 \cdot \dfrac{(n-1)n}{2} \\ &= 1 + 2n(n-1). \end{aligned}$

因此 $F_{20} = 1 + 2 \cdot 20 \cdot 19 = 761.$

所以正确答案是 E。

Going from $F_{n-1}$ to $F_n,$ the new outside square carries $4(n-1)$ diamonds. Starting from the single diamond of $F_1,$ $\begin{aligned} F_n &= 1 \\ &\quad {}+ 4\big(1 + 2 + \cdots + (n-1)\big) \\ &= 1 + 4 \cdot \dfrac{(n-1)n}{2} \\ &= 1 + 2n(n-1). \end{aligned}$

Therefore $F_{20} = 1 + 2 \cdot 20 \cdot 19 = 761.$

Thus, the correct answer is E.

2009 AMC 10A 第 15 题

答案：E

难度评级：1400

解答：

其他年份的第 15 题