AMC10 2009 A

AMC10 2009 A · Q15

AMC10 2009 A · Q15. It mainly tests Sequences & recursion (algebra), Basic counting (rules of product/sum).

The figures F₁, F₂, F₃ and F₄ shown are the first in a sequence of figures. For n ≥ 3, Fₙ is constructed from Fₙ₋₁ by surrounding it with a square and placing one more diamond on each side of the new square than Fₙ₋₁ had on each side of its outside square. For example, figure F₃ has 13 diamonds. How many diamonds are there in figure F₂₀?

所示的图形F₁, F₂, F₃和F₄是图形序列中的前几个。对于n ≥ 3，Fₙ由Fₙ₋₁构成，通过在其周围围上一个正方形，并在新正方形的每边放置比Fₙ₋₁外正方形每边多一个菱形。例如，图形F₃有13个菱形。图形F₂₀有多少个菱形？

(A) 401 401

(B) 485 485

(D) 626 626

(E) 761 761

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): The outside square for $F_n$ has 4 more diamonds on its boundary than the outside square for $F_{n-1}$. Because the outside square of $F_2$ has 4 diamonds, the outside square of $F_n$ has $4(n-2)+4=4(n-1)$ diamonds. Hence the number of diamonds in figure $F_n$ is the number of diamonds in $F_{n-1}$ plus $4(n-1)$, or \[ 1+4+8+12+\cdots+4(n-2)+4(n-1) \] \[ =1+4(1+2+3+\cdots+(n-2)+(n-1)) \] \[ =1+4\frac{(n-1)n}{2} \] \[ =1+2(n-1)n. \] Therefore figure $F_{20}$ has $1+2\cdot19\cdot20=761$ diamonds.

答案（E）：$F_n$ 的外层正方形边界上的菱形数比 $F_{n-1}$ 的外层正方形多 4 个。因为 $F_2$ 的外层正方形有 4 个菱形，所以 $F_n$ 的外层正方形有 $4(n-2)+4=4(n-1)$ 个菱形。因此，图形 $F_n$ 中的菱形总数等于 $F_{n-1}$ 中的菱形数加上 $4(n-1)$，即 \[ 1+4+8+12+\cdots+4(n-2)+4(n-1) \] \[ =1+4(1+2+3+\cdots+(n-2)+(n-1)) \] \[ =1+4\frac{(n-1)n}{2} \] \[ =1+2(n-1)n. \] 因此，图形 $F_{20}$ 有 $1+2\cdot19\cdot20=761$ 个菱形。

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