AMC10 2000 A

AMC10 2000 A · Q12

AMC10 2000 A · Q12. It mainly tests Sequences & recursion (algebra).

Figure 0, 1, 2, and 3 consist of 1, 5, 13, and 25 nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?

图0、1、2和3分别由1、5、13和25个不重叠的单位正方形组成。如果图案继续，如何多不重叠的单位正方形在图100中？

(A) 10401 10401

(B) 19801 19801

(D) 39801 39801

(E) 40801 40801

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Calculating the number of squares in the first few figures uncovers a pattern. Figure 0 has $2(0)+1=2(0^2)+1$ squares, figure 1 has $2(1)+3=2(1^2)+3$ squares, figure 2 has $2(1+3)+5=2(2^2)+5$ squares, and figure 3 has $2(1+3+5)+7=2(3^2)+7$ squares. In general, the number of unit squares in figure $n$ is $$2(1+3+5+\cdots+(2n-1))+2n+1=2(n^2)+2n+1.$$ Therefore, the figure 100 has $2(100^2)+2\cdot 100+1=20201$.

答案（C）：计算前几个图形中的小正方形数量可以发现规律。图形0有 $2(0)+1=2(0^2)+1$ 个正方形，图形1有 $2(1)+3=2(1^2)+3$ 个正方形，图形2有 $2(1+3)+5=2(2^2)+5$ 个正方形，图形3有 $2(1+3+5)+7=2(3^2)+7$ 个正方形。一般地，第 $n$ 个图形中的单位正方形数为 $$2(1+3+5+\cdots+(2n-1))+2n+1=2(n^2)+2n+1.$$ 因此，第100个图形有 $2(100^2)+2\cdot 100+1=20201$ 个正方形。

Topics

AL.012 Sequences & recursion (algebra)