Problem:
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?
Answer Choices:
A. 10401
B. 19801
C. 20201
D. 39801
E. 40801
Solution:
Calculating the number of squares in the first few figures uncovers a pattern. Figure 0 has 2(0)+1=2(02)+1 squares, figure 1 has 2(1)+3= 2(12)+3 squares, figure 2 has 2(1+3)+5=2(22)+5 squares, and figure 3 has 2(1+3+5)+7=2(32)+7 squares. In general, the number of unit squares in figure n is
2(1+3+5+⋯+(2n−1))+2n+1=2(n2)+2n+1
Therefore, the figure 100 has 2(1002)+2⋅100+1=20201.
OR
Each figure can be considered as a large square with identical small pieces deleted from each of the four corners. Figure 1 has 32−4(1) unit squares, figure 2 has 52−4(1+2) unit squares, and figure 3 has 72−4(1+2+3) unit squares. In general, figure n has
(2n−1)2−4(1+2+⋯+n)=(2n+1)2−2n(n+1) unit squares.
Thus figure 100 has 2012−200(101)=20201 unit squares.
OR
The number of unit squares in figure n is the sum of the first n positive odd integers plus the sum of the first n+1 positive odd integers. Since the sum of the first k positive odd integers is k2, figure n has n2+(n+1)2 unit squares. So figure 100 has 1002+1012=20201 unit squares.
Answer: C.
The problems on this page are the property of the MAA's American Mathematics Competitions
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