Problem:
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​?
F1​ F2​ F3​
Answer Choices:
A. 401
B. 485
C. 585
D. 626
E. 761
Solution:
The outside square for Fn​ has 4 more diamonds on its boundary than the outside square for Fn−1​. Because the outside square of F2​ has 4 diamonds, the outside square of Fn​ has 4(n−2)+4=4(n−1) diamonds. Hence the number of diamonds in figure Fn​ is the number of diamonds in Fn−1​ plus 4(n−1), or
1+4+8+12+⋯+4(n−2)+4(n−1)=1+4(1+2+3+⋯+(n−2)+(n−1))=1+42(n−1)n​=1+2(n−1)n.​
Therefore figure F20​ has 1+2⋅19⋅20=761​ diamonds.
The problems on this page are the property of the MAA's American Mathematics Competitions
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