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| − | ==Problem==
| + | #REDIRECT [[2000 AMC 12 Problems/Problem 8]] |
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| − | Figures <math>0</math>, <math>1</math>, <math>2</math>, and <math>3</math> consist of <math>1</math>, <math>5</math>, <math>13</math>, and <math>25</math> nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
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| − | <asy>
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| − | unitsize(8);
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| − | draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
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| − | draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);
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| − | draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);
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| − | draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);
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| − | draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);
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| − | draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);
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| − | draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);
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| − | draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);
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| − | draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);
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| − | draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);
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| − | label("Figure",(0.5,-1),S);
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| − | label("$0$",(0.5,-2.5),S);
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| − | label("Figure",(9.5,-1),S);
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| − | label("$1$",(9.5,-2.5),S);
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| − | label("Figure",(19.5,-1),S);
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| − | label("$2$",(19.5,-2.5),S);
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| − | label("Figure",(32.5,-1),S);
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| − | label("$3$",(32.5,-2.5),S);
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| − | </asy>
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| − | <math>\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801</math>
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| − | ==Solution==
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| − | We have a recursion:
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| − | <math>A_n=A_{n-1}+4(n-1)</math>.
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| − | I.E. we add increasing multiples of <math>4</math> each time we go up a figure.
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| − | So, to go from Figure 0 to 100, we add
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| − | <math>4 \cdot 1+4 \cdot 2+...+4 \cdot 99=4 \cdot 4950=19800</math>.
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| − | <math>19801</math>
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| − | B.
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| − | ==See Also==
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| − | {{AMC10 box|year=2000|num-b=11|num-a=13}}
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