View source for 2000 AMC 10 Problems/Problem 12

==Problem==

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?

<asy>
unitsize(8);
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);
draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);
draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);
draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);
draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);
draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);
draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);
draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);
draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);
label("Figure",(0.5,-1),S);
label("$0$",(0.5,-2.5),S);
label("Figure",(9.5,-1),S);
label("$1$",(9.5,-2.5),S);
label("Figure",(19.5,-1),S);
label("$2$",(19.5,-2.5),S);
label("Figure",(32.5,-1),S);
label("$3$",(32.5,-2.5),S);
</asy>


<math>\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801</math>

==Solution==

===Solution 1===

We have a recursion:

<math>A_n=A_{n-1}+4n</math>.

I.E. we add increasing multiples of <math>4</math> each time we go up a figure.

So, to go from Figure 0 to 100, we add

<math>4 \cdot 1+4 \cdot 2+...+4 \cdot 99+4\cdot 100=4 \cdot 5050=20200</math>.

<math>20201</math>

<math>\boxed{\text{C}}</math>

===Solution 2===

We can divide up figure <math>n</math> to get the sum of the sum of the first <math>n+1</math> odd numbers and the sum of the first <math>n</math> odd numbers. If you do not see this, here is the example for <math>n=3</math>:

<asy>
draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);
draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);
draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);
draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);
draw((3,0)--(3,7),linewidth(1.5));
</asy>

The sum of the first <math>n</math> odd numbers is <math>n^2</math>, so for figure <math>n</math>, there are <math>(n+1)^2+n^2</math> unit squares. We plug in <math>n=100</math> to get <math>20201</math>, which is choice <math>\boxed{\text{C}}</math>


===Solution 3===

Using the recursion from solution 1, we see that the first differences of <math>4, 8, 12, ...</math> form an arithmetic progression, and consequently that the second differences are constant and all equal to <math>4</math>.  Thus, the original sequence can be generated from a quadratic function.

If <math>f(n) = an^2 + bn + c</math>, and <math>f(0) = 1</math>, <math>f(1) = 5</math>, and <math>f(2) = 13</math>, we get a system of three equations in three variables:

<math>f(0) = 0</math> gives <math>c = 1</math>

<math>f(1) = 5</math> gives <math>a + b + c = 5</math>

<math>f(2) = 13</math> gives <math>4a + 2b + c = 13</math>

Plugging in <math>c=1</math> into the last two equations gives 

<math>a + b = 4</math>

<math>4a + 2b = 12</math>

Dividing the second equation by 2 gives the system:

<math>a + b = 4</math>

<math>2a + b = 6</math>

Subtracting the first equation from the second gives <math>a = 2</math>, and hence <math>b = 2</math>.  Thus, our quadratic function is:

<math>f(n) = 2n^2 + 2n + 1</math>

Calculating the answer to our problem, <math>f(100) = 20000 + 200 + 1 = 20201</math>, which is choice <math>\boxed{\text{C}}</math>

==See Also==

{{AMC10 box|year=2000|num-b=11|num-a=13}}