Difference between revisions of "2006 AIME A Problems/Problem 15"
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== Solution == | == Solution == | ||
| − | {{ | + | If you take the recursion formula and square both sides, you get <math>x_k^2=x_{k-1}^2+6x_{k-1}+9</math>, or <math>x_k^2-x_{k-1}^2=6x_{k-1}+9</math>. If you take the sum of this from 1 to 2007, you see that |
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| + | <center><math>\begin{eqnarray*}\sum_{k=1}^{2007}x_k^2-x_{k-1}^2=\sum_{k=1}^{2007}6x_{k-1}+9\\ | ||
| + | x_{2007}^2-x_0^2=6(x_0+x_1+x_2+\cdots +x_{2006})+9*2007\\ | ||
| + | x_{2007}^2=6(S)+9*2007\\ | ||
| + | S=\frac{x_{2007}^2-9*2007}{6}\end{eqnarray*}</math></center> | ||
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| + | where <math>S=x_0+x_1+x_2+\cdots +x_{2006}=x_1+x_2+\cdots +x_{2006}</math>. | ||
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| + | Note that x_{2007} is a multiple of 3, and the closest square multiple of 3 to 2*2007 is <math>(3*45)^2=9*2025</math>. Therefore the smallest value of the absolute value of <math>S</math> is | ||
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| + | <math>|S|=|\frac{9*2025-9*2007}{6}=\frac{9*18}{6}=\boxed{027}</math>. | ||
== See also == | == See also == | ||
Revision as of 21:43, 4 January 2009
Problem
Given that a sequence satisfies 
and 
for all integers 
find the minimum possible value of 
Solution
If you take the recursion formula and square both sides, you get 
, or 
. If you take the sum of this from 1 to 2007, you see that
x_{2007}^2-x_0^2=6(x_0+x_1+x_2+\cdots +x_{2006})+9*2007\\ x_{2007}^2=6(S)+9*2007\\
S=\frac{x_{2007}^2-9*2007}{6}\end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)where 
.
Note that x_{2007} is a multiple of 3, and the closest square multiple of 3 to 2*2007 is 
. Therefore the smallest value of the absolute value of 
is
.
See also
| 2006 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Final Problem | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
