Difference between revisions of "2001 AIME II Problems/Problem 3"
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== Problem == | == Problem == | ||
Given that | Given that | ||
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<cmath> | <cmath> | ||
\begin{align*}x_{1}&=211,\\ | \begin{align*}x_{1}&=211,\\ | ||
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x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*} | x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*} | ||
</cmath> | </cmath> | ||
+ | |||
find the value of <math>x_{531}+x_{753}+x_{975}</math>. | find the value of <math>x_{531}+x_{753}+x_{975}</math>. | ||
== Solution == | == Solution == | ||
− | <math>x_5= | + | We find that <math>x_5 = 267</math> by the recursive formula. Summing the [[recursion]]s |
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− | < | + | <cmath>\begin{align*} |
+ | x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4} \\ | ||
+ | x_{n-1}&=x_{n-2}-x_{n-3}+x_{n-4}-x_{n-5} | ||
+ | \end{align*}</cmath> | ||
− | <math> | + | yields <math>x_{n} = -x_{n-5}</math>. Thus <math>x_n = (-1)^k x_{n-5k}</math>. Since <math>531 = 106 \cdot 5 + 1,\ 753 = 150 \cdot 5 + 3,\ 975 = 194 \cdot 5 + 5</math>, it follows that |
− | + | <cmath>x_{531} + x_{753} + x_{975} = (-1)^{106} x_1 + (-1)^{150} x_3 + (-1)^{194} x_5 = 211 + 420 + 267 = \boxed{898}.</cmath> | |
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− | <cmath> | ||
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− | </cmath> | ||
== See also == | == See also == | ||
{{AIME box|year=2001|n=II|num-b=2|num-a=4}} | {{AIME box|year=2001|n=II|num-b=2|num-a=4}} |
Revision as of 22:18, 25 July 2008
Problem
Given that
find the value of .
Solution
We find that by the recursive formula. Summing the recursions
yields . Thus
. Since
, it follows that
See also
2001 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |