Difference between revisions of "2002 AMC 10P Problems/Problem 5"
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== Solution 1== | == Solution 1== | ||
| + | The recursive rule is equal to <math>a_{n+1}=\frac{1}{3}+a_{n}</math> for all <math>n \geq 1.</math> By recursion, <math>a_{n+2}=\frac{1}{3}+a_{n+1}=</math>a_{n+2}=\frac{1}{3}+\frac{1}{3}+a_n.<math> If we set </math>n=1<math> and repeat this process </math>2001<math> times, we will get </math>a_{2001+1}=\frac{1}{3}(2001) + a_1=667+1=668.<math> | ||
| − | + | Thus, our answer is </math>\boxed{\textbf{(C) } 668}.$ | |
== See also == | == See also == | ||
{{AMC10 box|year=2002|ab=P|num-b=4|num-a=6}} | {{AMC10 box|year=2002|ab=P|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:49, 14 July 2024
Problem
Let 
be a sequence such that 
and 
for all 
Find 
Solution 1
The recursive rule is equal to 
for all By recursion, 
a_{n+2}=\frac{1}{3}+\frac{1}{3}+a_n.
n=1
2001
a_{2001+1}=\frac{1}{3}(2001) + a_1=667+1=668.
\boxed{\textbf{(C) } 668}.$
See also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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