Difference between revisions of "2002 AMC 10B Problems/Problem 23"
(added See also section and AMC10 box) |
(→See also) |
||
Line 28: | Line 28: | ||
== See also == | == See also == | ||
− | {{AMC10 box|year=2002|ab=B|num-b= | + | {{AMC10 box|year=2002|ab=B|num-b=22|num-a=24}} |
Revision as of 18:24, 28 December 2011
Problem 23
Let be a sequence of integers such that and for all positive integers and Then is
Solution
First of all, write and in terms of
can be represented by in different ways.
Since both are equal to you can set them equal to each other.
Substitute the value of back into and substitute that into
See also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |