2002 AMC 10B Problems/Problem 23
Contents
[hide]Problem 23
Let be a sequence of integers such that
and
for all positive integers
and
Then
is
Solution 1
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
Solution 2
Substituting into
:
. Since
,
. Therefore,
, and so on until
. Adding the Left Hand Sides of all of these equations gives
; adding the Right Hand Sides of these equations gives
. These two expressions must be equal; hence
and
. Substituting
:
. Thus we have a general formula for
and substituting
:
.
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 |
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