Difference between revisions of "1992 AIME Problems/Problem 8"
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Thus, <math>a_1=\frac{1}{2!}(1-19)(1-92)=\boxed{819}</math>. | Thus, <math>a_1=\frac{1}{2!}(1-19)(1-92)=\boxed{819}</math>. | ||
+ | == Alternate Solution == | ||
+ | Note that in every sequence of <math>a_i</math>, <math>a_n=\binom{n-1}{1}\Delta a_n + \binom{n-1}{2}\Delta^2 a_n +\binom{n-1}{3}\Delta^3 a_n + ...</math> | ||
+ | |||
+ | Then <math>a_n=a_1 +\binom{n-1}{1}\Delta a_1 +\binom{n-1}{2}\Delta^2 a_1 +\binom{n-1}{3}\Delta^3 a_1 + ...</math> | ||
+ | |||
+ | Since <math>\Delta a_1 =a_2 -a_1</math>, <math>a_n=a_1 +\binom{n-1}{1}(a_2-a_1) +\binom{n-1}{2}\cdot 1=a_1 +n(a_2-a_1) +\binom{n-1}{2}</math> | ||
+ | |||
+ | <math>a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153</math> | ||
+ | |||
+ | <math>a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095</math> | ||
+ | |||
+ | Solving, <math>a_1=\boxed{819}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=1992|num-b=7|num-a=9}} | {{AIME box|year=1992|num-b=7|num-a=9}} |
Revision as of 21:23, 15 March 2009
Contents
[hide]Problem
For any sequence of real numbers , define to be the sequence , whose $n^\mbox{th}_{}$ (Error compiling LaTeX. Unknown error_msg) term is . Suppose that all of the terms of the sequence are , and that . Find .
Solution
Since the second differences are all and , can be expressed explicitly by the quadratic: .
Thus, .
Alternate Solution
Note that in every sequence of ,
Then
Since ,
Solving,
See also
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |