Difference between revisions of "1992 AIME Problems/Problem 8"
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== Solution == | == Solution == | ||
− | Since the second differences | + | Since the second differences are all <math>1</math> and <math>a_{19}=a_{92}^{}=0</math>, <math>a_n</math> can be expressed explicitly by the quadratic: <math>a_n=\frac{1}{2!}(n-19)(n-92)</math>. |
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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>. |
Revision as of 23:39, 27 February 2008
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, .
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 |