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>. | ||
− | == | + | == Solution 2 == |
Let <math>\Delta^1 A=\Delta A</math>, and <math>\Delta^n A=\Delta(\Delta^{(n-1)}A)</math>. | Let <math>\Delta^1 A=\Delta A</math>, and <math>\Delta^n A=\Delta(\Delta^{(n-1)}A)</math>. | ||
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Solving, <math>a_1=\boxed{819}</math>. | Solving, <math>a_1=\boxed{819}</math>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | Write out and add first <math>k-1</math> terms of the second finite difference sequence: | ||
+ | |||
+ | <math>a_3+a_1-2*a_2=1</math> | ||
+ | |||
+ | <math>a_4+a_2-2*a_3=1</math> | ||
+ | |||
+ | … | ||
+ | |||
+ | … | ||
+ | |||
+ | … | ||
+ | |||
+ | <math>a_k + a_{k-2} - 2*a_{k-1} = 1</math> | ||
+ | |||
+ | <math>a_{k+1} + a_{k-1} - 2*a_k = 1</math> | ||
+ | |||
+ | Adding the above <math>k-1</math> equations we get: | ||
+ | |||
+ | <math>\boxed{(a_{k+1} - a_k) = k-1 + (a_2-a_1)} --- (1)</math> | ||
+ | |||
+ | Now taking sum <math>k = 1</math> to <math>18</math> in equation <math>(1)</math> we get: | ||
+ | <math>18*(a_1-a_2) - a_1 = 153 --- (2)</math> | ||
+ | |||
+ | Now taking sum <math>k = 1</math> to <math>91</math> in equation <math>(1)</math> we get: | ||
+ | <math>91*(a_1-a_2) - a_1 = 4095 --- (3)</math> | ||
+ | |||
+ | <math>18* (3) - 91*(2)</math> gives <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 14:15, 10 December 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, .
Solution 2
Let , and .
Note that in every sequence of ,
Then
Since ,
Solving, .
Solution 3
Write out and add first terms of the second finite difference sequence:
…
…
…
Adding the above equations we get:
Now taking sum to in equation we get:
Now taking sum to in equation we get:
gives .
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 |