Difference between revisions of "1992 AIME Problems/Problem 8"
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== Solution 4 == | == Solution 4 == | ||
− | Since all terms of <math>\Delta(\Delta A)</math> are 1, we know that <math>\Delta A</math> looks like <math>(k,k+1,k+2,...)</math> for some <math>k</math>. This means <math>A</math> looks like <math>(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)</math>. More specifically, <math>A_n=a_1+k(n-1)+\frac{( | + | Since all terms of <math>\Delta(\Delta A)</math> are 1, we know that <math>\Delta A</math> looks like <math>(k,k+1,k+2,...)</math> for some <math>k</math>. This means <math>A</math> looks like <math>(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)</math>. More specifically, <math>A_n=a_1+k(n-1)+\frac{(n-1)(n-2)}{2}</math>. Plugging in <math>a_{19}=a_{92}=0</math>, we have the following linear system: <cmath>a_1+91k=-4095</cmath> <cmath>a_1+18k=-153</cmath> From this, we can easily find that <math>k=-54</math> and <math>a_1=\boxed{819}</math>. |
Solution by Zeroman | Solution by Zeroman | ||
+ | |||
== See also == | == See also == | ||
{{AIME box|year=1992|num-b=7|num-a=9}} | {{AIME box|year=1992|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:42, 23 March 2018
Problem
For any sequence of real numbers , define to be the sequence , whose term is . Suppose that all of the terms of the sequence are , and that . Find .
Solution 1
Note that the s are reminiscent of differentiation; from the condition , we are led to consider the differential equation This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; as we must have roots at and .
Thus, .
Solution 2
Let , and .
Note that in every sequence of ,
Then
Since ,
Solving, .
Solution 3
The sequence is the second finite difference sequence, and the first terms of this sequence can be computed in terms of the original sequence as shown below.
Adding the above equations we find that
We can sum equation from to , finding
We can also sum equation from to , finding Finally, gives .
Kris17
Solution 4
Since all terms of are 1, we know that looks like for some . This means looks like . More specifically, . Plugging in , we have the following linear system: From this, we can easily find that and . Solution by Zeroman
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.