Difference between revisions of "1992 AIME Problems/Problem 8"
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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>. | 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 | ||
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+ | Since the result of two finite differences of some sequence is a constant sequence, we know that sequence is a quadratic. Furthermore, we know that <math>f(19) = f(92) = 0</math> so the quadratic is <math>f(x) = a(x-19)(x-92)</math> for some constant <math>a.</math> Now we use the conditions that the finite difference is <math>1</math> to find <math>a.</math> We know <math>f(19) = 0</math> and <math>f(20) = -72a</math> and <math>f(18) = 74a.</math> Therefore applying finite differences once yields the sequence <math>-74a,-72a</math> and then applying finite differences one more time yields <math>2a</math> so <math>a = \displaystyle\frac{1}{2}.</math> Therefore <math>f() = 9 \cdot 91 = \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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:29, 8 September 2020
Contents
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 (uses calculus)
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
Solution 5
Since the result of two finite differences of some sequence is a constant sequence, we know that sequence is a quadratic. Furthermore, we know that so the quadratic is for some constant Now we use the conditions that the finite difference is to find We know and and Therefore applying finite differences once yields the sequence and then applying finite differences one more time yields so Therefore
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 |
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