Difference between revisions of "1992 AIME Problems/Problem 8"
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<cmath> \frac{d^2 A}{dn^2} = 1 </cmath> | <cmath> \frac{d^2 A}{dn^2} = 1 </cmath> | ||
This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; | This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; | ||
− | <cmath> a_{n} = \frac{1}{2}(n-19)(n- | + | <cmath> a_{n} = \frac{1}{2}(n-19)(n-92) </cmath> |
− | as we must have roots at <math>n = 19</math> and <math>n = | + | as we must have roots at <math>n = 19</math> and <math>n = 92</math>. |
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 17:54, 13 July 2014
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
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.