Difference between revisions of "1989 AIME Problems/Problem 8"
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Now consider the polynomial given by <math> f(z) := \sum_{i=1}^7 (z+i)^2x_i </math> (we are only treating the <math>x_i</math> as coefficients). | Now consider the polynomial given by <math> f(z) := \sum_{i=1}^7 (z+i)^2x_i </math> (we are only treating the <math>x_i</math> as coefficients). | ||
− | Notice that <math>f</math> is in fact a quadratic. We are given <math>f(0), \ f(1), \ f(2)</math> as <math>c_1, \ c_2, \ c_3</math> and are asked to find <math>c_4</math>. Using the concept of [[finite differences]] (a prototype of [[differentiation]]) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find <math>c_4=334</math>. | + | |
+ | Notice that <math>f</math> is in fact a quadratic. We are given <math>f(0), \ f(1), \ f(2)</math> as <math>c_1, \ c_2, \ c_3</math> and are asked to find <math>c_4</math>. Using the concept of [[finite differences]] (a prototype of [[differentiation]]) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find <math>c_4=334</math>. | ||
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+ | Alternatively, applying finite differences, one obtains <math>c_4 = {3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =334</math>. | ||
== See also == | == See also == |
Revision as of 19:24, 5 August 2009
Problem
Assume that are real numbers such that
Find the value of .
Solution
Solution 1
Notice that because we are given a system of equations with unknowns, the values are not fixed; indeed one can take any four of the variables and assign them arbitrary values, which will in turn fix the last three.
Given this, we suspect there is a way to derive the last expression as a linear combination of the three given expressions. Let the coefficent of in the first equation be ; then its coefficients in the second equation is and the third as . We need to find a way to sum these to make [this is in fact a specific approach generalized by the next solution below].
Thus, we hope to find constants satisfying . FOILing out all of the terms, we get
Comparing coefficents gives us the three equation system:
Subtracting the second and third equations yields that , so and . It follows that the desired expression is .
Solution 2
Notice that we may rewrite the equations in the more compact form as:
and
where and is what we're trying to find.
Now consider the polynomial given by (we are only treating the as coefficients).
Notice that is in fact a quadratic. We are given as and are asked to find . Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find .
Alternatively, applying finite differences, one obtains .
See also
1989 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 |