Difference between revisions of "1989 AIME Problems/Problem 8"
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Find the value of <math>16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7^{}</math>. | Find the value of <math>16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7^{}</math>. | ||
− | == Solution == | + | == Solution 1== |
Let us try to derive a way to find the last expression in terms of the three given expressions. The coefficients of <math>x_i</math> in the first equation can be denoted as <math>y^2</math>, making its coefficients in the second equation as <math>(y+1)^{2}</math> and the third as <math>(y+2)^2</math>. We need to find a way to sum them up to make <math>(y+3)^2</math>. | Let us try to derive a way to find the last expression in terms of the three given expressions. The coefficients of <math>x_i</math> in the first equation can be denoted as <math>y^2</math>, making its coefficients in the second equation as <math>(y+1)^{2}</math> and the third as <math>(y+2)^2</math>. We need to find a way to sum them up to make <math>(y+3)^2</math>. | ||
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Subtracting the second and third equations yields that <math>e = -3</math>, so <math>f = 3</math> and <math>d = 1</math>. Thus, we have to add <math>d \cdot 1 + e \cdot 12 + f \cdot 123 = 1 - 36 + 369 = 334</math>. | Subtracting the second and third equations yields that <math>e = -3</math>, so <math>f = 3</math> and <math>d = 1</math>. Thus, we have to add <math>d \cdot 1 + e \cdot 12 + f \cdot 123 = 1 - 36 + 369 = 334</math>. | ||
+ | == Solution 2== | ||
+ | Notice that we may rewrite the equations in the more compact form: | ||
+ | |||
+ | <math>\sum_{i=1}^{7}(2k_1+1)^2x_i=c_1, \sum_{i=1}^{7}(2k_2+1)^2x_i=c_2, \sum_{i=1}^{7}(2k_3+1)^2x_i=c_3,</math> and <math>\sum_{i=1}^{7}(2k_4+1)^2x_i=c_4</math> | ||
+ | |||
+ | , where <math>k_1=i-1, \ k_2=i, \ k_3=i+1, \ k_4=i+2</math> and <math>c_1=1, c_2=12, c_3=123,</math> and <math>c_4</math> is what we're trying to find. | ||
+ | |||
+ | Now undergo a paradigm shift: consider the polynomial in <math>k: \ f(k):= \sum_{i=1}^7 (2(k+i)-1)^2x_i </math> (we are only treating the <math>x_i</math> as coefficients). | ||
+ | Notice that the degree of <math>f</math> must be <math>2</math>; it is a quadratic. We are given <math>f(1), \ f(2), \ f(3)</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>. | ||
== See also == | == See also == | ||
{{AIME box|year=1989|num-b=7|num-a=9}} | {{AIME box|year=1989|num-b=7|num-a=9}} |
Revision as of 00:29, 27 February 2007
Contents
Problem
Assume that are real numbers such that
Find the value of .
Solution 1
Let us try to derive a way to find the last expression in terms of the three given expressions. The coefficients of in the first equation can be denoted as , making its coefficients in the second equation as and the third as . We need to find a way to sum them up to make .
Thus, we can write that . FOILing out all of the terms, we get . We can set up the three equation system:
Subtracting the second and third equations yields that , so and . Thus, we have to add .
Solution 2
Notice that we may rewrite the equations in the more compact form:
and
, where and and is what we're trying to find.
Now undergo a paradigm shift: consider the polynomial in (we are only treating the as coefficients). Notice that the degree of must be ; it is 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 .
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 |