1989 AIME Problems/Problem 8

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Problem

Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that

$x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7=1$ $4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7=12$ $9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7=123$

Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7$ .

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 $x_i$ in the first equation can be denoted as $y^2$ , making its coefficients in the second equation as $(y+1)^{2}$ and the third as $(y+2)^2$ . We need to find a way to sum them up to make $(y+3)^2$ .

Thus, we can write that $ay^2 + b(y+1)^2 + c(y+2)^2 = (y + 3)^2$ . FOILing out all of the terms, we get $ay^2 + by^2 + cy^2 + 2by + 4cy + b + 4c = y^2 + 6y + 9$ . We can set up the three equation system:

$a + b + c = 1$

$2b + 4c = 6$

$b + 4c = 9$

Subtracting the second and third equations yields that $e = -3$ , so $f = 3$ and $d = 1$ . Thus, we have to add $d \cdot 1 + e \cdot 12 + f \cdot 123 = 1 - 36 + 369 = \boxed{334}$ .

Solution 2

Notice that we may rewrite the equations in the more compact form:

$\sum_{i=1}^{7}(2k_1+1)^2x_i=c_1, \sum_{i=1}^{7}(2k_2+1)^2x_i=c_2, \sum_{i=1}^{7}(2k_3+1)^2x_i=c_3,$ and $\sum_{i=1}^{7}(2k_4+1)^2x_i=c_4$

, where $k_1=i-1, \ k_2=i, \ k_3=i+1, \ k_4=i+2$ and $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we're trying to find.

Now undergo a paradigm shift: consider the polynomial in $k: \ f(k):= \sum_{i=1}^7 (2(k+i)-1)^2x_i$ (we are only treating the $x_i$ as coefficients). Notice that the degree of $f$ must be $2$ ; it is a quadratic. We are given $f(1), \ f(2), \ f(3)$ as $c_1, \ c_2, \ c_3$ and are asked to find $c_4$ . 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 $c_4=334$ .

1989 AIME (Problems • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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