Difference between revisions of "1989 AIME Problems/Problem 8"

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

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 $d \cdot 1 + e \cdot 12 + f \cdot 123 = 1 - 36 + 369 = 334$.