1989 AIME Problem 8

Question

Assume that

x_1,

x_2,

\ldots,

x_7

are real numbers such that

\begin{aligned}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.\end{aligned}

Find the value of

\begin{aligned}16x_1+25x_2+36x_3&\\+49x_4+64x_5&\\+81x_6+100x_7.&\end{aligned}

Answer

Let

S(t)=\sum_{i=1}^7(i+t)^2x_i

Because

S(t)

is quadratic in

t,

its second finite differences are constant

Define

S(t)=\sum_{i=1}^7(i+t)^2x_i.

This is a quadratic polynomial in

t,

and the equations say

S(0)=1,

S(1)=12,

and

S(2)=123.

Its first two differences are

11

and

111,

so the constant second difference is

100.

The next first difference is therefore

211,

giving

S(3)=123+211=334.

Problem 8 in other years