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TOMATO Objective 44

Suppose that $$\mathbf{ x_1 , \cdots , x_n}$$ (n> 2) are real numbers such that x$$\mathbf{x_i = -x_{n-i+1}}$$ for $$\mathbf{1\le i \le n}$$ . Consider the sum $$\mathbf{ S = \sum \sum \sum x_i x_j x_k }$$ where the summations are taken over all i, j, k: $$\mathbf{ 1\le i, j, k \le n }$$ and i, j, k are all distinct. Then S equals:

(A) $$\mathbf{n!x_1 x_2 \cdots x_m }$$ ; (B) (n-3)(n-4); (C) (n-3)(n-4)(n-5); (D) none of the foregoing expressions;

Discussion:

$$\mathbf {( x_1 + x_2 + … + x_n )^3}$$

$$\mathbf{= \sum \sum \sum {x_i x_j x_k }+ \sum x_i ^2 ( \sum x_j ) + \sum x_i^3}$$

Since $$\mathbf {x_1 = – x_n}$$

Hence $$\mathbf {x_1 ^3 = -x_n ^3}$$

Since $$\mathbf{\sum {x_i} = 0 }$$ and $$\mathbf{\sum {x_i}^3 = 0}$$

Therefore $$\mathbf{\sum \sum \sum {x_i x_j x_k } = 0}$$

Hence option D

May 4, 2014