Math Help - Difficult Question!

1. Difficult Question!

Given p even integers and q odd integers x1 ,x2 ,x3,… xn (n=p+q), find the sum of all products,three at a time, form from, (-1)^ x1, (-1)^ x2 ,…. (-1)^xn , and show that this sum is 1/6[3(q2-p2)-(q-p)3-2(q-p)].

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3. Originally Posted by rafilv
Given p even integers and q odd integers x1 ,x2 ,x3,… xn (n=p+q), find the sum of all products,three at a time, form from, (-1)^ x1, (-1)^ x2 ,…. (-1)^xn , and show that this sum is 1/6[3(q2-p2)-(q-p)3-2(q-p)].
Hi rafilv,

Allow me to restate what I think you are asking:

Given p even integers and q odd integers $x_1, x_2, \cdots ,x_n \text { where } n=p+q$, show that the sum of all products taken three at a time from $(-1)^{x_1}, (-1)^{x_2}, \cdots , (-1)^{x_n} \text{ is } (1/6)[3(q^2-p^2) - (q-p)^3-2(q-p)]$.

Consider $(-1)^{x_i} (-1)^{x_j} (-1)^{x_k}$.
This product is 1 if all the x's are even, and there are $\binom{p}{3}$ such terms;
-1 if there is one odd x and two even x's, and there are $\binom {q}{1} \binom{p}{2}$ such terms;
1 if there are two odd x's and one even x, and there are $\binom {q}{2} \binom{p}{1}$ such terms;
-1 if there are three odd x's, and there are $\binom {q}{3}$ such terms.

So the sum is
$\binom{p}{3}(1) + \binom{q}{1}\binom{p}{2}(-1) + \binom{q}{2}\binom{p}{1}(1) + \binom{q}{3}(-1)$,
and that expression simplifies (after a little algebra) to the desired result.

jw