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)].

You need to study the available material on posting mathematical notation on the web,
The best way is to learn LaTeX notation.
But if that is not possible, then at least learn some standard notation.
I have to tell you that what you posted is gibberish.

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