Given the following polynomial:
${a_n}{x^n}+{a_{n-1}}{x^{n-1}}+...+{a_1}{x}+{a_0}=0 \ \mathbf{(1)}$
which is equivalent to
${a_n}(x-\alpha_1)(x-\alpha_2)...(x-\alpha_n)=0 \ \mathbf{(2)}$
How do you prove from (1) and (2) that:
$${\sum_{j=1}^n}{\sum_{k>j}^n}\alpha_j\alpha_k= \frac{a_{n-2}}{a_n}$$
I tried starting with cubic, then quartic then quintic to try and find a number pattern, but I am not having any luck with it.
Is there another way to solve this?
Thanks in advance!
WLOG, $a_n=1$ to simplify (divide all coefficients by $a_n$).
When you expand the product to get the coefficient of $x^{n-2}$, the relevant terms $x^{n-2}$ are those that "consume" $n-2$ $x$'s and two $\alpha$'s among the factors. The sign of every term is positive, and the coefficients correspond to all possible unordered pairs of indexes (the product is commutative), without repetitions. There are $\dfrac{n(n-1)}2$ such pairs.
By a similar reasoning, you establish that $-\dfrac{a_{n-1}}{a_n}$ is the sum of the roots, and other generalizations with sums of products, that are in fact the Vieta's relations.