Generalize This!

Haha , I used the same method too . But being the clumsy person that I am , I clicked the wrong option .

$(x-1) .(x-2)...(x-n) = x^{n} - (1+2+ \dots + n)x^{n-1} + (\text{Sum of product of two the two integers taken two at a time})x^{n-2} + \dots$

Now what we are interested in is "Sum of product of two the two integers taken two at a time"

We know that for two numbers $a,b$ , $ab= \frac{ (a+b)^{2} - ( a^{2} + b^{2})}{2}$

So we use analogy to apply it to n.

"Sum of product of two the two integers taken two at a time" = $\dfrac{ (\sum n)^{2} - \sum n^{2}}{2}$

Always remember... (square of sums-sum of squares) by 2. Upvoted

I'd have posted this as a solution, but since I'm too late to the party, let me present my approach which is far more simpler and easier to grasp. We'll make use of Vieta's formulas and Newton's Identities.

As usual, we denote $P_i$ and $e_i$ as the $i^{\textrm{th}}$ power sum of the roots and elementary symmetric polynomial respectively for the given polynomial. For the given problem, we have,

$e_1=P_1=\sum n~~~;~~~P_2=\sum n^2$

From Vieta's formulas, we know that the required coefficient is simply $e_2$ .

We use the following identity from Newton's Identities:

$P_2=e_1P_1-2e_2\implies e_2=\frac{1}{2}\left((P_1)^2-P_2\right)\\ \implies \boxed{e_2=\frac{1}{2}\left\{\left(\sum n\right)^2-\sum n^2\right\}}$

Great solution this is