How to find the coefficient of $x^{98}$ in $(x-1)(x-2)(x-3).....(x-100)$ Anyone please teach me how to solve  such questions... I am thinking of solving it using binomial theory but still no idea how to do it.. 
Plz help

$$(x-1)(x-2)(x-3)…(x-100)=x^{100} (1-\frac{1}{x})(1-\frac{2}{x})...(1-\frac{100}{x})$$
Denote $y=\frac{1}{x}$. Then 
$$x^{100} (1-\frac{1}{x})(1-\frac{2}{x})...(1-\frac{100}{x}) =x^{100} (1-y)(1-2y)...(1-100y)$$
and the problem becomes: Find the coefficient of $y^2$ in 
$$P(y)=(1-y)(1-2y)...(1-100y)$$
This coefficient is 
$$\frac{P''(0)}{2!}$$
Derivating twice and plugging $x=0$ we get
$$P''(0)=\sum_{k=1}^{100}\sum_{j=1, j\neq k}^{100} kj$$
thus your coefficient is 
$$\frac{1}{2} \sum_{k,j =1 , k\neq j}^{100} kj= \frac{1}{2} \left( \left( \sum_{k=1}^{100} k \right)^2 - \left( \sum_{k=1}^{100} k^2 \right) \right)$$ 
