Find the coefficient of x^{98} in the expansion of (x - 1)(x - 2)(x - 3) ... (x - 100).

Guest Jun 18, 2020

#1**0 **

By combinatorics, if we want to get a term in x^98, we need to choose 98 x's from all these brackets, and for the remaining two brackets, we choose its constant term.

Therefore, the coefficient of x^98 is

\(\displaystyle \sum_{\substack{1 \leqslant i < j \leqslant 100\\i\in \mathbb Z\\j\in\mathbb Z}} ij\)

This sum is \(\dfrac{(1 + 2 +\cdots + 100)^2 - (1^2 + 2^2 + \cdots + 100^2)}2\) (why?), which evaluates to **12,582,075**.

MaxWong Jun 18, 2020