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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.