The coefficient of $x^{100}$ in the expansion of $(1-x)^{-3}$. Please solve it and tell me the technique so that I can solve it in examination in multiple choice questions.

This is an application of the generalized binomial theorem, that
$$(x+y)^s = \sum_{k=0}^\infty {s \choose k} x^{s-k}y^k$$
where
$${s \choose k} = \frac{s(s-1)(s-2)\cdots(s-k+1)}{k!}$$
is the generalized binomial coefficient. Hence in this case,
$$(1-x)^{-3} = \sum_{k=0}^\infty {-3 \choose k} (1)^{-3-k}(-x)^{k}$$
and so taking $k = 100$, the coefficient of the $x^{100}$ term will be
$${-3 \choose 100} = \frac{(-3)(-4)\cdots(-102)}{(100)(99)\cdots(1)} = \frac{(102)(101)}{2} = 5151$$
