Lesson 5

\[\binom{-1}{n}=\frac{(-1)(-2)\dotsb(-n)}{n!}=(-1)^{n} \ .\] Therefore, \[\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n}x^{n}\] and \[\frac{1}{1-x}=\sum_{n=0}^{\infty}(-1)^{n}(-x)^{n} = \sum_{n=0}^{\infty}x^{n} \ .\]