Generating Functions · Maclaurin Series
Lesson 7
\[\binom{-2}{n}=\frac{(-2)(-3)\dotsb(-n-1)}{n!}=(-1)^{n}(n+1) \ .\] Therefore, \[\frac{1}{(1+x)^2}=\sum_{n=0}^{\infty}(-1)^{n}(n+1)x^{n} \ .\]
Generating Functions · Maclaurin Series
\[\binom{-2}{n}=\frac{(-2)(-3)\dotsb(-n-1)}{n!}=(-1)^{n}(n+1) \ .\] Therefore, \[\frac{1}{(1+x)^2}=\sum_{n=0}^{\infty}(-1)^{n}(n+1)x^{n} \ .\]