# How do you use the binomial series to expand f(x)= 1/((1+x)^2)?

Apr 20, 2017

$\setminus {\sum}_{n = 0}^{\setminus} \infty {\left(- 1\right)}^{n} \left(n + 1\right) {x}^{n}$

#### Explanation:

Recall (Binomial Series)

(1+x)^\alpha=\sum_{n=0}^\infty\frac{\alpha\cdot(\alpha-1)\cdot(\alpha-2)\cdot\cdots\cdot(\alpha-n+1)}{n!}x^n

Let us look at $f \left(x\right)$.

$f \left(x\right) = \setminus \frac{1}{{\left(1 + x\right)}^{2}} = {\left(1 + x\right)}^{- 2}$

By Binomial Series with $\setminus \alpha = - 2$,

=\sum_{n=0}^\infty\frac{(-2)\cdot(-3)\cdot(-4)\cdot\cdots\cdot(-n)\cdot(-(n+1))}{n!}x^n

By factoring out the negative signs,

$= \setminus {\sum}_{n = 0}^{\setminus} \infty {\left(- 1\right)}^{n} \setminus \frac{\setminus \cancel{2} \setminus \cdot \setminus \cancel{3} \setminus \cdot \setminus \cancel{4} \setminus \cdot \setminus \cdots \setminus \cdot \setminus \cancel{n} \setminus \cdot \left(n + 1\right)}{1 \setminus \cdot \setminus \cancel{2} \setminus \cdot \setminus \cancel{3} \setminus \cdot \setminus \cancel{4} \setminus \cdot \setminus \cdots \setminus \cdot \setminus \cancel{n}} {x}^{n}$

By cleaning up,

$= \setminus {\sum}_{n = 0}^{\setminus} \infty {\left(- 1\right)}^{n} \left(n + 1\right) {x}^{n}$

I hope that this was clear.