Question

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

Answer 1

`\sum_{n=0}^\infty(-1)^n(n+1)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(x)`.

`f(x)=\frac{1}{(1+x)^2}=(1+x)^{-2}`

By Binomial Series with `\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,

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

By cleaning up,

`=\sum_{n=0}^\infty(-1)^n(n+1)x^n`

I hope that this was clear.