Understanding the equality $x^k(1-x)^{-k} = \sum_{n = k}^{\infty}{{n-1}\choose{k-1}}x^n$

Question

Can anyone explain me why this equality is true?

$x^k(1-x)^{-k} = \sum_{n = k}^{\infty}{{n-1}\choose{k-1}}x^n$

I really don't see how any manipulation could give me this result.

Thanks!

Answer 1

The Binomial Theorem says $$(1-x)^{-k}=\sum_0^{\infty}{-k\choose r}(-x)^r$$ Now $${-k\choose r}={(-k)(-k-1)\cdots(-k-r+1)\over r!}=(-1)^r{k+r-1\choose r}$$

Can you take it from there?

Answer 2

Note that $$\dfrac{1}{1-x}=1+x+x^2+x^3+\cdots+x^n+\cdots$$ for $|x|\lt 1.$ Multiplying this equation by $x$ we have that, $$\dfrac{x}{1-x}=x+x^2+x^3+x^4+\cdots+x^n+\cdots.$$ Now $$\left(\dfrac{x}{1-x}\right)^2=(x+x^2+x^3+\cdots)(x+x^2+x^3+\cdots)=x^2+2x^3+3x^4+\cdots+nx^{n+1}+\cdots.$$ Now you can continue this with mathematical induction using suitable identities of binomial coefficients. Good Luck.