How to prove that $\dfrac{1}{1+kx}=\sum_{n=0}^{\infty}(-x)^nk^n$

Question

I have recently discovered that:

$\dfrac{1}{1+x}=1-x+x^2-x^3+x^4-x^5+O(x^6)$

$\dfrac{1}{1+2x}=1-2x+2^2x^2-2^3x^3+2^4x^4-2^5x^5+O(x^6)$

$\dfrac{1}{1+3x}=1-3x+3^2x^2-3^3x^3+3^4x^4-3^5x^5+O(x^6)$

and so on

I wish to prove this result by induction but got stuck.

Prove first for $n=1$

$\dfrac{1}{1+x}=1-x+x^2-x^3+x^4-x^5+O(x^6)$ (Why is it true?)

Assume that it is true for $n=k$

Prove that it is true for $n=k+1$

$\dfrac{1}{1+(k+1)x}=1-(k+1)x+(k+1)^2x^2-(k+1)^3x^3+(k+1)^4x^4-(k+1)^5x^5+O(x^6)$

I am stuck at this step, what should I do here?

Answer 1

By geometric series we have

$$\sum_{n=0}^{\infty}(-x)^nk^n=\sum_{n=0}^{\infty}(-xk)^n=\dfrac{1}{1+kx}$$

for $|xk|<1$.