Induction proof I'm having trouble with: $1+x+x^2+x^3+...+x^n = \frac{1-x^{n+1}}{1-x}$

Question

So I'm being asked to use induction to prove that for every $x\in\{a\ |\ a\in R, a\neq 1\}$ and for every $n\in N$

$$ 1+x+x^2+x^3+...+x^n = \frac{1-x^{n+1}}{1-x} $$

I have no trouble proving it for $n=1$ :

$$ 1+x = \frac{1-x^2}{1-x} $$ Factor the polynomial: $$ 1+x = \frac{(1-x)(1+x)}{1-x} $$ Divide by $(1-x)$ $$ 1+x=1+x $$

And there you have it. The trouble I'm running into is with the induction step. If we assume that our claim is true for $n=k$ then $$ 1+x+x^2+x^3+...+x^k+x^{k+1} = \frac{1-x^{k+2}}{1-x} $$ Or in other words, $$ \frac{1-x^{k+1}}{1-x} + x^{k+1} = \frac{1-x^{k+2}}{1-x} $$

Can someone help with this? I'm having some trouble with the factoring and the book I'm studying from isn't very clear on how they proved the last equation is true.

Thanks in advance :)

Answer 1

From where you are stuck, there is only one little step left: reduce both terms of the LHS to the same denominator. That is, $$\frac{1-x^{k+1}}{1-x} + x^{k+1} = \frac{1-x^{k+1}+(1-x) x^{k+1}}{1-x}=\frac{1-x^{k+1}+x^{k+1} - x\cdot x^{k+1}}{1-x} = \frac{1-x^{k+2}}{1-x}.$$

Answer 2

$$\frac{1-x^{k+1}}{1-x} +x^{k+1}= \frac{1-x^{k+1}}{1-x} + \frac{1-x}{1-x}x^{k+1} = \frac{1-x^{k+1} + (1-x)x^{k+1}}{1-x} = \dots$$