1. ## Proof By Induction

Hey, I was wondering if someone could help me with a question I have.

Prove by induction

x^n-1 + x^n-2 + .... + x + 1 = (x^n)-1 / x - 1

Thank you

2. ## Re: Proof By Induction

You need grouping symbols.

Prove by induction for $\displaystyle n \ge 1$.

x^(n - 1) + x^(n - 2) + ... + x + 1 = (x^n - 1)/(x - 1)

Have you done the base case, n = 1?

3. ## Re: Proof By Induction

Originally Posted by jkd98
Prove by induction
x^n-1 + x^n-2 + .... + x + 1 = (x^n)-1 / x - 1
Thank you
It is very clear that this is the case if $n=1$

So suppose that $J>1~\&~\dfrac{{{x^J} - 1}}{{x - 1}} = \sum\limits_{k = 0}^{J - 1} {{x^k}}$

\begin{align*}\frac{{{x^{J + 1}} - 1}}{{x - 1}} &= {x^{J}} + \frac{{{x^J} - 1}}{{x - 1}}\\&= {x^{J + 1}} + \sum\limits_{k = 0}^J {{x^k}}\\&= \sum\limits_{k = 0}^{J + 1} {{x^k}} \end{align*}

4. ## Re: Proof By Induction

\displaystyle \begin{align*}x^n + x^{n-1} + x^{n-2} + \ldots + 1 &= x^n + (x^{n-1} + x^{n-2} + \ldots + 1) \\ &=x^n + \frac{x^n - 1}{x-1} &\text{by the induction hypothesis} \\ &= \frac{x-1}{x-1}x^n + \frac{x^n - 1}{x-1}\end{align*}
I'm sure you can finish from there.