$\forall x \in \mathbb{Z^+}, x \neq 1 \implies \left(\forall n \in \mathbb{N}, \sum_{i=0}^{n-1}{x^i}=\frac{x^n-1}{x-1}\right) $

Question

Can someone please prove this?

$$\forall x \in \mathbb{Z^+}, x \neq 1 \implies \left(\forall n \in \mathbb{N}, \sum_{i=0}^{n-1}{x^i}=\frac{x^n-1}{x-1}\right) $$

I'm confused over whether to add $x^k$ or $x^{k+1}$ to the right hand side and have gotten nowhere in both cases. Just can't factor it out like the rest of them. It would be much appreciated if someone could point me in the right direction.

Note: This is cleary a different one from Proving the geometric sum formula by induction so please stop down-voting.

Answer 1

\begin{align*} \text{We want to prove:} \ \forall x \in \mathbb{Z^+}, x \neq 1 \implies \left(\forall n \in \mathbb{N}, \sum_{i=0}^{n-1}{x^i}=\frac{x^n-1}{x-1}\right) \end{align*} Let $x \in \mathbb{R^+}$. Assume that $x\neq1$.

Let $P(n)$ be the claim: $$\sum_{i=0}^{n-1}\frac{x^n-1}{x-1}$$ For the base case, $n=0$:

The $L.H.S.=\sum_{i=0}^{0-1}{x^i}=\sum_{i=0}^{-1}{x^i}=0$.

The $R.H.S.=\frac{x^0-1}{x-1}=\frac{1-1}{x-1}=0$

Therefore, $P(0)$ holds.

Assuming $P(k)$ holds for $n=k$, we have: \begin{align*} P(k):\sum_{i=0}^{k-1}{x^i}=\frac{x^k-1}{x-1} \end{align*} We need to prove $P(k+1)$ which should be: $$\sum_{i=0}^{k}{x^i}=\frac{x^{k+1}-1}{x-1} $$ Now, we know \begin{align*} \sum_{i=0}^{k}{x^i} &= \sum_{i=0}^{k-1}{x^i} +x^k\\ &= \frac{x^k-1}{x-1}+x^k\\ &= \frac{x^k-1+x^k(x-1)}{x-1}\\ &= \frac{x^k-1+x^k.x-x^k)}{x-1}\\ &= \frac{x^{k+1}-1}{x-1}\\ \end{align*} And so $P(k+1)$ holds.