-1

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.

Community
  • 1
HKT
  • 723
  • 1
    This the classical sum of a limited geometrical series: you find it in all textbooks and web resources e;g.,(http://mathworld.wolfram.com/GeometricSeries.html). Usually it is presented for any real $x$ different from $1$. It is true a fortiori if you restrict $x$ to be a positive integer... – Jean Marie Mar 08 '17 at 10:44
  • $x$ doesn't have to lie in $\mathbf Z^+$ for this formula to hold. To prove it, just start from the high-school formula $a^n-b^n=\dotsm$. – Bernard Mar 08 '17 at 10:45
  • 1
    What do you want more than a site where what you ask is proven (and even in the more general context of reals : do you agree that integers are included into reals ?). – Jean Marie Mar 08 '17 at 10:55
  • I am simply asking how to proceed from $ \frac{x^k -1 + x^{k+1}(x-1)}{x-1} $ as I cannot seem to end up with $\frac{x^{k+1}-1}{x-1}$. Please stop over-complicating everything and refrain from commenting or down-voting when you choose not to be helpful or constructive. – HKT Mar 08 '17 at 11:28
  • Possible duplicate of Proving the geometric sum formula by induction – Did Mar 08 '17 at 13:45
  • So... you are asking why $$\frac{x^n-1}{x-1}+x^n=\frac{x^{n+1}-1}{x-1}$$ right? But this is what is done on the page you mention (approx. 2.5 times), no? – Did Mar 08 '17 at 13:49
  • @Did No, allow me to demonstrate in my own answer. – HKT Mar 09 '17 at 10:16
  • I fail to imagine how your "own answer" to come will show that the linked page does not fully answer your question... – Did Mar 09 '17 at 10:21
  • And now we can see that, unsurprisingly, your "own answer" closely mimicks some answers on the page this one duplicates. Useful? – Did Mar 09 '17 at 14:42

1 Answers1

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.

HKT
  • 723