Sum of geometric series $a^{n-1} + a^{n-2} + \cdots + a^{1} + a^{0}$?

Question

Apparently the sum of $a^{n-1} + a^{n-2} + \cdots + a^{1} + a^{0}$ is $\frac{a^n-1}{a-1}$, but I'm not sure how.

The series is a geometric series with $a=a^n$ and $r=a^{-1}$, right? So just using the formula for the sum of a geometric series, we have $$s = a\left(\frac{1-r^n}{1-r}\right)=a^n\left(\frac{1-\frac1a^n}{1-\frac1a}\right)$$

and I don't see a way to simplify that to $\frac{a^n-1}{a-1}$. If there's a way to do so I'd appreciate you pointing it out!

Another line of reasoning I followed, not involving the formula, was this (again using $s$ to denote the sum):

$$s=a^{n-1} + a^{n-2} + \cdots + a^{1} + a^{0} = \frac{a^n}{a} + \frac{a^n}{a^2} + \cdots +\frac{a^n}{a^{n-1}} + \frac{a^n}{a^n}$$

And so

$$\frac{s}{a}=\frac{a^n}{a^2}+\frac{a^n}{a^3}+\cdots+\frac{a^n}{a^{n-1}} + \frac{a^n}{a^n}$$

Which tells us that

$$s- \frac{s}{a} = \frac{a^n}{a}$$

Which tells us that $s=\frac{a^n}{a-1}$, which contradicts the answer of $\frac{a^n-1}{a-1}$...

Can anyone see what was wrong with my latter approach or how to use the formula here? Any help is appreciated!

Answer 1

Let $S=1+a+a^2+\dots+a^{n-1}$. Then $aS=a+a^2+\dots+a^n$. Then, $S-aS=(1-a)S=1-a^n$. Thus, if $a\neq 1$, $S=\frac{1-a^n}{1-a}$.

Answer 2

You made a mistake in the first displayed line, it should be $$a^{n-1}\left(\frac{1-(\frac{1}{a})^n}{1-\frac{1}{a}}\right) = \frac{a^n}{a}\left(\frac{1-\frac{1}{a^n}}{1-\frac{1}{a}}\right) = \left(\frac{a^n-1}{a-1}\right)$$ as desired.

Answer 3

It simply results from the high-school identity: $$a^n-1=(a-1)(a^{n-1}+a^{n-2}+\dots +a+1),$$ which is often used as an example of an easy, but not trivial, proof by induction.