How to show $1+r+r^2 + \cdots + r^n = \frac{r^{n+1} - 1}{r-1}$ where $r \neq 1$

Question

I want to show that the geometric series $$1+r+r^2 + \cdots + r^n = \frac{r^{n+1} - 1}{r-1}, r \neq 1$$

I first started with the following using Gauss method:

$$S = 1+r+r^2 + \cdots + r^n$$

$$S=r^n + r^{n-1} + r^{n-2} + ... + 1$$

Adding both S's:

$$2S = (1+r^n) + (r+r^{n-1})+(r^2+r^{n-2})+...+(1+r^n)$$

Am I on the right track?

PS: I initially tried to follow this question, however, there was not enough to go on, so I attempted to do this on my own.

Update:

I originally assumed that I can use Gauss method which proved (when he was 10 years old!) $$\sum_{k=1}^nk= \frac{n(n+1)}{2}$$

but from user comments below, Gauss method cannot be applied here, as my questions is regarding a geometric sum not an arithmetic sum.

Based on your argument I'm assuming the "Gauss method" is the one attributed to him for proving $\sum_{k=1}^nk=n(n+1)/2$. But that is an arithmetic sum, not a geometric sum. — anon, Apr 19 '15 at 16:41
The method you try to use works for arithmetic series, but unfortunately not for geometric series. As André says, consider $rS - S$. — David E, Apr 19 '15 at 16:41
@DavidE: Ahh, okay that's why it never works. So, where did $rS-S$ come from? How do I decide to subtract? — lucidgold, Apr 19 '15 at 16:42
@lucidgold, try: \begin{align} S &= 1 + r + r^2 + \cdots + r^n \ rS &= r + r^2 + \cdots + r^{n+1} \end{align} So... \begin{align} rS - S &= (0 - 1) + (r - r) + (r^2 - r^2) + \cdots + (r^{n+1} - 0) \ \dots & \quad \dots \end{align} — David E, Apr 19 '15 at 16:45
see this http://math.stackexchange.com/questions/1241222/gauss-method-to-show/1241228#1241228 — abel, Apr 19 '15 at 16:45
We decide to subtract because $S$ and $rS$ have all but two terms in common. Elimination is a powerful technique in mathematics. — anon, Apr 19 '15 at 16:47
@abel: That's the first question I found however, I am now trying to workout the algebra on my own. I really need to do this on my own to figure it out, as I did not understand how/why they deduced that way. — lucidgold, Apr 19 '15 at 16:47
I was literally 2 seconds away from submitting my answer, and it closed. — Akiva Weinberger, Apr 19 '15 at 17:17

score 2 · Accepted Answer · 2015-04-19T16:54:24.033

2

If $S = 1 + r + r^2 + \ldots + r^n$ then $rS$ = $r+r^2 + \ldots + r^n + r^{n+1}$, so you just have to subtract them.

We are trying to find a formula for $S_n=\sum_{k=0}^{n}r^k$, where $r \neq 1$. We have: $$S_n+r^{n+1}=1+\sum_{k=1}^{n+1}r^k=1+\sum_{k=0}^{n}r^{k+1}= 1+r\sum_{k=0}^{n}r^{k}=1+rS_n,$$ so $(q-1)S_n=q^{n+1}-1$ and $S_n=\frac{q^{n+1}-1}{q-1}$.

Here is another example of this method: we want to find out what $G_n=\sum_{k=0}^{n}k2^k$ really is. It turns out that

\begin{align} L & = G_n+(n+1)2^{n+1}=0\cdot 2^0+\sum_{k=1}^{n+1}k2^k = \sum_{k=0}^n(k+1)2^{k+1} \\ & =2 \sum_{k=0}^n k 2^k + 2 \sum_{k=0}^n 2^k = 2G_n + 2(2^{n+1}-1), \end{align} which we can transform into $$G_n=(n+1)2^{n+1}-2\cdot (2^{n+1}-1)=(n-1)2^{n+1}+2.$$

edited Apr 19 '15 at 16:54

answered Apr 19 '15 at 16:41

Where did $rS$ come from? How do I decide to subtract $S$ from $rS$? what lead you to that idea? – lucidgold Apr 19 '15 at 16:43
3

You ask how to decide what to subtract? Well, there is no general method to say what will work. I've tried to multiply $S$ by $r$ and noticed, that this will lead to the solution. – Apr 19 '15 at 16:44
@lucidgold: when you multiply the series by $r$, you see that all terms but the two extreme remain unchanged (you "earn" $r^{n+1}$ and "lose" $1$). The method just exploits this property. – Apr 19 '15 at 17:17

How to show $1+r+r^2 + \cdots + r^n = \frac{r^{n+1} - 1}{r-1}$ where $r \neq 1$

Update:

1 Answers1