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The way I know, for a sequence $a_1 = 1 + r + r^2 + ... + r^n$, is to create another sequence $a_2 = r \cdot a_1 = r + r^2 + ... + r^{n+1}$, then subtract $a_1$ from $a_2$, to end up with $$a_2 - a_1 = r^{n+1}-1 = ra_1-a_1 = a_1(r-1)$$ so that $$a_1 = \frac{r^{n+1} - 1}{r-1}$$

However, even though I don't have a problem "believing" the algebra, it isn't very intuitive to me.

Is there a different way to derive that formula without resorting to this algebraic trick? Or do you have a way to think about it that makes it more intuitive?

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    I'm in the same boat. I sometimes wonder how mathematicians come up with these algebraic (valid of course, but not at all obvious) manipulations – 2017-01-10
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    Do you have an example for what you mean by "intuitive"? – 2017-01-10
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    @Jack, I think what I mean is a level of understanding such that I could rederive it on my own without remembering the trick. For example, with the sum of integers from 1 to $n$ = $\frac{n \cdot (n+1)}{2}$, there's a visual/geometric interpretation that makes a lot of sense. I guess I'd like to understand how you can come to seeing that solution without just throwing every trick in the Algebra book at it until something works. – 2017-01-10
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    $\frac{r^{n+1}-1}{r-1}$ is by **definition** the unique number $a$ such that $(r-1)a=r^{n+1}-1$. So the question is whether it is even possible to completely avoid the "trick" – 2017-01-10
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    @HagenvonEitzen, but what would make you go for that trick in the first place? For example, with the sum of integers from 1 to $n$, you'd notice (possibly) the famous pairing of numbers, and that would lead you to derive that formula. – 2017-01-10
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    Why is "pairing of numbers" not a trick but "multiplying by the constant ratio" is? – 2017-01-10
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    @Jack, I say trick to imply the application of a seemingly arbitrary process not stemming from a lower level intuition. I'm sure for whomever came up with that manipulation, it wasn't a trick. Don't get me wrong, I understand the point you're making and I understand that the distinction I make is arbitrary based on certain assumptions. What I'm really asking is what thought process could bring you to attempt this specific algebraic manipulation as opposed to another one? What do you see that makes you try that? – 2017-01-10

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You could take a look at the sides lengths, which may be easier. Notice that the length from $O$ to $N$ is given by the geometric series. Similarly, the triangle shows the length from $N$ to $P$ is one less than the geometric series.

Then use a small bit of similar triangles.

enter image description here

Once you have that, it follows that your finite geometric series is the above triangle minus the tip of it.

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    Thank you, this is very helpful, I'm much better visually/geometrically. – 2017-01-10
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    @jeremyradcliff Then if only I could make all things visual :D – 2017-01-10
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Do you remember fat fit identity ?$$(x-y)=(x-y).1\\ (x^2-y^2)=(x-y)(x+y)\\ (x^3-y^3)=(x-y)(x^2+xy+y^2)\\ (x^4-y^4)=(x^3+x^2y+xy^2+y^3)\\...\\ (x^n-y^n)=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+...x^{2}y^{n-3}+xy^{n-2}+y^{n-1})$$ so now :take $x=r ,y=1$ you will have $$(x^n-y^n)=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+...x^{2}y^{n-3}+xy^{n-2}+y^{n-1})\\(r^n-1^n)=(r-1)(r^{n-1}+r^{n-2}1+r^{n-3}1^2+...r^{2}1^{n-3}+r1^{n-2}+1^{n-1})\\ (r^n-1)=(r-1)(r^{n-1}+r^{n-2}+...+r^2+r+1)\\ (r^{n-1}+r^{n-2}+...+r^2+r+1)=\frac{(r^n-1)}{r-1}$$ In the case of $|r|<1$ there is some visual proof too like below enter image description here

enter image description here

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    No, I don't remember the [fat fit identity](https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=fat+fit+identity+math) (and don't mind me linking to google for this one, since I can't find what your talking about). Not to mention, I think the identity you use stems from the geometric series... – 2017-01-10
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    See this page :https://artofproblemsolving.com/wiki/index.php/Proofs_without_words – 2017-01-10
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    But where is the fat fit identity? – 2017-01-10
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    It is not important that $(x^n-y^n)=(x-y)(x^{n-1}+...)$ has a "name"! ,but it's function (performance ) is important. – 2017-01-10
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    Hm... but now your saying "Oh, this problem that seems complicated for you is just a special case of an even more complicated thing, which I am simply going to present and say plug in numbers without any proof that it works". – 2017-01-10
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    visaul proof for algebric ,geometric ,...relations :http://mathoverflow.net/questions/8846/proofs-without-words – 2017-01-10
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    First of all, that's MO. (But still, awesome post, I use it often) Secondly, how does this relate here? The idea is to make it simpler, not more complicated, and I don't see where this is here. – 2017-01-10