Other way to derive/prove $a_n = \frac{r^{n+1} - 1}{r-1}$ for sum of geometric sequence?

Question

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?

Answer 1

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.

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

Answer 2

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