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


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