Find sum of $ar^0 + ar^1 + ar^2 + \dotsm + ar^n$

Question

I am trying to deduce the formula of sum of $n$ terms of a GP in a way not described in the book and hence after taking $a$ as common factor, we are left behind with $r^0 + r + r^2 + r^3$ ( I took $n = 4$ ). I am stuck so as to how to find the sum of $r^0 + r + r^2 + r^3$ for any value of $n$. Is there any formula to find the sum of the series $\sum_{i=0}^{n-1} a^i$ ?

Answer 1

As $(1 + r + r^2 + ......... + r^n)(1-r) = $

$1\times (1 + r + r^2 + ......... + r^n) - r\times(1 + r + r^2 + ......... + r^n)=$

$(1 + r + r^2 + ......... + r^n)-(r + r^2 + r^3........ + r^{n+1})=$

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

$1 - r^{n+1}$

Then so long as $r \ne 1$ we have $(1 + r+ r^2 + ..... + r^n) = \frac {1-r^n}{1-r}$.

And if $r$ does equal $1$ we have $1 + r + .... + r^n = 1+1+..... + 1 = n+1$.