Deriving Sum of a Geometric Progression

Solution 1:

Call

$$ S = \sum_{i = 0}^{n-1}ar^i = a + ar + ar^2 + \cdots + ar^{n-1} \tag{1} $$

if you multiply both sides by $r$ you get

$$ r S = ar + a r^2 + ar^3 + \cdots + ar^n \tag{2} $$

When you subtract Eq. (1) from Eq. (2) you will see that most terms cancel out

\begin{eqnarray} S - rS &=& S(1- r) = (a + ar + ar^2 + \cdots + ar^{n-1}) - (ar + ar^2 + ar^3 + \cdots + ar^n) \\ &=& a - ar^n = a(1 - r^n) \end{eqnarray}

So that

$$ S = a \frac{1 - r^n}{1 - r} \tag{3} $$

Therefore

$$ \sum_{i=0}^{n-1} ar^i = a \frac{1-r^n}{1 - r} ~~~~~\mbox{if}~~~r\ne 1 $$

Solution 2:

Without words:

$$\begin{align}S=&a+ar^1+ar^2+...\ \ \ \ \ \ \ ar^{n-1}\\ rS=&\ \ \ \ \ \ \ ar^1+ar^2+ar^3+...\ \ \ \ \ \ \ \ \ ar^{n}\end{align}$$

$$rS-S=ar^n-a$$ $$S=a\frac{r^n-1}{r-1}$$

Solution 3:

Using the difference of powers identity:

$$ a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2 + \cdots + a b^{n-2}+b^{n-1}) $$

with $\,a=1, b=r\,$ gives $\;1-r^n=(1-r)(1+r+\cdots+ r^{n-1})\;$ therefore:

$$ 1+r+\cdots+r^{n-1} = \cfrac{1-r^n}{1-r} $$