# Sum of Geometric Series

Until now, I can’t really memorize the formula of sum of geometric series. But don’t worry I know how to derive the formula. Don’t be like me; memorizing the formula by heart is always an advantage for you. If you are doing a constant practice about series and progression you will earn one very important skill. That is to read what is going on inside the dots.

Consider the geometric progression $a, ar,ar^2,ar^3,\dots,ar^{n-2},ar^{n-1}$

Expressing the series to its sum(Sn) we have,

$S_n=a+ar+ar^2+ar^3+\dots+ar^{n-2}+ar^{n-1}$  (1)

Multiplying r to both sides of equation we have,

$(S_n=a+ar+ar^2+ar^3+\dots+ar^{n-1})r$

$rS_n=ar+ar^2+ar^3+\dots+ar^{n-1}+ar^n$        (2)

(2)-(1):

$rS_n=ar+ar^2+ar^3+\dots+ar^{n-2}+ar^{n-1}+ar^n$

$S_n=a+ar+ar^2+ar^3+\dots+ar^{n-2}+ar^{n-1}$

From the term $ar \to ar^{n-1}$ of (1) cancels out leaving the following

$rS_n-S_n=ar^n-a$

By common factor,

$S_n(r-1)=a(r^n-1)$ $S_n=\displaystyle\frac{a(r^n-1)}{r-1}$

Sample Problem 1:

Find the sum of the following $2+1+\frac{1}{2}+\dots+\frac{1}{32}+\frac{1}{64}$

Solution:

Let $S_n=2+1+\frac{1}{2}+\dots+\frac{1}{32}+\frac{1}{64}$ (1)

$2\cdot S_n=4+2+1+\frac{1}{2}+\dots+\frac{1}{32}$              (2)

(2)-(1)

$2\cdot S_n=4+2+1+\frac{1}{2}+\dots+\frac{1}{32}$

$S_n=2+1+\frac{1}{2}+\dots+\frac{1}{32}+\frac{1}{64}$

$2S_n-S_n=4-\frac{1}{64}$ $S_n=\displaystyle\frac{255}{64}$

Sample Problem 2:

Find the sum of the series $3+3^2+3^3+\dots + 3^8+3^9$

This time let’s use our formula:

$S_n=\displaystyle\frac{a(r^n-1)}{r-1}$

Where a=3, r=3, n=9

$S_9=\displaystyle\frac{3(3^9-1)}{3-1}$ $S_9=29523$

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.

### 2 Responses

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