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,


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




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


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}



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



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:


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

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

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