Example

Example: Finding the Sum of a Geometric Series Given in Summation Notation

Find the sum i=1152(3)i\displaystyle\sum_{i=1}^{15} 2(3)^i.

Because the series is expressed in sigma notation, start by identifying the values needed for the sum formula Sn=a1(1rn)1rS_n = \frac{a_1(1 - r^n)}{1 - r}: the first term a1a_1, the common ratio rr, and the number of terms nn. Write out the first few terms by substituting successive values of ii:

2(3)1,  2(3)2,  2(3)3,  2(3)^1,\; 2(3)^2,\; 2(3)^3,\; \dots

which gives 6,  18,  54,  6,\; 18,\; 54,\; \dots

Identify a1a_1: the first term is a1=6a_1 = 6.

Find the common ratio by dividing consecutive terms: 186=3\frac{18}{6} = 3 and 5418=3\frac{54}{18} = 3, so r=3r = 3.

The upper limit of the summation gives n=15n = 15. Substitute a1=6a_1 = 6, r=3r = 3, and n=15n = 15 into the formula:

S15=6(1315)13S_{15} = \frac{6(1 - 3^{15})}{1 - 3}

Simplify:

S15=43,046,718S_{15} = 43{,}046{,}718

This example illustrates how to evaluate a geometric series that is presented in summation notation: expand the first few terms to determine a1a_1 and rr, read nn from the upper limit, and then apply the partial-sum formula.

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Updated 2026-05-25

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