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A geometric series is the sum of the terms of a geometric sequence.

$GeometricÂÂSequenceÂÂ1,2,4,8,â€¦ÂÂâ€‹ÂÂGeometricÂÂSeriesÂÂ1+2+4+8+â€¦â€‹$

The explicit rule of the geometric sequence above is $a_{n}=2_{nâˆ’1}.$ By using this rule, the series can be written using sigma notation. $SumÂofInfiniteÂTermsi=1âˆ‘âˆžâ€‹2_{iâˆ’1}â€‹â€‹SumÂofnÂTermsi=1âˆ‘nâ€‹2_{iâˆ’1}â€‹$

The sum of an infinite or a finite geometric series can be found by using their corresponding formulas.
What follows is a derivation of the formula for the sum of a geometric series.

Consider a geometric series of n terms, whose first term is a1 and whose common ratio is r. The sum of the series can be written as
Notice that the series is a polynomial, and can be written in standard form.
Since all of the terms contain a factor of a1, it can be factored out of the right-hand side.
Since the sum above is a polynomial, polynomial division can be used to rewrite it. In fact, one polynomial identity gives
Thus, the formula for the sum of a geometric series can be written as follows.

Consider a geometric series of n terms, whose first term is a1 and whose common ratio is r. The sum of the series can be written as

Q.E.D.

Calculate the sum of the geometric series.

Show Solution

To find the sum, we can use the formula for a geometric series
where a1 is the first term, r is the common ratio, and n is the number of terms. From the given sigma notation, we can see that n=4 and r=1.5. To find a1, we can substitute n=1 into the expression $80â‹…1.5_{nâˆ’1}.$
Thus, a1=80. We can substitute the noted values into the formula for $S_{n}.$
The sum of the given geometric series is 650.

$80â‹…1.5_{nâˆ’1}$

Substitute

n=1

$80â‹…1.5_{1âˆ’1}$

SubTerm

Subtract term

80â‹…1.50

CalcPow

Calculate power

80â‹…1

Multiply

Multiply

80

$S_{n}=1âˆ’ra(1âˆ’r_{n})â€‹$

SubstituteValues

Substitute values

$S_{4}=1âˆ’1.580(1âˆ’1.5_{4})â€‹$

SubTerm

Subtract term

$S_{4}=-0.580(1âˆ’1.5_{4})â€‹$

SimpQuot

Simplify quotient

$S_{4}=-160(1âˆ’1.5_{4})$

UseCalc

Use a calculator

$S_{4}=650$

When the common ratio of a geometric series is greater than 1, each term becomes larger and larger as the series continues. Similarly, when the common ratio is less than -1, each term becomes more and more negative. In these cases, and when r=Â±1, the infinite series has no sum. On the other hand, when r is a fraction such that
Thus, the standard formula for the sum of a geometric series, $S_{n}=1âˆ’ra_{1}(1âˆ’r_{n})â€‹,$ becomes the following for an infinite geometric series with -1<r<1.

-1<r<1,

$r_{n}$ becomes very small as the number of terms, n, increases. In fact, If the common ratio of an infinite geometric series is less than or equal to -1, or greater than or equal to 1, the sum of the series does not exist. However, it's possible to find a partial sum, or the sum of the first several terms in the series. The partial series can be thought of as a finite series. Thus, its sum can be found using the formula for a finite geometric series.

Find the sum of the geometric series.

Show Solution

To begin, recall that the sum of an infinite geometric series can only be found if
Thus, the common ratio is $r=31â€‹.$ Since r<1, it's possible to find the sum of the series. We can substitute a1 and r into the following formula.
The sum of the series is 1.5.

-1<r<1.

Thus, we must first determine r, which can be done by dividing the second term, a2, by a1. $S=1âˆ’ra_{1}â€‹$

SubstituteII

a1=1, $r=31â€‹$

$S=1âˆ’31â€‹1â€‹$

OneToFrac

Rewrite 1 as $33â€‹$

$S=33â€‹âˆ’31â€‹1â€‹$

SubFrac

Subtract fractions

$S=2/31â€‹$

DivByFracD

$b/caâ€‹=baâ‹…câ€‹$

$S=23â€‹$

WriteDec

Write as a decimal

S=1.5

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