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{{ printedBook.courseTrack.name }} {{ printedBook.name }} When the terms in a geometric sequence are written as a sum of terms rather than a list of numbers, it is called a geometric series. If the sequence is short enough, such as $1,2,4,8,$ writing and calculating the sum is straightforward. $s=1+2+4+8=15$ Sigma notation can also be used. For the series written above, the explicit rule of the related geometric sequence is $a_{n}=2_{n−1}.$ Thus, the series can be written as $n=1∑4 2_{n−1}=15.$ If the series had been infinite, the infinity symbol would be above the sigma, instead of $4.$ When a series is finite, its sum can be found using the formula for a sum of a geometric series.

It can also be used when calculating partial sums of an infinite series.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 $a_{1}$ and whose common ratio is $r.$ The sum of the series can be written as $S_{n}=a_{1}+a_{1}r+a_{1}r_{2}+a_{1}r_{3}+…+a_{1}r_{n−1}$
Notice that the series is a polynomial, and can be written in standard form.
$S_{n}=a_{1}r_{n−1}+…+a_{1}r_{3}+a_{1}r_{2}+a_{1}r+a$
Since all of the terms contain a factor of $a_{1},$ it can be factored out of the right-hand side. $S_{n}=a_{1}(r_{n−1}+…+r_{3}+r_{2}+r+1)$
Since the sum above is a polynomial, polynomial division can be used to rewrite it. In fact, one polynomial identity gives $1−r1−r_{n} =r_{n−1}+…+r_{3}+r_{2}+r+1.$
Thus, the formula for the sum of a geometric series can be written as follows.

$S_{n}=1−ra_{1}(1−r_{n}) $

Q.E.D.

Calculate the sum of the geometric series. $n=1∑4 80⋅1.5_{n−1}$

Show Solution

To find the sum, we can use the formula for a geometric series
$S_{n}=1−ra_{1}(1−r_{n}) ,$
where $a_{1}$ 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 $a_{1},$ we can substitute $n=1$ into the expression $80⋅1.5_{n−1}.$
Thus, $a_{1}=80.$ We can substitute the noted values into the formula for $S_{n}.$
The sum of the given geometric series is $650.$

$80⋅1.5_{n−1}$

Substitute

$n=1$

$80⋅1.5_{1−1}$

SubTerm

Subtract term

$80⋅1.5_{0}$

CalcPow

Calculate power

$80⋅1$

Multiply

Multiply

$80$

$S_{n}=1−ra(1−r_{n}) $

SubstituteValues

Substitute values

$S_{4}=1−1.580(1−1.5_{4}) $

SubTerm

Subtract term

$S_{4}=-0.580(1−1.5_{4}) $

SimpQuot

Simplify quotient

$S_{4}=-160(1−1.5_{4})$

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 $r=±1,$ the infinite series has no sum. On the other hand, when $r$ is a fraction such that $-1<r<1,$ $r_{n}$ becomes very small as the number of terms, $n,$ increases. In fact, $r_{n}→0.$ Thus, the standard formula for the sum of a geometric series, $S_{n}=1−ra_{1}(1−r_{n}) ,$ becomes the following for an infinite geometric series with $-1<r<1.$

$S_{∞}=1−ra_{1}(1−0) ⇔S_{∞}=1−ra_{1} $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.

$S_{n}=1−ra_{1}(1−r_{n}) $Find the sum of the geometric series. $1+31 +91 +271 +…$

Show Solution

To begin, recall that the sum of an infinite geometric series can only be found if $-1<r<1.$
Thus, we must first determine $r,$ which can be done by dividing the second term, $a_{2},$ by $a_{1}.$ $a_{1}a_{2} =11/3 =31 .$
Thus, the common ratio is $r=31 .$ Since $r<1,$ it's possible to find the sum of the series. We can substitute $a_{1}$ and $r$ into the following formula.
The sum of the series is $1.5.$

$S=1−ra_{1} $

SubstituteII

$a_{1}=1$, $r=31 $

$S=1−31 1 $

OneToFrac

Rewrite $1$ as $33 $

$S=33 −31 1 $

SubFrac

Subtract fractions

$S=2/31 $

DivByFracD

$b/ca =ba⋅c $

$S=23 $

WriteDec

Write as a decimal

$S=1.5$

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