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Calculus 2 : Series in Calculus

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #11 : Arithmetic And Geometric Series

Calculate the sum, rounded to the nearest integer, of the first 20 terms of the following geometric series: 4+6+9+13.5+20.25...

Possible Answers:

26597

26596

26594

26595

Correct answer:

26594

Explanation:

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

We have $a_{1}$ and n and we just need to find r before calculating the sum.

Solution:

$a_{1}=4$

$r=\frac{a_{2}}{a_{1}}$

$r=\frac{6}{4} = 1.5$

$S_{20}=\frac{a_{1}*(1-r^n)}{1-r}$

$S_{20}=\frac{4*(1-1.5^{20})}{1-1.5}$

$S_{20}=26594$

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Example Question #201 : Series In Calculus

Calculate the sum, rounded to the nearest integer, of the first 16 terms of the following geometric series: 6+12.6+26.46+116.6886+245.04606...

Possible Answers:

780305

780300

780310

780295

Correct answer:

780305

Explanation:

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

We have $a_{1}$ and n and we just need to find r before calculating the sum.

Solution:

$a_{1}=6$

$r=\frac{a_{2}}{a_{1}}$

$r=\frac{12.6}{6} = 2.1$

$S_{16}=\frac{a_{1}*(1-r^n)}{1-r}$

$S_{16}=\frac{6*(1-2.1^{16})}{1-2.1}$

$S_{16}=780305$

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Example Question #2984 : Calculus Ii

Calculate the sum, rounded to the nearest integer, of the first 9 terms of the following geometric series: 2+9.2+42.32+895.4921+4119.25952...

Possible Answers:

512327

512329

512330

512328

Correct answer:

512327

Explanation:

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

We have $a_{1}$ and n and we just need to find r before calculating the sum.

Solution:

$a_{1}=2$

$r=\frac{a_{2}}{a_{1}}$

$r=\frac{9.2}{2} = 4.6$

$S_{9}=\frac{a_{1}*(1-r^n)}{1-r}$

$S_{9}=\frac{2*(1-4.6^{9})}{1-4.6}$

$S_{9}=512327$

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Example Question #21 : Types Of Series

Calculate the sum, rounded to the nearest integer, of the first 100 terms of the following geometric series: 16+13.76+11.8336+10.176896+8.75213056...

Possible Answers:

116

115

117

114

Correct answer:

114

Explanation:

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

We have $a_{1}$ and n and we just need to find r before calculating the sum.

Solution:

$a_{1}=16$

$r=\frac{a_{2}}{a_{1}}$

$r=\frac{13.76}{16} = .86$

$S_{100}=\frac{a_{1}*(1-r^n)}{1-r}$

$S_{100}=\frac{16*(1-.86^{100})}{1-.86}$

$S_{100}=144$

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Example Question #2986 : Calculus Ii

Calculate the sum of the following infinite geometric series:

$\sum_{k=0}^{\infty}19\left(\frac{6}{15}\right)^k$

Possible Answers:

$\frac{283}{9}$

$\frac{284}{9}$

$\frac{286}{9}$

$\frac{95}{3}$

Correct answer:

$\frac{95}{3}$

Explanation:

This is an infinite geometric series.

The sum of an infinite geometric series can be calculated with the following formula,

$S = \frac{a_{1}}{1-r}$ , where $a_{1}$ is the first value of the summation, and r is the common ratio.

Solution:

Value of $a_{1}$ can be found by setting k = 0

$a_{1}=19$

r is the value contained in the exponent

$r=\frac{6}{15}$

$S=\frac{a_{1}}{1-r}$

$S=\frac{19}{1-\frac{6}{15}} = \frac{19}{\frac{9}{15}}$

$S=\frac{285}{9}=\frac{95}{3}$

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Example Question #2987 : Calculus Ii

Calculate the sum of a geometric series with the following values:

n=10 , $a_{1}=12$ , $r=\frac{1}{4}$ ,

rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

For this question, we are given all of the information we need.

Solution:

$S_{10}=\frac{a_{1}*(1-r^n)}{1-r}$

$S_{10}=\frac{12*(1-(\frac{1}{4})^{10})}{1-\frac{1}{4}}$

$S_{10}=\frac{12*(1-\frac{1}{4^{10}}}{\frac{3}{4}} = \frac{48}{3}(1-\frac{1}{4^{10}})$

$S_{10}=16*(1-\frac{1}{4^{10}})$

$S_{10}=15.999$

Rounding, $S_{10}=16$

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Example Question #2988 : Calculus Ii

Calculate the sum of a geometric series with the following values:

n=40 , $a_{1}=5,$ $r=\frac{2}{7}$ ,

rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

For this question, we are given all of the information we need.

Solution:

$S_{40}=\frac{a_{1}*(1-r^n)}{1-r}$

$S_{40}=\frac{5*(1-(\frac{2}{7})^{40})}{1-\frac{2}{7}}$

$S_{40}=\frac{5*(1-(\frac{2}{7})^{40})}{\frac{5}{7}} = \frac{5*7(1-(\frac{2}{7})^{40})}{5}$

$S_{40}=7*(1-(\frac{2}{7})^{40})$

$S_{40}=6.99999...$

Rounding, $S_{40}=7$

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Example Question #3 : Series Of Constants

Calculate the sum of a geometric series with the following values:

n=10 , $a_{1}=11$ , $r=\frac{2}{20}$ ,

rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

For this question, we are given all of the information we need.

Solution:

$S_{10}=\frac{a_{1}*(1-r^n)}{1-r}$

$S_{10}=\frac{11*(1-(\frac{2}{20})^{10})}{1-\frac{2}{20}}$

$S_{10}=\frac{11*(1-(\frac{1}{10})^{10}}{\frac{9}{10}} = \frac{11*10*(1-\frac{1}{10^{10}})}{9}$

$S_{10}=\frac{110}{9}(1-\frac{1}{10^{10}})$

$S_{10}=12.2222...$

Rounding, $S_{10}=12$

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Example Question #31 : Types Of Series

Calculate the sum of the following infinite geometric series:

$\sum_{k=0}^{\infty}17\left(\frac{1}{6}\right)^k$

Possible Answers:

$\frac{102}{5}$

$\frac{104}{5}$

$\frac{103}{5}$

$\frac{101}{5}$

Correct answer:

$\frac{102}{5}$

Explanation:

This is an infinite geometric series.

The sum of an infinite geometric series can be calculated with the following formula,

$S = \frac{a_{1}}{1-r}$ , where $a_{1}$ is the first value of the summation, and r is the common ratio.

Solution:

Value of $a_{1}$ can be found by setting k = 0

$a_{1}=17$

r is the value contained in the exponent

$r=\frac{1}{6}$

$S=\frac{a_{1}}{1-r}$

$S=\frac{17}{1-\frac{1}{6}} = \frac{17*6}{5}$

$S=\frac{102}{5}$

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Example Question #31 : Types Of Series

Calculate the sum of the following infinite geometric series:

$\sum_{k=0}^{\infty}3\left(\frac{3}{7}\right)^k$

Possible Answers:

$\frac{27}{4}$

$\frac{21}{4}$

$\frac{25}{4}$

$\frac{23}{4}$

Correct answer:

$\frac{21}{4}$

Explanation:

This is an infinite geometric series.

The sum of an infinite geometric series can be calculated with the following formula,

$S = \frac{a_{1}}{1-r}$ , where $a_{1}$ is the first value of the summation, and r is the common ratio.

Solution:

Value of $a_{1}$ can be found by setting k = 0

$a_{1}=3$

r is the value contained in the exponent

$r=\frac{3}{7}$

$S=\frac{a_{1}}{1-r}$

$S=\frac{3}{1-\frac{3}{7}} = \frac{3}{\frac{4}{7}}$

$S=\frac{21}{4}$

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Calculus 2 : Series in Calculus

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : Arithmetic And Geometric Series

Example Question #201 : Series In Calculus

Example Question #2984 : Calculus Ii

Example Question #21 : Types Of Series

Example Question #2986 : Calculus Ii

Example Question #2987 : Calculus Ii

Example Question #2988 : Calculus Ii

Example Question #3 : Series Of Constants

Example Question #31 : Types Of Series

Example Question #31 : Types Of Series

All Calculus 2 Resources

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