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Algebra II : Infinite Series

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Example Questions

Example Question #21 : Infinite Series

Determine the sum, if possible: $4-\frac{2}{3}+\frac{1}{9}-...$

Possible Answers:

$\frac{24}{7}$

$\frac{71}{21}$

$\textup{The sum cannot be determined.}$

$\frac{17}{5}$

$\frac{64}{19}$

Correct answer:

$\frac{24}{7}$

Explanation:

Determine the common ratio of the infinite series by dividing the second term with the first term and the third term with the second term. The common ratios should be similar.

$\frac{-\frac{2}{3}}{4} = -\frac{2}{3} \div 4= -\frac{2}{3}\times\frac{1}{4} = -\frac{1}{6}$

$\frac{\frac{1}{9}}{-\frac{2}{3}} = \frac{1}{9} \div -\frac{2}{3} = \frac{1}{9} \times -\frac{3}{2} = -\frac{1}{6}$

Write the formula for infinite series, and substitute the terms.

$\textup{Sum} =\frac{\textup{first term}}{1-r} = \frac{4}{1-(-\frac{1}{6})} = \frac{4}{\frac{7}{6}}$

Simplify the complex fraction.

$\frac{4}{\frac{7}{6}} = 4\div \frac{7}{6} = 4 \times \frac{6}{7} = \frac{24}{7}$

The answer is: $\frac{24}{7}$

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Example Question #22 : Infinite Series

Determine the sum of: $1+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+...$

Possible Answers:

$\textup{The series will diverge.}$

$\frac{7}{6}$

$\frac{1}{6}$

$\frac{13}{6}$

Correct answer:

$\frac{7}{6}$

Explanation:

Notice that there is a pattern only after the second term. Rewrite the infinite series.

$1+(\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+...)$

We can find the common difference among the second and third term, and with the third term and fourth term, but there is no relation with the first and second term.

To determine the common difference, we will need to divide the third term with the second term or the fourth term with the third.

$r= \frac{1}{24} \div\frac{1}{12} =\frac{1}{24} \times\frac{12} {1} = \frac{1}{2}$

Write the formula for the infinite series.

$S=\frac{\textup{first term}}{1-r}$

The first term is NOT 1! The first term is the starting value of the series, which is $\frac{1}{12}$ . Substitute the terms.

$S=\frac{\frac{1}{12}}{1-\frac{1}{2}} =\frac{1}{12}\div \frac{1}{2} = \frac{1}{12}\times \frac{2}{1} = \frac{1}{6}$

This means that:

$1+(\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+...) = 1+\frac{1}{6} = \frac{7}{6}$

The answer is: $\frac{7}{6}$

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Example Question #21 : Infinite Series

Determine the sum, if possible: $9+1+\frac{1}{6}+\frac{1}{36}+...$

Possible Answers:

$\frac{33}{2}$

$\frac{51}{5}$

$\frac{15}{2}$

$\frac{61}{6}$

$\textup{The series will diverge.}$

Correct answer:

$\frac{51}{5}$

Explanation:

Notice that the series start after the nine. This means that:

$9+[1+\frac{1}{6}+\frac{1}{36}+...]$

Write the formula to determine the sum of the infinite series.

$S=\frac{a_1}{1-r}$

The first term is: a_1 =1

The common ratio for each term is: $r= \frac{1}{6}$

Substitute the terms.

The expression becomes:

$9+[1+\frac{1}{6}+\frac{1}{36}+...] = 9+[\frac{1}{1-\frac{1}{6}}] = 9+\frac{1}{\frac{5}{6}}$

Simplify the complex fraction.

$\frac{1}{\frac{5}{6}} = 1\div \frac{5}{6} = 1\times \frac{6}{5}=\frac{6}{5}$

The terms become:

$9+\frac{6}{5} = \frac{45}{5}+\frac{6}{5} = \frac{51}{5}$

The answer is: $\frac{51}{5}$

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Example Question #2781 : Algebra Ii

Evaluate:

$\sum_{k = 1}^{\infty} \frac{1}{k^{2}+k}$

Possible Answers:

$\frac{1}{2}$

The series diverges.

Correct answer:

Explanation:

A trick to working this is to rewrite the expression

$\frac{1}{k^{2}+k}$

by, first, factoring the denominator:

$= \frac{1}{k (k+1)}$

Using the method of partial fractions, we can rewrite this as

$= \frac{A}{k } + \frac{B}{k+1}$

$= \frac{A (k+1)}{k (k+1) } + \frac{B k}{k (k+1)}$

$= \frac{A k+A}{k (k+1) } + \frac{B k}{k (k+1)}$

$= \frac{(A+B) k+A}{k (k+1) }$

Comparing numerators, we get

(A+B) k+A = 1 ,

A = 1

and

A+ B = 0

1+ B = 0

B = -1

and the series can be restated as

$\sum_{k = 1}^{\infty} \left ( \frac{1}{k} - \frac{1}{k+1} \right )$

This can be rewritten by substituting the integers from 1 to 10, in turn, and adding:

$=\left ( 1 - \frac{1}{2}\right ) + \left ( \frac{1}{2} - \frac{1}{3}\right )+ \left ( \frac{1}{3} - \frac{1}{4}\right )+ ... .$

Regrouping, we see this is a telescoping series, in which all numbers after 1 cancel out:

$=1+\left ( - \frac{1}{2} + \frac{1}{2} \right ) +\left ( - \frac{1}{3} + \frac{1}{3} \right )+\left ( - \frac{1}{4} + \frac{1}{4} \right ) +...$

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Example Question #2782 : Algebra Ii

Evaluate: $\sum_{k = 0}^{\infty} (-1)^{k} \frac{3}{\pi ^{k}}$

Possible Answers:

$\frac{3 \pi }{ \pi+1 }$

$\frac{ \pi }{3 \pi-1 }$

The series diverges.

$\frac{ \pi }{3 \pi+1 }$

$\frac{3 \pi }{ \pi-1 }$

Correct answer:

$\frac{3 \pi }{ \pi+1 }$

Explanation:

Rewrite the general term so that the sigma notation is as follows:

$\sum_{k = 0}^{\infty} (-1)^{k} \frac{3}{\pi ^{k}}$

$= \sum_{k = 0}^{\infty} (-1)^{k} \cdot 3 \cdot \frac{1}{\pi ^{k}}$

$= \sum_{k = 0}^{\infty} 3 \cdot (-1)^{k} \cdot \left ( \frac{1}{\pi } \right )^{k}$

$= \sum_{k = 0}^{\infty} 3 \left (- \frac{1}{\pi } \right )^{k}$

A series of the form $\sum_{k = 0}^{\infty}a r^{k}$ is an infinite geometric series with common ratio . If |r| < 1 and the initial term is $a_{0}$ , the sum of the series is

$S = \frac{a_{0}}{1-r}$

Here, $r = - \frac{1}{\pi}$ , a = 3 , and

$a_{0}= (-1)^{0} \frac{3}{\pi ^{0}} = 1 \cdot \frac{3}{1} = 3$

Substitute these values in the formula:

$S = \frac{a_{0}}{1-r} = \frac{3}{1- \left ( -\frac{1}{\pi}\right )} = \frac{3}{1+\frac{1}{\pi}}$

Simplifying:

$S = \frac{3}{1+\frac{1}{\pi}}$

$= \frac{3 \cdot \pi }{\left (1+\frac{1}{\pi} \right ) \cdot \pi}$

$= \frac{3 \pi }{1\cdot \pi+\frac{1}{\pi}\cdot \pi }$

$= \frac{3 \pi }{ \pi+1 }$

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Example Question #2783 : Algebra Ii

Evaluate: $\sum_{k = 0}^{\infty}7\left ( -\frac{6}{5} \right )^{k}$

Possible Answers:

$3\frac{4}{11}$

$3\frac{9}{11}$

The series diverges.

Correct answer:

The series diverges.

Explanation:

A series of the form $\sum_{k = 0}^{\infty}a r^{k}$ is an infinite geometric series with common ratio . The series converges to a sum if and only if |r|< 1 . However, in the given expression, $r = -\frac{6}{5}$ , so $|r| =\left | -\frac{6}{5} \right | = \frac{6}{5} > 1$ , so the series diverges rather than converges.

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Example Question #2784 : Algebra Ii

Evaluate:

$\sum_{k = 0}^{\infty} \frac{4}{e^{k}}$

Possible Answers:

$\frac{e+1}{4 e }$

$\frac{4 e }{e+1}$

$\frac{e-1}{4 e }$

$\frac{4 e }{e-1}$

Cannot be determined

Correct answer:

$\frac{4 e }{e-1}$

Explanation:

Rewrite the general term so that the sigma notation is as follows:

$\sum_{k = 0}^{\infty} \frac{4}{e^{k}} = \sum_{k = 0}^{\infty}\left ( 4 \cdot \frac{1}{e^{k}} \right )= \sum_{k = 0}^{\infty}4 \cdot \left ( \frac{1}{e} \right )^{k}$

A series of the form $\sum_{k = 0}^{\infty}a r^{k}$ is an infinite geometric series with common ratio . If |r| < 1 and the initial term is $a_{0}$ , the sum of the series is

$S = \frac{a_{0}}{1-r}$

Here, $r = \frac{1}{e}$ , a = 4 , and

$a_{0}= \frac{4}{e^{0}} = \frac{4}{1} = 4$

Substitute these values in the formula:

$S = \frac{a_{0}}{1-r}$

$= \frac{4}{1- \frac{1}{e}}$

Simplifying:

$S = \frac{4 \cdot e }{\left (1- \frac{1}{e} \right )\cdot e}$

$S = \frac{4 e }{e-1}$

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Example Question #2785 : Algebra Ii

Evaluate:

$\sum_{k = 0}^{\infty} \frac{5}{3^{k}}$

Possible Answers:

$7\frac{1}{2}$

$6\frac{2}{3}$

Cannot be determined

$3\frac{1}{3}$

$3\frac{3}{4}$

Correct answer:

$7\frac{1}{2}$

Explanation:

Rewrite the general term so that the sigma notation is as follows:

$\sum_{k = 0}^{\infty} \frac{5}{3^{k}} = \sum_{k = 0}^{\infty}\left ( 5 \cdot \frac{1}{3^{k}} \right )= \sum_{k = 0}^{\infty}5 \cdot \left ( \frac{1}{3} \right )^{k}$

A series of the form $\sum_{k = 0}^{\infty}a r^{k}$ is an infinite geometric series with common ratio . If |r| < 1 and the initial term is $a_{0}$ , the sum of the series is

$S = \frac{a_{0}}{1-r}$

Here, $r = \frac{1}{3}$ , a =5 , and

$a_{0}= \frac{5}{3^{0}} = \frac{5}{1} = 5$

Substitute these values in the formula:

$S = \frac{a_{0}}{1-r}= \frac{5}{1- \frac{1}{3}} = \frac{5}{ \frac{2}{3}}= 5 \div \frac{2}{3} = \frac{5}{1} \cdot\frac{3}{2} = \frac{15}{2}= 7\frac{1}{2}$ .

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Algebra II : Infinite Series

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #21 : Infinite Series

Example Question #22 : Infinite Series

Example Question #21 : Infinite Series

Example Question #2781 : Algebra Ii

Example Question #2782 : Algebra Ii

Example Question #2783 : Algebra Ii

Example Question #2784 : Algebra Ii

Example Question #2785 : Algebra Ii

All Algebra II Resources

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