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Algebra II : Mathematical Relationships and Basic Graphs

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Example Questions

Example Question #111 : Mathematical Relationships And Basic Graphs

Calculate the sum of the following series, if it exists:

$6+2+\frac{2}{3}+\frac{2}{9}+...$

Possible Answers:

Correct answer:

Explanation:

Each successive term in this series is determined by multiplying the previous term by $\frac{1}{3}$ ; hence, the series is geometric and its common ratio is $r=\frac{1}{3}$ .

Since $|r|=\frac{1}{3}<1$ , the sum of this series exists and we can calculate it via the following formula:

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

where is the desired sum of the geometric series, is the common ratio of the geometric series, and a_1 is the first term appearing in the series.

Here, $r=\frac{1}{3}$ and a_1=6 . Hence, the sum of this geometric series is

$S=\frac{6}{1-\frac{1}{3}}=\frac{6}{\frac{2}{3}}=\frac{(6)(3)}{2}=9$ .

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Example Question #11 : Infinite Series

Determine the sum: $1-\frac{1}{11}+\frac{1}{121}-...$

Possible Answers:

$\frac{121}{12}$

$\frac{12}{11}$

$\frac{11} {12}$

$\frac{6}{5}$

$\frac{11}{10}$

Correct answer:

$\frac{11} {12}$

Explanation:

Write the formula for the sum of an infinite series.

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

The is the common ratio between the terms in the series. This can be found by dividing the second term with the first term, the third term with the second, and so forth.

$\frac{-\frac{1}{11}}{1} =-\frac{1}{11}$

$\frac{\frac{1}{121}}{-\frac{1}{11}}=\frac{1}{121} \div -\frac{1}{11} = \frac{1}{121} \times -\frac{11}{1}=-\frac{1}{11}$

Substitute the known terms into the equation.

$\textup{Sum} = \frac{1}{1-(-\frac{1}{11})} = \frac{1}{1+\frac{1}{11}} =\frac{1}{\frac{11}{11}+\frac{1}{11}} =\frac{1}{\frac{12}{11}}$

Simplify the complex fraction.

$\frac{1}{\frac{12}{11}} = 1\div \frac{12}{11} = 1\times \frac{11} {12} = \frac{11} {12}$

The answer will converge to: $\frac{11} {12}$

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Example Question #17 : Infinite Series

Determine the sum, if possible: $2-\frac{1}{3}+\frac{1}{18}-...$

Possible Answers:

$\frac{12}{7}$

$\textup{The series will diverge.}$

$\frac{5}{3}$

$-\frac{1}{6}$

$\frac{8}{5}$

Correct answer:

$\frac{12}{7}$

Explanation:

Write the formula for the sum of an infinite series.

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

The first term is two. To determine the common ratio, we will need to divide the second term by the first, third term with the second, and so forth. The common ratio should be same for each term.

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

$r=\frac{\frac{1}{18}}{-\frac{1}{3}} = \frac{1}{18}\div -\frac{1}{3} = \frac{1}{18}\times -\frac{3}{1} =-\frac{1}{6}$

The common ratios are verified to be the same. Substitute the into the formula. This value must be between negative one and one or the series will diverge!

$\textup{Sum} =\frac{\textup{first term}}{1-r} = \frac{2}{1-(-\frac{1}{6})} = \frac{2}{\frac{7}{6}}$

Simplify this complex fraction.

$2\div \frac{7}{6} =2\times \frac{6}{7} =\frac{12}{7}$

The series will converge to $\frac{12}{7}$ .

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Example Question #12 : Infinite Series

Determine the sum, if possible: $1+\frac{1}{8}+\frac{1}{64}+...$

Possible Answers:

$\textup{The series will diverge.}$

$\frac{12}{11}$

$\frac{4}{3}$

$\frac{8}{7}$

$\frac{8}{5}$

Correct answer:

$\frac{8}{7}$

Explanation:

Write the formula for the sum of an infinite series.

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

Determine the common ratio, . Divide the second term by the first. This ratio should be the same if the third term was divided by the second term, and so forth.

$\frac{\frac{1}{8}}{1} = \frac{1}{8}$

The common ratio is $\frac{1}{8}$ .

Substitute the known terms into the equation.

$\textup{Sum} =\frac{1}{1-\frac{1}{8}} = \frac{1}{\frac{7}{8}}$

Solve the complex fraction.

$\frac{1}{\frac{7}{8}} = 1\div \frac{7}{8} = 1\times \frac{8}{7} = \frac{8}{7}$

The answer is: $\frac{8}{7}$

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Example Question #111 : Mathematical Relationships And Basic Graphs

Find the sum, if possible: $5+\frac{3}{5}+\frac{9}{125}+...$

Possible Answers:

$\frac{125}{3}$

$\frac{709}{125}$

$\frac{125}{22}$

$\frac{57}{10}$

$\textup{The series will diverge.}$

Correct answer:

$\frac{125}{22}$

Explanation:

Write the formula for infinite series.

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

Determine the common ratio by dividing the second term with the first term.

$\frac{3}{5}\div 5 = \frac{3}{5}\times \frac{1}{5} = \frac{3}{25}$

The ratio should also be similar if the third term is divided by the second term. Verify that the common ratios are similar.

$\frac{9}{125}\div \frac{3}{5} = \frac{9}{125}\times \frac{5}{3} = \frac{3}{25}$

Substitute the known terms into the equation.

$\textup{Sum} =\frac{\textup{first term}}{1-r} = \frac{5}{1-\frac{3}{25}} = \frac{5}{\frac{25}{25}-\frac{3}{25}}= \frac{5}{\frac{22}{25}}$

Simplify the complex fraction.

$5\div \frac{22}{25} = 5\times \frac{25}{22} = \frac{125}{22}$

The sum will converge to: $\frac{125}{22}$

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Example Question #20 : Infinite Series

Determine the sum, if possible: $3-\frac{1}{4}+\frac{1}{48}-...$

Possible Answers:

$\textup{The series will diverge.}$

$\frac{17}{6}$

$\frac{61}{23}$

$\frac{31}{11}$

$\frac{36}{13}$

Correct answer:

$\frac{36}{13}$

Explanation:

Find the common ratio of the infinite series. Divide the second term with the first, or the third with the second term. The common ratios should be similar to each other.

$-\frac{1}{4} \div 3 = -\frac{1}{4} \times \frac{1}{3} = -\frac{1}{12}$

$\frac{1}{48}\div -\frac{1}{4} =\frac{1}{48}\times- \frac{4}{1}=- \frac{1}{12}$

Write the formula for infinite series and substitute the numbers.

$\textup{Sum} =\frac{\textup{first term}}{1-r}=\frac{3}{1-(-\frac{1}{12})}$

Simplify the denominator.

$\frac{3}{\frac{12}{12}+\frac{1}{12}} = \frac{3}{\frac{13}{12}}$

$\frac{3}{\frac{13}{12}} =3\times \frac{12}{13} = \frac{36}{13}$

The sum converges to $\frac{36}{13}$ .

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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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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #111 : Mathematical Relationships And Basic Graphs

Example Question #11 : Infinite Series

Example Question #17 : Infinite Series

Example Question #12 : Infinite Series

Example Question #111 : Mathematical Relationships And Basic Graphs

Example Question #20 : Infinite Series

Example Question #21 : Infinite Series

Example Question #22 : Infinite Series

Example Question #21 : Infinite Series

Example Question #2781 : Algebra Ii

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