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Algebra II : Summations and Sequences

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Example Questions

Example Question #1 : Infinite Series

Evaluate: $\sum_{n=1}^{\infty }\left(\frac{3}{4}\right)^n$

Possible Answers:

$\frac{10}{3}$

$\frac{5}{2}$

$\frac{8}{3}$

Correct answer:

Explanation:

Write the formula for infinite geometric series.

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

The value of a_1 is the first term of the series, which is $\frac{3}{4}$ .

The value of the common ratio, , is also $\frac{3}{4}$ .

The ratio is $\frac{3}{4}$ because if we were to write out the first few terms in the series we would see,

$\frac{3}{4}^1+\frac{3}{4}^2+ \frac{3}{4}^3+...$

each term is three fourths more than the previous term therefore, giving us the ratio.

Substitute the values into the equation and evaluate.

$S=\frac{\frac{3}{4}}{1-\frac{3}{4}} = \frac{\frac{3}{4}}{\frac{1}{4}} =\frac{3}{4}\cdot4 = 3$

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Example Question #6 : Infinite Series

If a_1= 5 and $r=\frac{1}{4}$ , what will be the sum of the infinite series?

Possible Answers:

$\frac{20}{3}$

$\frac{15}{4}$

$\text{Diverges}$

$\frac{19}{2}$

Correct answer:

$\frac{20}{3}$

Explanation:

Write the infinite series formula.

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

Substitute the values of a_1 and .

$S=\frac{a_1}{1-r}= \frac{5}{1-\frac{1}{4}} = \frac{5}{\frac{3}{4}} = 5\cdot\frac{4}{3}=\frac{20}{3}$

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Example Question #7 : Infinite Series

Evaluate the infinite series for 10 - 9 + 8.1 - 7.29 + ...

Possible Answers:

5.263

0.19

100

Correct answer:

5.263

Explanation:

The first term of this sequence is 10. To find the common ratio r, we can just divide the second term by the first: $-9 \div 10 = - 0.9$ . So "r" is -0.9. We can find the infinite sum using the formula $\frac{a}{1-r}$ where a is the first term and r is the common ratio:

$\frac{10}{1-(-0.9)} = \frac{10}{1+0.9 } = \frac{10}{1.9} \approx 5.263$

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Example Question #2 : Infinite Series

Find the sum of the infinite series 10 + 11 + 12.1 + 13.31 + ...

Possible Answers:

Cannot be determined - the sum is infinite

100

Correct answer:

Cannot be determined - the sum is infinite

Explanation:

An infinite sum is only calculable if $\left | r \right | < 1$ where r is the common ratio. We can find the common ratio easily by dividing the second term by the first: $11 \div 10 = 1.1$ . This is greater than 1, so we can't find the infinite sum - it is infinite.

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Example Question #9 : Infinite Series

What is the sum? $\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...$

Possible Answers:

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{2}{5}$

$\textup{The series will diverge.}$

Correct answer:

$\frac{1}{4}$

Explanation:

Write the formula to find the sum of an infinite geometric series.

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

The first term is: $a_1 = \frac{1}{8}$ .

The common ratio is: $r=\frac{1}{2}$

Substitute the values into the formula.

$S=\frac{\frac{1}{8}}{1-\frac{1}{2}} = \frac{\frac{1}{8}}{\frac{1}{2}}$

Rewrite the complex fraction.

$S=\frac{1}{8}\div \frac{1}{2} = \frac{1}{8} \times 2 = \frac{1}{4}$

The sum will converge to $\frac{1}{4}$ .

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Example Question #10 : Infinite Series

What is the sum? $4-2+1-\frac{1}{2}+\frac{1}{4}-...$

Possible Answers:

$\frac{2}{3}$

$\frac{8}{3}$

$\frac{3}{8}$

$\textup{The series does not converge.}$

$\frac{1}{8}$

Correct answer:

$\frac{8}{3}$

Explanation:

Write the formula for the sum of an infinite series.

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

The value a_1 is the first term, and is the common ratio.

a_1=4

Divide the second term with the first term, third term and the second, and so forth, and we will get a common ratio of:

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

Substitute the values into the formula.

$S=\frac{a_1}{1-r}= \frac{4}{1-(-\frac{1}{2})} = \frac{4}{\frac{3}{2}}$

Rewrite the complex fraction using a division sign.

$\frac{4}{\frac{3}{2}} = 4 \div \frac{3}{2}$

Change the division sign to a multiplication and take the reciprocal of the second term.

$4 \times \frac{2}{3} = \frac{8}{3}$

The series will converge to $\frac{8}{3}$ .

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Example Question #2761 : Algebra Ii

Find the sum if the series converges: $10-6+\frac{18}{5}-\frac{54}{25}+...$

Possible Answers:

$\frac{19}{3}$

$\textup{The series will diverge.}$

$\frac{28}{3}$

$\frac{25}{4}$

Correct answer:

$\frac{25}{4}$

Explanation:

Write the formula for finding the sum of an infinite geometric series.

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

The first term a_1 is ten.

Find the common ratio by dividing the second term with the first term, third term with the second term, or the fourth term with the third term, and so forth.

The common ratio should all be the same after dividing each term.

$r=\frac{-6}{10} = -\frac{3}{5}$

As long as is between negative one and one, we can use the formula to find the sum. Substitute the givens into the equation and solve.

$S=\frac{10}{1-(-\frac{3}{5})} = \frac{10}{\frac{8}{5}} = 10\times \frac{5}{8}= \frac{50}{8}= \frac{25}{4}$

The answer is: $\frac{25}{4}$

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Example Question #2762 : Algebra Ii

Determine the sum, if the series converges: $2+\frac{2}{9}+\frac{2}{81}+...$

Possible Answers:

$\frac{9}{4}$

$\frac{11}{4}$

$\frac{7}{2}$

$\frac{5}{2}$

$\frac{7}{4}$

Correct answer:

$\frac{9}{4}$

Explanation:

Write the formula for an infinite series.

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

The term a_1 represents the first term. The value of the common ratio must be between negative one and positive one, and can be determined by dividing the second term with the first term, third term with the second term, and so forth.

$\frac{2}{9}\div 2 = \frac{2}{9} \times \frac{1}{2} =\frac{1}{9}$

$\frac{2}{81}\div\frac{2}{9} = \frac{2}{81}\times\frac{9}{2} = \frac{9}{81} = \frac{1}{9}$

We can see that the common ratio is: $r=\frac{1}{9}$

Substitute the known values into the formula.

$S=\frac{2}{1-\frac{1}{9}} = \frac{2}{\frac{8}{9}} = 2\div \frac{8}{9} = 2\times \frac{9}{8}= \frac{9}{4}$

The sum will converge to: $\frac{9}{4}$

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Example Question #2771 : Algebra Ii

Determine the sum of: $9+3+1+\frac{1}{3}+\frac{1}{9}+...$

Possible Answers:

$\frac{40}{3}$

$\frac{29}{2}$

$\frac{27}{2}$

Correct answer:

$\frac{27}{2}$

Explanation:

Notice that this an infinite series.

$9+3+1+\frac{1}{3}+\frac{1}{9}+...$

Find the common ratio by dividing the second term with the first term, third term with the second term, and so forth. The common ratio should be same for each term divided.

$\frac{3}{9}=\frac{\frac{1}{3}}{1} = \frac{1}{3}$

The common ratio is: $r=\frac{1}{3}$

Write the formula for the sum of an infinite series.

$S=\frac{\textup{first term}}{1-r} = \frac{9}{1-\frac{1}{3}} = \frac{9}{\frac{2}{3}} = \frac{27}{2}$

The answer is: $\frac{27}{2}$

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Example Question #14 : Infinite Series

Determine the sum of the infinite series: $1+\frac{5}{6}+\frac{25}{36}+...$

Possible Answers:

$\frac{50}{3}$

$\textup{The series will diverge.}$

$\frac{75}{9}$

Correct answer:

Explanation:

To find the sum, write the formula for infinite series.

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

Determine the common ratio by dividing the second term with the first term, or third term with the second term. They should have similar common ratios.

Simplify the complex fractions to verify that both have similar common ratios.

$\frac{(\frac{5}{6})}{1} = \frac{5}{6}$

$\frac{\frac{25}{36}}{\frac{5}{6}} = \frac{25}{36}\div \frac{5}{6} = \frac{25}{36}\times \frac{6}{5} = \frac{5}{6}$

Substitute the known terms into the formula.

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

Simplify the complex fraction.

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

The sum will converge to .

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Algebra II : Summations and Sequences

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Infinite Series

Example Question #6 : Infinite Series

Example Question #7 : Infinite Series

Example Question #2 : Infinite Series

Example Question #9 : Infinite Series

Example Question #10 : Infinite Series

Example Question #2761 : Algebra Ii

Example Question #2762 : Algebra Ii

Example Question #2771 : Algebra Ii

Example Question #14 : Infinite Series

All Algebra II Resources

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