Geometric Series - AP Calculus BC

Card 0 of 14

Question

Consider: $\sum_{n=0}^{\infty} \left (\frac{\sqrt 2}{2}\right )^n$ . Will the series converge or diverge? If converges, where does this coverge to?

Answer

This is a geometric series. Use the following formula, where is the first term of the series, and is the ratio that must be less than 1. If is greater than 1, the series diverges.

$\frac{a}{1-r} = \frac{1}{1-\frac{\sqrt2}{2}}= \frac{1}{\frac{2-\sqrt2}{2}} = \frac{2}{2-\sqrt2}$

Rationalize the denominator.

$\frac{2}{2-\sqrt2} \cdot \frac{2+\sqrt2}{2+\sqrt2} = \frac{2(2+\sqrt2)}{4-2}= 2+\sqrt2$

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Question

Consider the following summation: $1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...$ . Does this converge or diverge? If it converges, where does it approach?

Answer

The problem can be reconverted using a summation symbol, and it can be seen that this is geometric.

$1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...=\sum_{n=0}^{\infty} \left ( \frac{1}{4} \right )^n$

Since the ratio is less than 1, this series will converge. The formula for geometric series is:

$\frac{a}{1-r}$

where is the first term, and is the common ratio. Substitute these values and solve.

$\frac{a}{1-r} = \frac{1}{1-\frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$

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Question

A worm crawls up a wall during the day and slides down slowly during the night. The first day the worm crawls one meter up the wall. The first night the worm slides down a third of a meter. The second day the worm regains one third of the lost progress and slides down one third of that distance regained on the second night. This pattern of motion continues...

Which of the following is a geometric sum representing the distance the worm has travelled after 12-hour periods of motion? (Assuming day and night are both 12 hour periods).

Answer

The sum must be alternating, and after one period you should have the worm at 1m. After two periods, the worm should be at 2/3m. There is only one sum for which that is true.

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Question

Determine whether the following series converges or diverges. If it converges, what does it converge to?

$\sum_{n=0}^{\infty}(2)^{-2n}$

Answer

First, we reduce the series into a simpler form.

$\sum_{n=0}^{\infty}(2)^{-2n}= \sum_{n=0}^{\infty}(\frac{1}{2})^{2n}= \sum_{n=0}^{\infty}(\frac{1}{4})^{n}$

We know this series converges because

$\lim_{n \rightarrow \infty}(\frac{1}{4})^{n}=0$

By the Geometric Series Theorem, the sum of this series is given by

$\sum_{n=0}^{\infty}(\frac{1}{4})^{n}= \frac{(\frac{1}{4})^0}{1-\frac{1}{4}}= \frac{4}{3}$

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Question

Calculate the sum of a geometric series with the following values:, $a_{1}=4$ , $r=\frac{1}{3}$ . Round the answer to the nearest integer.

Answer

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:

$S_{15}=\frac{a_{1}(1-r^n)}{1-r}$

$S_{15}=\frac{4(1-(\frac{1}{3})^{15})}{1-\frac{1}{3}}$

$S_{15}=\frac{43(1-\frac{1}{3^{15}})}{2}$

$S_{15}=6*(1-\frac{1}{3^{15}})$

$S_{15}=5.999...$

Rounding, $S_{15}=6$

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Question

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

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

rounded to the nearest integer.

Answer

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:

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

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

$S_{10}=\frac{11(1-(\frac{1}{10})^{10}}{\frac{9}{10}} = \frac{1110(1-\frac{1}{10^{10}})}{9}$

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

$S_{10}=12.2222...$

Rounding, $S_{10}=12$

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Question

Calculate the sum, rounded to the nearest integer, of the first 16 terms of the following geometric series:

Answer

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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Question

Consider: $\sum_{n=0}^{\infty} \left (\frac{\sqrt 2}{2}\right )^n$ . Will the series converge or diverge? If converges, where does this coverge to?

Answer

This is a geometric series. Use the following formula, where is the first term of the series, and is the ratio that must be less than 1. If is greater than 1, the series diverges.

$\frac{a}{1-r} = \frac{1}{1-\frac{\sqrt2}{2}}= \frac{1}{\frac{2-\sqrt2}{2}} = \frac{2}{2-\sqrt2}$

Rationalize the denominator.

$\frac{2}{2-\sqrt2} \cdot \frac{2+\sqrt2}{2+\sqrt2} = \frac{2(2+\sqrt2)}{4-2}= 2+\sqrt2$

Compare your answer with the correct one above

Question

Consider the following summation: $1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...$ . Does this converge or diverge? If it converges, where does it approach?

Answer

The problem can be reconverted using a summation symbol, and it can be seen that this is geometric.

$1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...=\sum_{n=0}^{\infty} \left ( \frac{1}{4} \right )^n$

Since the ratio is less than 1, this series will converge. The formula for geometric series is:

$\frac{a}{1-r}$

where is the first term, and is the common ratio. Substitute these values and solve.

$\frac{a}{1-r} = \frac{1}{1-\frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$

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Question

A worm crawls up a wall during the day and slides down slowly during the night. The first day the worm crawls one meter up the wall. The first night the worm slides down a third of a meter. The second day the worm regains one third of the lost progress and slides down one third of that distance regained on the second night. This pattern of motion continues...

Which of the following is a geometric sum representing the distance the worm has travelled after 12-hour periods of motion? (Assuming day and night are both 12 hour periods).

Answer

The sum must be alternating, and after one period you should have the worm at 1m. After two periods, the worm should be at 2/3m. There is only one sum for which that is true.

Compare your answer with the correct one above

Question

Determine whether the following series converges or diverges. If it converges, what does it converge to?

$\sum_{n=0}^{\infty}(2)^{-2n}$

Answer

First, we reduce the series into a simpler form.

$\sum_{n=0}^{\infty}(2)^{-2n}= \sum_{n=0}^{\infty}(\frac{1}{2})^{2n}= \sum_{n=0}^{\infty}(\frac{1}{4})^{n}$

We know this series converges because

$\lim_{n \rightarrow \infty}(\frac{1}{4})^{n}=0$

By the Geometric Series Theorem, the sum of this series is given by

$\sum_{n=0}^{\infty}(\frac{1}{4})^{n}= \frac{(\frac{1}{4})^0}{1-\frac{1}{4}}= \frac{4}{3}$

Compare your answer with the correct one above

Question

Calculate the sum of a geometric series with the following values:, $a_{1}=4$ , $r=\frac{1}{3}$ . Round the answer to the nearest integer.

Answer

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:

$S_{15}=\frac{a_{1}(1-r^n)}{1-r}$

$S_{15}=\frac{4(1-(\frac{1}{3})^{15})}{1-\frac{1}{3}}$

$S_{15}=\frac{43(1-\frac{1}{3^{15}})}{2}$

$S_{15}=6*(1-\frac{1}{3^{15}})$

$S_{15}=5.999...$

Rounding, $S_{15}=6$

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Question

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

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

rounded to the nearest integer.

Answer

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:

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

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

$S_{10}=\frac{11(1-(\frac{1}{10})^{10}}{\frac{9}{10}} = \frac{1110(1-\frac{1}{10^{10}})}{9}$

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

$S_{10}=12.2222...$

Rounding, $S_{10}=12$

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Question

Calculate the sum, rounded to the nearest integer, of the first 16 terms of the following geometric series:

Answer

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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