Geometric Series - AP Calculus BC

Card 0 of 70

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?

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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$

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?

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

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).

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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.

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

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

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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

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}$$

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

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This is a geometric series.

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

, where n is the number of terms to sum up, r is the common ratio, and is the value of the first term.

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

Solution:

$S_${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}$$})$

Rounding,

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

,, $r=\frac{2}{20}$$ ,

rounded to the nearest integer.

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This is a geometric series.

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

, where n is the number of terms to sum up, r is the common ratio, and is the value of the first term.

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

Solution:

$S_${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}$$})$

Rounding,

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

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This is a geometric series.

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

, where n is the number of terms to sum up, r is the common ratio, and is the value of the first term.

We have and n and we just need to find r before calculating the sum.

Solution:

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

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

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?

Tap to see back →

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$

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?

Tap to see back →

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

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).

Tap to see back →

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.

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

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

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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

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}$$

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

Tap to see back →

This is a geometric series.

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

, where n is the number of terms to sum up, r is the common ratio, and is the value of the first term.

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

Solution:

$S_${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}$$})$

Rounding,

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

,, $r=\frac{2}{20}$$ ,

rounded to the nearest integer.

Tap to see back →

This is a geometric series.

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

, where n is the number of terms to sum up, r is the common ratio, and is the value of the first term.

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

Solution:

$S_${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}$$})$

Rounding,

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

Tap to see back →

This is a geometric series.

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

, where n is the number of terms to sum up, r is the common ratio, and is the value of the first term.

We have and n and we just need to find r before calculating the sum.

Solution:

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

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

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?

Tap to see back →

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$

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?

Tap to see back →

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

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).

Tap to see back →

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.

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

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

Tap to see back →

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

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}$$

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

Tap to see back →

This is a geometric series.

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

, where n is the number of terms to sum up, r is the common ratio, and is the value of the first term.

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

Solution:

$S_${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}$$})$

Rounding,

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

,, $r=\frac{2}{20}$$ ,

rounded to the nearest integer.

Tap to see back →

This is a geometric series.

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

, where n is the number of terms to sum up, r is the common ratio, and is the value of the first term.

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

Solution:

$S_${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}$$})$

Rounding,