Question 1

For the given geometric sequence, find the limit of the infinite series, if it exists.
-$2 \cdot \left( \frac { 1 } { 2 } \right) ^ { \mathrm { n } - 1 }$

Accepted Answer

The answer of For the given geometric sequence, find the...

Question 2

For the given geometric sequence, find the limit of the infinite series, if it exists.
-$$45,15,5 , \frac { 5 } { 3 } , \ldots$$

Accepted Answer

The given sequence is a geometric series with the first term $a = 45$ and common ratio $r = \frac{1}{3}$. The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$ if $|r| < 1$. Substituting $a = 45$ and $r = \frac{1}{3}$, we get $S = \frac{45}{1 - \frac{1}{3}} = \frac{45}{\frac{2}{3}} = \frac{45 \times 3}{2} = \frac{135}{2}$.

Question 3

Solve the problem.
-A town has a population of 1000 people and is increasing by 9% every year. What will the population be at the end of 8 years? Round to the nearest unit.

Accepted Answer

The answer of Solve the problem.
-A town has a population...

Question 4

For the given geometric sequence, find the limit of the infinite series, if it exists.
-$$a _ { n } = 5 \cdot ( 2 ) ^ { n }$$

Accepted Answer

The series diverges to infinity because as n approaches infinity, the terms of the sequence grow without bound.

Question 5

For the given geometric sequence, find the limit of the infinite series, if it exists.
-$\frac { 2 } { 3 } \cdot \left( - \frac { 1 } { 3 } \right) ^ { n - 1 }$

Accepted Answer

The answer of For the given geometric sequence, find the...

Question 6

Simplify.
-$\frac { 8 ! } { 5 ! \cdot 3 ! }$

Accepted Answer

The expression simplifies to $\frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$, since the $5!$ in the denominator cancels out the $5 \times 4 \times 3 \times 2 \times 1$ part of $8!$ in the numerator.

Question 7

Solve the problem.
-A ball is dropped from a height of 53 feet. On each bounce, it bounces to 72% of its previous height. How high does it get after the 4th bounce? (Round to the nearest hundredth, if necessary.)

Accepted Answer

The height after each bounce can be calculated by multiplying the height of the previous bounce by 0.72. After the 4th bounce, the height is $53 \times 0.72^4 \approx 14.24$ feet.

Question 8

Find the partial sum of the geometric sequence with the given first term and common ratio.
-$9 , a _ { 1 } = \frac { 1 } { 4 } , r = - 3$

Accepted Answer

The formula for the sum of the first $n$ terms of a geometric sequence is $S_n = \frac{a_1(1 - r^n)}{1 - r}$, where $a_1$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. Given $a_1 = \frac{1}{4}$ and $r = -3$, and since we're looking for the sum of the first 9 terms, we substitute $n = 9$, $a_1 = \frac{1}{4}$, and $r = -3$ into the formula to get $S_9 = \frac{\frac{1}{4}(1 - (-3)^9)}{1 - (-3)}$. Simplifying this gives $S_9 = \frac{\frac{1}{4}(1 - (-19683))}{4}$, which simplifies further to $S_9 = \frac{4921}{4}$.

Question 9

Simplify.
-10!

Accepted Answer

10 factorial (10!) means 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, which equals 3,628,800.

Question 10

For the given geometric sequence, find the limit of the infinite series, if it exists.
-$$- 15 , - 5 , - \frac { 5 } { 3 } , - \frac { 5 } { 9 } , \ldots$$

Accepted Answer

The given sequence is a geometric series with the first term $a = -15$ and common ratio $r = \frac{1}{3}$. The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$ if $|r| < 1$. Substituting $a = -15$ and $r = \frac{1}{3}$, we get $S = \frac{-15}{1 - \frac{1}{3}} = \frac{-15}{\frac{2}{3}} = -\frac{45}{2}$.

Question 11

Find the indicated partial sum for the given geometric sequence.
-s8

Accepted Answer

The answer of Find the indicated partial sum for the...

Question 12

Find the indicated partial sum for the given geometric sequence.
-$\mathrm { s } _ { 6 } , \frac { 1 } { 2 } , \frac { 1 } { 4 } , \frac { 1 } { 8 } , \frac { 1 } { 16 } , \ldots$

Accepted Answer

The formula for the sum of a finite geometric sequence is 
$$S_n = \frac{a(1-r^n)}{1-r},$$
where $a$ is the first term, $r$ is the common ratio and $n$ is the number of terms.

In this case, we have $a = \mathrm{s}_1 = 1$, $r = \frac12$ and $n = 6$, so the sum of the first six terms is 
$$S_6 = \frac{1(1-1/2^6)}{1-1/2} = \frac{63}{64}.$$

Therefore, the answer is (C).

Question 13

Simplify.
-$\frac { 8 ! } { 6 ! }$

Accepted Answer

Simplifying the expression, we get:

\begin{align*}
\frac{8!}{6!}&=\frac{8	imes 7	imes 6	imes 5	imes 4	imes 3	imes 2	imes 1}{6	imes 5	imes 4	imes 3	imes 2	imes 1}\
&=\frac{8	imes 7}{1	imes 2}\
&=56
\end{align*}

Therefore, the answer is D) 56.

Question 14

Simplify.
-$\frac { 4 ! } { 6 ! }$

Accepted Answer

We can simplify the numerator as follows: 
$$4!=4\cdot3\cdot2\cdot1=24$$
We can simplify the denominator as follows:
$$6!=6\cdot5\cdot4\cdot3\cdot2\cdot1=720$$
Therefore, 
$$\frac{4!}{6!}=\frac{24}{720}=\frac{1}{30}$$
So the best choice is B, which simplifies the expression correctly.

Question 15

Find the partial sum of the geometric sequence with the given first term and common ratio.
-$\mathrm { s } 5 , \mathrm { a } 1 = 1 , \mathrm { r } = \frac { 1 } { 2 }$

Accepted Answer

The formula for the sum of the first $n$ terms of a geometric sequence is $S_n = \frac{a_1(1 - r^n)}{1 - r}$, where $a_1$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. Plugging in $a_1 = 1$, $r = \frac{1}{2}$, and $n = 5$, we get $S_5 = \frac{1(1 - (\frac{1}{2})^5)}{1 - \frac{1}{2}} = \frac{1 - \frac{1}{32}}{\frac{1}{2}} = \frac{31}{32} \times 2 = \frac{31}{16}$.

Question 16

For the given geometric sequence, find the limit of the infinite series, if it exists.
-$$4 , - \frac { 4 } { 3 } , \frac { 4 } { 9 } , - \frac { 4 } { 27 } , \ldots$$

Accepted Answer

The given sequence is a geometric series with the first term $a = 4$ and common ratio $r = -\frac{1}{3}$. The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$ if $|r| < 1$. Substituting $a = 4$ and $r = -\frac{1}{3}$, we get $S = \frac{4}{1 - (-\frac{1}{3})} = \frac{4}{\frac{4}{3}} = 3$.

Question 17

Solve the problem.
-Suppose you invest $9197 at the end of each year into an account earning 8% interest (compounded annually). At the end of 8 years, how much money will be in the account?

Accepted Answer

To solve this problem, we will use the formula for the future value of an annuity:

FV = PMT x ((1 + r)^n - 1) / r

Where:
- PMT = the periodic payment (or deposit)
- r = the interest rate per period (in this case, per year)
- n = the number of periods (in this case, the number of years)

Plugging in the given values, we get:

FV = $9,197 x ((1 + 0.08)^8 - 1) / 0.08
FV = $9,197 x (1.08^8 - 1) / 0.08
FV = $9,197 x (2.158924 - 1) / 0.08
FV = $9,197 x 16.98655
FV = $156,235.85

Therefore, the amount of money in the account at the end of 8 years will be $156,235.85.

The correct answer choice is C ($97,825.06) which is the nearest value to our answer after rounding it to two decimal places.

Question 18

Find the indicated partial sum for the given geometric sequence.
-s4, 16, 32, 64, 128, ...

Accepted Answer

The answer of Find the indicated partial sum for the...

Question 19

For the given geometric sequence, find the limit of the infinite series, if it exists.
--2, 2, -2, 2, . . .

Accepted Answer

The sequence alternates between -2 and 2, so the partial sums oscillate between -2 and 0. Since the partial sums do not converge to a single value, the infinite series does not have a limit.

Question 20

For the given geometric sequence, find the limit of the infinite series, if it exists.
-$5 , \frac { 5 } { 3 } , \frac { 5 } { 9 } , \frac { 5 } { 27 } , \ldots$

Accepted Answer

Notice that the common ratio between two consecutive terms in the sequence is $\frac {1}{3}$. Let the infinite series be denoted as S, then 
$S = 5 + \frac{5}{3} + \frac{5}{9} + \frac{5}{27} + \cdots$
Multiplying both sides by the common ratio:
$ \frac{1}{3} S = \frac{5}{3} + \frac{5}{9} + \frac{5}{27} + \frac{5}{81} + \cdots$
Subtracting these equations gives:
$ S - \frac{1}{3}S = 5$
Solving for S, we get:
$S = \boxed{\frac{15}{2}}$
Thus, the answer is choice C).

Quiz 14: Sequences, Series, and the Binomial Theorem

For the given geometric sequence, find the limit of the infinite series, if it exists. - $2 \cdot \left( \frac { 1 } { 2 } \right) ^ { \mathrm { n } - 1 }$

For the given geometric sequence, find the limit of the infinite series, if it exists. - $45,15,5 , \frac { 5 } { 3 } , \ldots$

Solve the problem. -A town has a population of 1000 people and is increasing by 9% every year. What will the population be at the end of 8 years? Round to the nearest unit.

For the given geometric sequence, find the limit of the infinite series, if it exists. - $a _ { n } = 5 \cdot ( 2 ) ^ { n }$

For the given geometric sequence, find the limit of the infinite series, if it exists. - $\frac { 2 } { 3 } \cdot \left( - \frac { 1 } { 3 } \right) ^ { n - 1 }$

Simplify. - $\frac { 8 ! } { 5 ! \cdot 3 ! }$

Solve the problem. -A ball is dropped from a height of 53 feet. On each bounce, it bounces to 72% of its previous height. How high does it get after the 4th bounce? (Round to the nearest hundredth, if necessary.)

Find the partial sum of the geometric sequence with the given first term and common ratio. - $9 , a _ { 1 } = \frac { 1 } { 4 } , r = - 3$

Simplify. -10!

For the given geometric sequence, find the limit of the infinite series, if it exists. - $- 15 , - 5 , - \frac { 5 } { 3 } , - \frac { 5 } { 9 } , \ldots$

Find the indicated partial sum for the given geometric sequence. -s8

Find the indicated partial sum for the given geometric sequence. - $\mathrm { s } _ { 6 } , \frac { 1 } { 2 } , \frac { 1 } { 4 } , \frac { 1 } { 8 } , \frac { 1 } { 16 } , \ldots$

Simplify. - $\frac { 8 ! } { 6 ! }$

Simplify. - $\frac { 4 ! } { 6 ! }$

Find the partial sum of the geometric sequence with the given first term and common ratio. - $\mathrm { s } 5 , \mathrm { a } 1 = 1 , \mathrm { r } = \frac { 1 } { 2 }$

For the given geometric sequence, find the limit of the infinite series, if it exists. - $4 , - \frac { 4 } { 3 } , \frac { 4 } { 9 } , - \frac { 4 } { 27 } , \ldots$

Solve the problem. -Suppose you invest $9197 at the end of each year into an account earning 8% interest (compounded annually) . At the end of 8 years, how much money will be in the account?

Find the indicated partial sum for the given geometric sequence. -s4, 16, 32, 64, 128, ...

For the given geometric sequence, find the limit of the infinite series, if it exists. --2, 2, -2, 2, . . .

For the given geometric sequence, find the limit of the infinite series, if it exists. - $5 , \frac { 5 } { 3 } , \frac { 5 } { 9 } , \frac { 5 } { 27 } , \ldots$

Quiz 14: Sequences, Series, and the Binomial Theorem

For the given geometric sequence, find the limit of the infinite series, if it exists. - 2⋅(12) n−12 \cdot \left( \frac { 1 } { 2 } \right) ^ { \mathrm { n } - 1 }2⋅(21​) n−1

For the given geometric sequence, find the limit of the infinite series, if it exists. - 45,15,5,53,…45,15,5 , \frac { 5 } { 3 } , \ldots45,15,5,35​,…

Solve the problem. -A town has a population of 1000 people and is increasing by 9% every year. What will the population be at the end of 8 years? Round to the nearest unit.

For the given geometric sequence, find the limit of the infinite series, if it exists. - an=5⋅(2) na _ { n } = 5 \cdot ( 2 ) ^ { n }an​=5⋅(2) n

For the given geometric sequence, find the limit of the infinite series, if it exists. - 23⋅(−13) n−1\frac { 2 } { 3 } \cdot \left( - \frac { 1 } { 3 } \right) ^ { n - 1 }32​⋅(−31​) n−1

Simplify. - 8!5!⋅3!\frac { 8 ! } { 5 ! \cdot 3 ! }5!⋅3!8!​