Algebra II - Other Sequences and Series

A sequence begins as follows:

$\frac{1}{5} , \frac{1}{6} , \frac{1}{7} , \frac{1}{8} , ...$

Which statement is true?

The sequence cannot be arithmetic or geometric.

The sequence may be geometric.

The sequence may be arithmetic.

All of these

None of these

Explanation

An arithmetic sequence is one in which each term is generated by adding the same number - the common difference - to the previous term. As can be seen here, the difference between each term and the previous term varies from term to term:

$\frac{1}{6} - \frac{1}{5} = \frac{5}{30} - \frac{6}{30} = -\frac{1}{30}$

$\frac{1}{7} - \frac{1}{6} = \frac{6}{42} - \frac{7}{42} = -\frac{1}{42}$

The sequence cannot be arithmetic.

A geometric sequence is one in which each term is generated by multiplying the previous term by the same number - the common ratio. As can be seen here, the ratio of each term to the previous one also varies from term to term:

$\frac{1}{6} \div \frac{1}{5} =\frac{1}{6} \times \frac{5} {1}= \frac{5}{6}$

$\frac{1}{7} \div \frac{1}{6} =\frac{1}{7} \times \frac{6} {1}= \frac{6}{7}$

The sequence cannot be geometric.

A sequence begins as follows:

Which statement is true?

The sequence cannot be arithmetic or geometric.

The sequence may be arithmetic.

The sequence may be geometric.

All of these

None of these

Explanation

The first difference:

The second difference:

The sequence cannot be arithmetic.

The first ratio: $\frac{4}{3}$

The second ratio: $\frac{6}{4}= \frac{3}{2}$

The sequence cannot be geometric.

A sequence begins as follows:

Which statement is true?

The sequence cannot be arithmetic or geometric.

The sequence may be geometric.

The sequence may be arithmetic.

All of these

None of these

Explanation

The sequence cannot be arithmetic.

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

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

The sequence cannot be geometric.

A sequence begins as follows:

Which statement is true?

The sequence cannot be arithmetic or geometric.

The sequence may be arithmetic.

The sequence may be geometric.

The sequence may be arithmetic and geometric.

None of these

Explanation

An arithmetic sequence is one in which each term is generated by adding the same number - the common difference - to the previous term. As can be seen here, the differences between each term and the previous term is not constant from term to term:

The sequence cannot be arithmetic.

A geometric sequence is one in which each term is generated by multiplying the previous term by the same number - the common ratio. As can be seen here, the ratios of each term to the previous one is not constant from term to term:

$\frac{-16}{8} = -2$

$\frac{24}{-16} = -\frac{3}{2}$

The sequence cannot be geometric.

Evaluate:

$\sum_{k = 1} ^{5} \frac{1}{k}$

$2\frac{17}{60}$

$2\frac{13}{60}$

$2\frac{19}{60}$

$2\frac{11}{60}$

$2\frac{71}{60}$

Explanation

$\sum_{k = 1} ^{5} \frac{1}{k}$ is equal to the sum of the expressions formed by substituting 1, 2, 3, 4, and 5, in turn, for in the expression $\frac{1}{k}$ . This is simply the sum of the reciprocals of these 5 integers, which is equal to

$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+ \frac{1}{5} = 2\frac{17}{60}$

The harmonic series is $1, \frac{1}{2} , \frac{1}{3} , \frac{1}{4} , ... , \frac{1}{n}$ where the nth term is the reciprocal of n. Which would work as a recursive formula $a _{n} = f(a_{n-1})$ where $a_{n}$ is the nth term?

$a_{n} =\frac{1}{\frac{1}{a_{n-1}}+1}$

$a_{n} = a_{n-1} + 1$

$a_{n} = a_{n-1} - \frac{1}{2}$

$a_{n} = \frac{1}{a_{n-1} + 1 }$

$a_{n} = \frac{1}{a_{n-1}} + 1$

Explanation

To go from $a_{n-1}$ to $a_{n}$ , we're adding 1 to the denominator. In words, we're flipping $a_{n-1}$ , adding 1, then flipping it again. For example, to get from $a_{4}=\frac{1}{4}$ to $a_{5 } = \frac{1}{5}$ we would have to flip $\frac{1}{4}$ to be 4, add 1 to get 5, then flip again to get $\frac{1}{5}$ .

The formula that shows this is

$a_{n} =\frac{1}{\frac{1}{a_{n-1}}+1}$

The sum of the first n integerss can be found using the formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ .

Find the sum of every number between 17 and 8,043, inclusive.

Explanation

To find the sum of all the integers in between 17 and 8,043, first we will find the sum of every integer from 8,043, and then we will subtract out the sum of the numbers 1-16, since those aren't between 17 and 8,043.

The sum of the first 8,043 integers is $\frac{(8043)(8044)}{2 } = 32, 348, 946$