5.1 Sequences (Page 2/25)

Calculus volume 2 Page 2 / 25

A graph in quadrant one containing the following points: (1, 2), (2, 4), (3, 8), (4, 16). — The plotted points are a graph of the sequence ${2^{n}} .$

Two types of sequences occur often and are given special names: arithmetic sequences and geometric sequences. In an arithmetic sequence , the difference between every pair of consecutive terms is the same. For example, consider the sequence

3, 7, 11, 15, 19,… .

You can see that the difference between every consecutive pair of terms is $4 .$ Assuming that this pattern continues, this sequence is an arithmetic sequence. It can be described by using the recurrence relation

{\begin{matrix} a_{1} = 3 \\ a_{n} = a_{n - 1} + 4 for n \geq 2 . \end{matrix}

Note that

\begin{array}{l} a_{2} = 3 + 4 \\ a_{3} = 3 + 4 + 4 = 3 + 2 \cdot 4 \\ a_{4} = 3 + 4 + 4 + 4 = 3 + 3 \cdot 4 . \end{array}

Thus the sequence can also be described using the explicit formula

\begin{matrix} a_{n} & = 3 + 4 (n - 1) \\ = 4 n - 1 . \end{matrix}

In general, an arithmetic sequence is any sequence of the form $a_{n} = c n + b .$

In a geometric sequence , the ratio of every pair of consecutive terms is the same. For example, consider the sequence

2, - \frac{2}{3}, \frac{2}{9}, - \frac{2}{27}, \frac{2}{81},… .

We see that the ratio of any term to the preceding term is $- \frac{1}{3} .$ Assuming this pattern continues, this sequence is a geometric sequence. It can be defined recursively as

\begin{matrix} a_{1} = 2 \\ a_{n} = - \frac{1}{3} \cdot a_{n - 1} for n \geq 2 . \end{matrix}

Alternatively, since

\begin{matrix} a_{2} = - \frac{1}{3} \cdot 2 \\ a_{3} = (- \frac{1}{3}) (- \frac{1}{3}) (2) = {(- \frac{1}{3})}^{2} \cdot 2 \\ a_{4} = (- \frac{1}{3}) (- \frac{1}{3}) (- \frac{1}{3}) (2) = {(- \frac{1}{3})}^{3} \cdot 2, \end{matrix}

we see that the sequence can be described by using the explicit formula

a_{n} = 2 {(- \frac{1}{3})}^{n - 1} .

The sequence ${2^{n}}$ that we discussed earlier is a geometric sequence, where the ratio of any term to the previous term is $2 .$ In general, a geometric sequence is any sequence of the form $a_{n} = c r^{n} .$

Finding explicit formulas

For each of the following sequences, find an explicit formula for the $n th$ term of the sequence.

$- \frac{1}{2}, \frac{2}{3}, - \frac{3}{4}, \frac{4}{5}, - \frac{5}{6},…$
$\frac{3}{4}, \frac{9}{7}, \frac{27}{10}, \frac{81}{13}, \frac{243}{16},…$

First, note that the sequence is alternating from negative to positive. The odd terms in the sequence are negative, and the even terms are positive. Therefore, the $n th$ term includes a factor of ${(−1)}^{n} .$ Next, consider the sequence of numerators ${1, 2, 3,…}$ and the sequence of denominators ${2, 3, 4,…} .$ We can see that both of these sequences are arithmetic sequences. The $n th$ term in the sequence of numerators is $n,$ and the $n th$ term in the sequence of denominators is $n + 1 .$ Therefore, the sequence can be described by the explicit formula

$a_{n} = \frac{{(−1)}^{n} n}{n + 1} .$
The sequence of numerators $3, 9, 27, 81, 243,…$ is a geometric sequence. The numerator of the $n th$ term is $3^{n}$ The sequence of denominators $4, 7, 10, 13, 16,…$ is an arithmetic sequence. The denominator of the $n th$ term is $4 + 3 (n - 1) = 3 n + 1 .$ Therefore, we can describe the sequence by the explicit formula $a_{n} = \frac{3^{n}}{3 n + 1} .$

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Find an explicit formula for the $n th$ term of the sequence ${\frac{1}{5}, - \frac{1}{7}, \frac{1}{9}, - \frac{1}{11},…} .$

$a_{n} = \frac{{(−1)}^{n + 1}}{3 + 2 n}$

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Defined by recurrence relations

For each of the following recursively defined sequences, find an explicit formula for the sequence.

$a_{1} = 2,$ $a_{n} = −3 a_{n - 1}$ for $n \geq 2$
$a_{1} = \frac{1}{2},$ $a_{n} = a_{n - 1} + {(\frac{1}{2})}^{n}$ for $n \geq 2$

Writing out the first few terms, we have

$\begin{matrix} a_{1} = 2 \\ a_{2} = −3 a_{1} = −3 (2) \\ a_{3} = −3 a_{2} = {(−3)}^{2} 2 \\ a_{4} = −3 a_{3} = {(−3)}^{3} 2 . \end{matrix}$

In general,

$a_{n} = 2 {(−3)}^{n - 1} .$
Write out the first few terms:

$\begin{matrix} a_{1} = \frac{1}{2} \\ a_{2} = a_{1} + {(\frac{1}{2})}^{2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \\ a_{3} = a_{2} + {(\frac{1}{2})}^{3} = \frac{3}{4} + \frac{1}{8} = \frac{7}{8} \\ a_{4} = a_{3} + {(\frac{1}{2})}^{4} = \frac{7}{8} + \frac{1}{16} = \frac{15}{16} . \end{matrix}$

From this pattern, we derive the explicit formula

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

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Find an explicit formula for the sequence defined recursively such that $a_{1} = −4$ and $a_{n} = a_{n - 1} + 6 .$

$a_{n} = 6 n - 10$

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Source: OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2

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