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9.1: Introduction to Sequences and Series

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Skills to Develop

• Find any element of a sequence given a formula for its general term.
• Use sigma notation and expand corresponding series.
• Distinguish between a sequence and a series.
• Calculate the \(n\)th partial sum of sequence.

A sequence 1 is a function whose domain is a set of consecutive natural numbers beginning with \(1\). For example, the following equation with domain \(\{1,2,3, \dots\}\) defines an infinite sequence 2 :

\(a(n)=5 n-3\) or \(a_{n}=5 n-3\)

The elements in the range of this function are called terms of the sequence. It is common to define the \(n\)th term, or the general term of a sequence 3 , using the subscripted notation \(a_{n}\), which reads “\(a\) sub \(n\).” Terms can be found using substitution as follows:

\(\begin{aligned}\color{Cerulean} { General\: term: } \quad &\color{black}{a_{n}=5 n-3} \\ \color{Cerulean} { First\: term (n=1) :}\quad & \color{black}{a_{1}=}5(\color{Cerulean}{1}\color{black}{)}-3=2 \\ \color{Cerulean}{Second\:term(n=2):} \quad& \color{black}{a_{2}=5}(\color{Cerulean}{2}\color{black}{)}-3=7 \\ \color{Cerulean}{Third\:term(n=3):} \quad& \color{black}{a_{3}=5}(\color{Cerulean}{3}\color{black}{)}-3=12 \\\color{Cerulean}{Fourth\:term(n=4):} \quad& \color{black}{a_{4}=5}(\color{Cerulean}{4}\color{black}{)}-3=17 \\ \color{Cerulean}{Fifth\:term(n=5):} \quad& \color{black}{a_{5}=5}(\color{Cerulean}{5}\color{black}{)}-3=22 \\ \vdots\end{aligned}\)

This produces an ordered list,

\(2,7,12,17,22, \ldots\)

The ellipsis \((…)\) indicates that this sequence continues forever. Unlike a set, order matters. If the domain of a sequence consists of natural numbers that end, such as \(\{1,2,3, \ldots, k\}\), then it is called a finite sequence 4 .

Example \(\PageIndex{1}\):

Given the general term of a sequence, find the first \(5\) terms as well as the \(100^{th}\) term: \(a_{n}=\frac{n(n-1)}{2}\).

To find the first \(5\) terms, substitute \(1, 2, 3, 4\), and \(5\) for \(n\) and then simplify.

\(\begin{array}{l}{a_{1}=\frac{\color{Cerulean}{1}\color{black}{(}\color{Cerulean}{1}\color{black}{-}1)}{2}=\frac{1(0)}{2}=\frac{0}{2}=0} \\ {a_{2}=\frac{\color{Cerulean}{2}\color{black}{(}\color{Cerulean}{2}\color{black}{-}1)}{2}=\frac{2(1)}{2}=\frac{2}{2}=1} \\ {a_{3}=\frac{\color{Cerulean}{3}\color{black}{(}\color{Cerulean}{3}\color{black}{-}1)}{2}=\frac{3(2)}{2}=\frac{6}{2}=3} \\ {a_{4}=\frac{\color{Cerulean}{4}\color{black}{(}\color{Cerulean}{4}\color{black}{-}1)}{2}=\frac{4(3)}{2}=\frac{12}{2}=6} \\ {a_{5}=\frac{\color{Cerulean}{5}\color{black}{(}\color{Cerulean}{5}\color{black}{-}1)}{2}=\frac{5(4)}{2}=\frac{20}{2}=10}\end{array}\)

Use \(n = 100\) to determine the \(100^{th}\) term in the sequence.

\(a_{100}=\frac{100(100-1)}{2}=\frac{100(99)}{2}=\frac{9,900}{2}=4,950\)

First five terms: \(0, 1, 3, 6, 10\); \(a_{100} = 4,950\)

Sometimes the general term of a sequence will alternate in sign and have a variable other than \(n\).

Example \(\PageIndex{2}\):

Find the first \(5\) terms of the sequence: \(a_{n}=(-1)^{n} x^{n+1}\).

Here we take care to replace \(n\) with the first \(5\) natural numbers and not \(x\).

\(\begin{array}{l}{a_{1}=(-1)\color{Cerulean}{^{1}}\color{black}{ x}^{\color{Cerulean}{1}\color{black}{+}1}=-x^{2}} \\ {a_{2}=(-1)^{\color{Cerulean}{2}}\color{black}{ x}^{\color{Cerulean}{2}\color{black}{+}1}=x^{3}} \\ {a_{3}=(-1)^{\color{Cerulean}{3}} \color{black}{x}^{\color{Cerulean}{3}\color{black}{+}1}=-x^{4}} \\ {a_{4}=(-1)^{\color{Cerulean}{4}}\color{black}{ x}^{\color{Cerulean}{4}\color{black}{+}1}=x^{5}} \\ {a_{5}=(-1)^{\color{Cerulean}{5}}\color{black}{ x}^{\color{Cerulean}{5}\color{black}{+}1}=-x^{6}}\end{array}\)

\(-x^{2}, x^{3},-x^{4}, x^{5},-x^{6}\)

Exercise \(\PageIndex{1}\)

Find the first \(5\) terms of the sequence: \(a_{n}=(-1)^{n+1} 2^{n}\).

\(2, −4, 8, −16, 32.\)

http://www.youtube.com/v/uuQ3jYL-g_I

One interesting example is the Fibonacci sequence. The first two numbers in the Fibonacci sequence are \(1\), and each successive term is the sum of the previous two. Therefore, the general term is expressed in terms of the previous two as follows:

\(F_{n}=F_{n-2}+F_{n-1}\)

Here \(F_{1} = 1, F_{2} = 1\), and \(n > 2\). A formula that describes a sequence in terms of its previous terms is called a recurrence relation 5 .

Example \(\PageIndex{3}\):

Find the first \(7\) Fibonacci numbers.

Given that \(F_{1} = 1\) and \(F_{2} = 1\), use the recurrence relation \(F_{n}=F_{n-2}+F_{n-1}\) where \(n\) is an integer starting with \(n = 3\) to find the next \(5\) terms:

\(\begin{array}{l}{F_{3}=F_{\color{Cerulean}{3}\color{black}{-}2}+F_{\color{Cerulean}{3}\color{black}{-}1}=F_{1}+F_{2}=1+1=2} \\ {F_{4}=F_{\color{Cerulean}{4}\color{black}{-}2}+F_{\color{Cerulean}{4}\color{black}{-}1}=F_{2}+F_{3}=1+2=3} \\ {F_{5}=F_{\color{Cerulean}{5}\color{black}{-}2}+F_{\color{Cerulean}{5}\color{black}{-}1}=F_{3}+F_{4}=2+3=5} \\ {F_{6}=F_{\color{Cerulean}{6}\color{black}{-}2}+F_{\color{Cerulean}{6}\color{black}{-}1}=F_{4}+F_{5}=3+5=8} \\ {F_{7}=F_{\color{Cerulean}{7}\color{black}{-}2}+F_{\color{Cerulean}{7}\color{black}{-}1}=F_{5}+F_{6}=5+8=13}\end{array}\)

\(1,1,2,3,5,8,13\)

Figure 9.1.1 : Leonardo Fibonacci (1170–1250)

Fibonacci numbers appear in applications ranging from art to computer science and biology. The beauty of this sequence can be visualized by constructing a Fibonacci spiral. Consider a tiling of squares where each side has a length that matches each Fibonacci number:

Figure 9.1.2

Connecting the opposite corners of the squares with an arc produces a special spiral shape.

Figure 9.1.3

This shape is called the Fibonacci spiral and approximates many spiral shapes found in nature.

A series 6 is the sum of the terms of a sequence. The sum of the terms of an infinite sequence results in an infinite series 7 , denoted \(S_{∞}\). The sum of the first \(n\) terms in a sequence is called a partial sum 8 , denoted \(S_{n}\). For example, given the sequence of positive odd integers \(1, 3, 5,…\) we can write:

\(\begin{array}{l}{S_{\infty}=1+3+5+7+9+\dots \quad\color{Cerulean} { Infinite\: series }} \\ {S_{5}=1+3+5+7+9=25 \quad\:\:\color{Cerulean} { 5th\: partial\: sum }}\end{array}\)

Example \(\PageIndex{4}\):

Determine the \(3^{rd}\) and \(5^{th}\) partial sums of the sequence: \(3,−6, 12,−24, 48,… \)

\(S_{3}=3+(-6)+12=9\) \(S_{5}=3+(-6)+12+(-24)+48=33\)

\(S_{3}=9 ; S_{5}=33\)

If the general term is known, then we can express a series using sigma 9 (or summation 10 ) notation :

\(\begin{aligned}S_{\infty}&=\sum_{n=1}^{\infty} n^{2}=1^{2}+2^{2}+3^{2}+\ldots & \color{Cerulean}{Infinite\:series} \\ S_{3}&=\sum_{n=1}^{3} n^{2}=1^{2}+2^{2}+3^{2} & \color{Cerulean}{3rd\:partial\:sum}\end{aligned}\)

The symbol \(\Sigma\) (upper case Greek letter sigma) is used to indicate a series. The expressions above and below indicate the range of the index of summation 11 , in this case represented by \(n\). The lower number indicates the starting integer and the upper value indicates the ending integer. The \(n\)th partial sum \(S_{n}\) can be expressed using sigma notation as follows:

\(S_{n}=\sum_{k=1}^{n} a_{k}=a_{1}+a_{2}+\cdots+a_{n}\)

This is read, “the sum of \(a_{k}\) as \(k\) goes from \(1\) to \(n\).” Replace \(n\) with \(∞\) to indicate an infinite sum.

Example \(\PageIndex{5}\):

Evaluate: \(\sum_{k=1}^{5}(-3)^{n-1}\).

\(\begin{aligned} \sum_{k=1}^{5}(-3)^{k-1} &=(-3)^{\color{Cerulean}{1}\color{black}{-}1}+(-3)^{\color{Cerulean}{2}\color{black}{-}1}+(-3)^{\color{Cerulean}{3}\color{black}{-}1}+(-3)^{\color{Cerulean}{4}\color{black}{-}1}+(-3)^{\color{Cerulean}{5}\color{black}{-}1} \\ &=(-3)^{0}+(-3)^{1}+(-3)^{2}+(-3)^{3}+(-3)^{4} \\ &=1-3+9-27+81 \\ &=61 \end{aligned}\)

When working with sigma notation, the index does not always start at \(1\).

Evaluate: \(\sum_{k=2}^{5}(-1)^{k}(3 k)\)

Here the index is expressed using the variable \(k\), which ranges from \(2\) to \(5\).

Exercise \(\PageIndex{2}\)

Evaluate: \(\sum_{n=1}^{5}(15-9 n)\).

http://www.youtube.com/v/aZ3sPd8N1TU

Infinity is used as the upper bound of a sum to indicate an infinite series.

Example \(\PageIndex{7}\):

Write in expanded form: \(\sum_{i=0}^{\infty} \frac{n}{n+1}\).

In this case we begin with \(n = 0\) and add three dots to indicate that this series continues forever.

\(\begin{aligned} \sum_{n=0}^{\infty} \frac{n}{n+1} &=\frac{\color{Cerulean}{0}}{\color{Cerulean}{0}\color{black}{+}1}+\frac{\color{Cerulean}{1}}{\color{Cerulean}{1}\color{black}{+}1}+\frac{\color{Cerulean}{2}}{\color{Cerulean}{2}\color{black}{+}1}+\frac{\color{Cerulean}{3}}{\color{Cerulean}{3}\color{black}{+}1}+\cdots \\ &=\frac{0}{1}+\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots \\ &=0+\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots \end{aligned}\)

\(0+\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots\)

When expanding a series, take care to replace only the variable indicated by the index.

Example \(\PageIndex{8}\):

Write in expanded form: \(\sum_{i=1}^{\infty}(-1)^{i-1} x^{2 i}\).

\(\begin{aligned} \sum_{i=1}^{\infty}(-1)^{i-1} x^{2 i} &=(-1)^{\color{Cerulean}{1}\color{black}{-}1} x^{2(\color{Cerulean}{1}\color{black}{)}}+(-1)^{\color{Cerulean}{2}\color{black}{-}1} x^{2(\color{Cerulean}{2}\color{black}{)}}+(-1)^{\color{Cerulean}{3}\color{black}{-}1} x^{2(\color{Cerulean}{3}\color{black}{)}}+\cdots \\ &=(-1)^{0} x^{2(1)}+(-1)^{1} x^{2(2)}+(-1)^{2} x^{2(3)}+\cdots \\ &=x^{2}-x^{4}+x^{6}-\cdots \end{aligned}\)

\(x^{2}-x^{4}+x^{6}-\cdots\)

Key Takeaways

• A sequence is a function whose domain consists of a set of natural numbers beginning with \(1\). In addition, a sequence can be thought of as an ordered list.
• Formulas are often used to describe the \(n\)th term, or general term, of a sequence using the subscripted notation \(a_{n}\).
• A series is the sum of the terms in a sequence. The sum of the first \(n\) terms is called the \(n\)th partial sum and is denoted \(S_{n}\).
• Use sigma notation to denote summations in a compact manner. The nth partial sum, using sigma notation, can be written \(S_{n}=\sum_{k=1}^{n} a_{k}\). The symbol \(\Sigma\) denotes a summation where the expression below indicates that the index \(k\) starts at \(1\) and iterates through the natural numbers ending with the value \(n\) above.

Exercise \(\PageIndex{3}\)

Find the first \(5\) terms of the sequence as well as the \(30^{th}\) term.

• \(a_{n}=2 n\)
• \(a_{n}=2 n+1\)
• \(a_{n}=\frac{n^{2}-1}{2}\)
• \(a_{n}=\frac{n}{2 n-1}\)
• \(a_{n}=(-1)^{n}(n+1)^{2}\)
• \(a_{n}=(-1)^{n+1} n^{2}\)
• \(a_{n}=3^{n-1}\)
• \(a_{n}=2^{n-2}\)
• \(a_{n}=\left(\frac{1}{2}\right)^{n}\)
• \(a_{n}=\left(-\frac{1}{3}\right)^{n}\)
• \(a_{n}=\frac{(-1)^{n-1}}{3 n-1}\)
• \(a_{n}=\frac{2(-1)^{n}}{n+5}\)
• \(a_{n}=1+\frac{1}{n}\)
• \(a_{n}=\frac{n^{2}+1}{n}\)

1. \(2,4,6,8,10 ; a_{30}=60\)

3. \(0, \frac{3}{2}, 4, \frac{15}{2}, 12 ; a_{30}=\frac{899}{2}\)

5. \(-4,9,-16,25,-36 ; a_{30}=961\)

7. \(1,3,9,27,81 ; a_{30}=3^{29}\)

9. \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32} ; a_{30}=\frac{1}{2^{30}}\)

11. \(\frac{1}{2},-\frac{1}{5}, \frac{1}{8},-\frac{1}{11}, \frac{1}{14} ; a_{30}=-\frac{1}{89}\)

13. \(2, \frac{3}{2}, \frac{4}{3}, \frac{5}{4}, \frac{6}{5} ; a_{30}=\frac{31}{30}\)

Exercise \(\PageIndex{4}\)

Find the first \(5\) terms of the sequence.

• \(a_{n}=2 x^{2 n-1}\)
• \(a_{n}=(2 x)^{n-1}\)
• \(a_{n}=\frac{x^{n}}{n+4}\)
• \(a_{n}=\frac{x^{2 n}}{x-2}\)
• \(a_{n}=\frac{n x^{2 n}}{n+1}\)
• \(a_{n}=\frac{(n+1) x^{n}}{n^{2}}\)
• \(a_{n}=(-1)^{n} x^{3 n}\)
• \(a_{n}=(-1)^{n-1} x^{n+1}\)

1. \(2 x, 2 x^{3}, 2 x^{5}, 2 x^{7}, 2 x^{9}\)

3. \(\frac{x}{5}, \frac{x^{2}}{6}, \frac{x^{3}}{7}, \frac{x^{4}}{8}, \frac{x^{5}}{9}\)

5. \(\frac{x^{2}}{2}, \frac{2 x^{4}}{3}, \frac{3 x^{6}}{4}, \frac{4 x^{8}}{5}, \frac{5 x^{10}}{6}\)

7. \(-x^{3}, x^{6},-x^{9}, x^{12},-x^{15}\)

Exercise \(\PageIndex{5}\)

Find the first 5 terms of the sequence defined by the given recurrence relation.

• \(a_{n}=a_{n-1}+5\) where \(a_{1}=3\)
• \(a_{n}=a_{n-1}-3\) where \(a_{1}=4\)
• \(a_{n}=3 a_{n-1}\) where \(a_{1}=-2\)
• \(a_{n}=-2 a_{n-1}\) where \(a_{1}=-1\)
• \(a_{n}=n a_{n-1}\) where \(a_{1}=1\)
• \(a_{n}=(n-1) a_{n-1}\) where \(a_{1}=1\)
• \(a_{n}=2 a_{n-1}-1\) where \(a_{1}=0\)
• \(a_{n}=3 a_{n-1}+1\) where \(a_{1}=-1\)
• \(a_{n}=a_{n-2}+2 a_{n-1}\) where \(a_{1}=-1\) and \(a_{2}=0\)
• \(a_{n}=3 a_{n-1}-a_{n-2}\) where \(a_{1}=0\) and \(a_{2}=2\)
• \(a_{n}=a_{n-1}-a_{n-2}\) where \(a_{1}=1\) and \(a_{2}=3\)
• \(a_{n}=a_{n-2}+a_{n-1}+2\) where \(a_{1}=-4\) and \(a_{2}=-1\)

1. \(3, 8, 13, 18, 23\)

3. \(−2, −6, −18, −54, −162\)

5. \(1, 2, 6, 24, 120\)

7. \(0, −1, −3, −7, −15\)

9. \(−1, 0, −1, −2, −5\)

11. \(1, 3, 2, −1, −3\)

Exercise \(\PageIndex{6}\)

Find the indicated term.

• \(a_{n}=2-7 n ; a_{12}\)
• \(a_{n}=3 n-8 ; a_{20}\)
• \(a_{n}=-4(5)^{n-4} ; a_{7}\)
• \(a_{n}=6\left(\frac{1}{3}\right)^{n-6} ; a_{9}\)
• \(a_{n}=1+\frac{1}{n}; a_{10}\)
• \(a_{n}=(n+1) 5^{n-3} ; a_{5}\)
• \(a_{n}=(-1)^{n} 2^{2 n-3} ; a_{4}\)
• \(a_{n}=n(n-1)(n-2) ; a_{6}\)
• An investment of $\(4,500\) is made in an account earning \(2\)% interest compounded quarterly. The balance in the account after \(n\) quarters is given by \(a_{n}=4500\left(1+\frac{0.02}{4}\right)^{n}\). Find the amount in the account after each quarter for the first two years. Round to the nearest cent.
• The value of a new car after \(n\) years is given by the formula \(a_{n}=18,000\left(\frac{3}{4}\right)^{n}\). Find and interpret \(a_{7}\). Round to the nearest whole dollar.
• The number of comparisons a computer algorithm makes to sort n names in a list is given by the formula \(a_{n}=n \log _{2} n\). Determine the number of comparisons it takes this algorithm to sort \(2 × 10^{6}\) (2 million) names.
• The number of comparisons a computer algorithm makes to search \(n\) names in a list is given by the formula \(a_{n}=n^{2}\) Determine the number of comparisons it takes this algorithm to search \(2 × 10^{6}\) (2 million) names.

3. \(-500\)

5. \(\frac{11}{10}\)

9. Year 1: QI: $\(4,522.50\); QII: $\(4,545.11\); QIII: $\(4,567.84\); QIV: $\(4,590.68\); Year 2: QI: $\(4,613.63\); QII: $\(4,636.70\); QIII: $\(4,659.88\); QIV: $\(4,683.18\)

11. Approximately \(4 \times 10^{7}\) comparisons

Exercise \(\PageIndex{7}\)

Find the indicated partial sum.

• \(3,5,9,17,33, \ldots ; S_{4}\)
• \(-5,7,-29,79,-245, \ldots ; S_{4}\)
• \(4,1,-4,-11,-20, \ldots ; S_{5}\)
• \(0,2,6,12,20, \ldots ; S_{3}\)
• \(a_{n}=2-7 n ; S_{5}\)
• \(a_{n}=3 n-8 ; S_{5}\)
• \(a_{n}=-4(5)^{n-4} ; S_{3}\)
• \(a_{n}=6\left(\frac{1}{3}\right)^{n-6} ; S_{3}\)
• \(a_{n}=1+\frac{1}{n}; S_{4}\)
• \(a_{n}=(n+1) 5^{n-3} ; S_{3}\)
• \(a_{n}=(-1)^{n} 2^{2 n-3} ; S_{5}\)
• \(a_{n}=n(n-1)(n-2) ; S_{4}\)

7. \(-\frac{124}{125}\)

9. \(\frac{73}{12}\)

11. \(-\frac{205}{2}\)

Exercise \(\PageIndex{8}\)

• \(\sum_{k=1}^{5} 3 k\)
• \(\sum_{k=1}^{6} 2 k\)
• \(\sum_{i=2}^{6} i^{2}\)
• \(\sum_{i=0}^{4}(i+1)^{2}\)
• \(\sum_{n=1}^{5}(-1)^{n+1} 2^{n}\)
• \(\sum_{n=5}^{10}(-1)^{n} n^{2}\)
• \(\sum_{k=-2}^{2}\left(\frac{1}{2}\right)^{k}\)
• \(\sum_{k=-4}^{0}\left(\frac{1}{3}\right)^{k}\)
• \(\sum_{k=0}^{4}(-2)^{k+1}\)
• \(\sum_{k=-1}^{3}(-3)^{k-1}\)
• \(\sum_{n=1}^{5} 3\)
• \(\sum_{n=1}^{7}-5\)
• \(\sum_{k=-2}^{3} k(k+1)\)
• \(\sum_{k=-2}^{2}(k-2)(k+2)\)

7. \(\frac{31}{4}\)

9. \(−22\)

Exercise \(\PageIndex{9}\)

Write in expanded form.

• \(\sum_{n=1}^{\infty} \frac{n-1}{n}\)
• \(\sum_{n=1}^{\infty} \frac{n}{2 n-1}\)
• \(\sum_{n=1}^{\infty}\left(-\frac{1}{2}\right)^{n-1}\)
• \(\sum_{n=0}^{\infty}\left(-\frac{2}{3}\right)^{n+1}\)
• \(\sum_{n=1}^{\infty} 3\left(\frac{1}{5}\right)^{n}\)
• \(\sum_{n=0}^{\infty} 2\left(\frac{1}{3}\right)^{n}\)
• \(\sum_{k=0}^{\infty}(-1)^{k} x^{k+1}\)
• \(\sum_{k=1}^{\infty}(-1)^{k+1} x^{k-1}\)
• \(\sum_{i=0}^{\infty}(-2)^{i+1} x^{i}\)
• \(\sum_{i=1}^{\infty}(-3)^{i-1} x^{3 i}\)
• \(\sum_{k=1}^{\infty}(2 k-1) x^{2 k}\)
• \(\sum_{k=1}^{\infty} \frac{k x^{k-1}}{k+1}\)

1. \(0+\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots\)

3. \(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots\)

5. \(\frac{3}{5}+\frac{3}{25}+\frac{3}{125}+\frac{3}{625}+\cdots\)

7. \(x-x^{2}+x^{3}-x^{4}+\cdots\)

9. \(-2+4 x-8 x^{2}+16 x^{3}-\dots\)

11. \(x^{2}+3 x^{4}+5 x^{6}+7 x^{8}+\cdots\)

Exercise \(\PageIndex{10}\)

Express the following series using sigma notation.

• \(x+2 x^{2}+3 x^{3}+4 x^{4}+5 x^{5}\)
• \(\frac{1}{2} x^{2}+\frac{2}{3} x^{3}+\frac{3}{4} x^{4}+\frac{4}{5} x^{5}+\frac{5}{6} x^{6}\)
• \(2+2^{2} x+2^{3} x^{2}+2^{4} x^{3}+2^{5} x^{4}\)
• \(3 x+3^{2} x^{2}+3^{3} x^{3}+3^{4} x^{4}+3^{5} x^{5}\)
• \(2 x+4 x^{2}+8 x^{3}+\dots+2^{n} x^{n}\)
• \(x+3 x^{2}+9 x^{3}+\dots+3^{n} x^{n+1}\)
• \(5+(5+d)+(5+2 d)+\dots+(5+n d)\)
• \(2+2 r^{1}+2 r^{2}+\dots+2 r^{n-1}\)
• \(\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\dots+3\left(\frac{1}{2}\right)^{n}\)
• \(\frac{8}{3}+\frac{16}{4}+\frac{32}{5}+\dots+\frac{2^{n}}{n}\)
• A structured settlement yields an amount in dollars each year, represented by \(n\), according to the formula \(p_{n}=10,000(0.70)^{n-1}\). What is the total amount gained from the settlement after \(5\) years?
• The first row of seating in a small theater consists of \(14\) seats. Each row thereafter consists of \(2\) more seats than the previous row. If there are \(7\) rows, how many total seats are in the theater?

1. \(\sum_{k=1}^{5} k x^{k}\)

3. \(\sum_{k=1}^{5} 2^{k} x^{k-1}\)

5. \(\sum_{k=1}^{n} 2^{k} x^{k}\)

7. \(\sum_{k=0}^{n}(5+k d)\)

9. \(\sum_{k=2}^{n} 3\left(\frac{1}{2}\right)^{k}\)

11. \(\$ 27,731\)

Exercise \(\PageIndex{11}\)

• Research and discuss Fibonacci numbers as they are found in nature.
• Research and discuss the life and contributions of Leonardo Fibonacci.
• Explain the difference between a sequence and a series. Provide an example of each.

1. Answer may vary

3. Answer may vary

1 A function whose domain is a set of consecutive natural numbers starting with \(1\).

2 A sequence whose domain is the set of natural numbers \(\{1,2,3, \dots\}\).

3 An equation that defines the nth term of a sequence commonly denoted using subscripts \(a_{n}\).

4 A sequence whose domain is \(\{1,2,3, \dots, k\}\) where \(k\) is a natural number.

5 A formula that uses previous terms of a sequence to describe subsequent terms.

6 The sum of the terms of a sequence.

7 The sum of the terms of an infinite sequence denoted \(S_{∞}\).

8 The sum of the first n terms in a sequence denoted \(S_{n}\).

9 A sum denoted using the symbol \(\Sigma\) (upper case Greek letter sigma).

10 Used when referring to sigma notation.

11 The variable used in sigma notation to indicate the lower and upper bounds of the summation.

Introduction to Sequences

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Unit 1: Algebra foundations

Unit 2: solving equations & inequalities, unit 3: working with units, unit 4: linear equations & graphs, unit 5: forms of linear equations, unit 6: systems of equations, unit 7: inequalities (systems & graphs), unit 8: functions, unit 9: sequences, unit 10: absolute value & piecewise functions, unit 11: exponents & radicals, unit 12: exponential growth & decay, unit 13: quadratics: multiplying & factoring, unit 14: quadratic functions & equations, unit 15: irrational numbers, unit 16: creativity in algebra.

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Algebra 1 - Arithmetic sequences

Algebra 1 – arithmetic sequences.

Arithmetic sequences is a part of syllabus in algebra 1 (second math course), which finds application in many algebra questions including algebra word problems. An arithmetic progression (AP) or arithmetic sequence is a sequence of numbers, whose consecutive numbers differ by a constant. When n terms of an arithmetic sequence are added, you get arithmetic series. Arithmetic series formula is given by : sum = n/2 (a + l) where a is the 1st term and l is the last term. Learn more about Arithmetic sequences using the resources on this page.

The apps, sample questions, videos and worksheets listed below will help you learn Arithmetic Sequences.

Sample questions related to arithmetic sequences, worksheets related to arithmetic sequences, educational apps related to arithmetic sequences.

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Related Topics

What is a recursive formula for an arithmetic sequence.

Recursive formula: This sequence is neither arithmetic nor geometric. It does, however, have a pattern of development based upon each previous term. Notice how the value of n is used as the exponent for the value (-1).

How do you find the nth term of a sequence?

Imagine we want to find a formula for the nth term of this sequence: 7, 11, 15, 19, 23, … We can see that the terms in this sequence go up by 4 each time, so 4n must appear in the formula. The sequence generated by the formula 4n is the four times table, but it isn’t quite the sequence we want: 4n sequence:

What is the recursive formula?

A recursive formula designates the starting term, a1, and the nth term of the sequence, an , as an expression containing the previous term (the term before it), an-1. The process of recursion can be thought of as climbing a ladder.

How do you find the common ratio of a geometric sequence?

To find the common ratio, divide the second term by the first term. Notice the non-linear nature of the scatter plot of the terms of a geometric sequence. The domain consists of the counting numbers 1, 2, 3, 4, … and the range consists of the terms of the sequence.

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Common Core State Standards Initiative

High School: Algebra

Standards in this domain:, ccss.math.content.hsa.introduction introduction, expressions..

An expression is a record of a computation with numbers, symbols that represent numbers, arithmetic operations, exponentiation, and, at more advanced levels, the operation of evaluating a function. Conventions about the use of parentheses and the order of operations assure that each expression is unambiguous. Creating an expression that describes a computation involving a general quantity requires the ability to express the computation in general terms, abstracting from specific instances.

Reading an expression with comprehension involves analysis of its underlying structure. This may suggest a different but equivalent way of writing the expression that exhibits some different aspect of its meaning. For example, p + 0.05 p can be interpreted as the addition of a 5% tax to a price p . Rewriting p + 0.05 p as 1.05 p shows that adding a tax is the same as multiplying the price by a constant factor.

Algebraic manipulations are governed by the properties of operations and exponents, and the conventions of algebraic notation. At times, an expression is the result of applying operations to simpler expressions. For example, p + 0.05 p is the sum of the simpler expressions p and 0.05 p . Viewing an expression as the result of operation on simpler expressions can sometimes clarify its underlying structure.

A spreadsheet or a computer algebra system (CAS) can be used to experiment with algebraic expressions, perform complicated algebraic manipulations, and understand how algebraic manipulations behave.

Equations and inequalities.

An equation is a statement of equality between two expressions, often viewed as a question asking for which values of the variables the expressions on either side are in fact equal. These values are the solutions to the equation. An identity, in contrast, is true for all values of the variables; identities are often developed by rewriting an expression in an equivalent form.

The solutions of an equation in one variable form a set of numbers; the solutions of an equation in two variables form a set of ordered pairs of numbers, which can be plotted in the coordinate plane. Two or more equations and/or inequalities form a system. A solution for such a system must satisfy every equation and inequality in the system.

An equation can often be solved by successively deducing from it one or more simpler equations. For example, one can add the same constant to both sides without changing the solutions, but squaring both sides might lead to extraneous solutions. Strategic competence in solving includes looking ahead for productive manipulations and anticipating the nature and number of solutions.

Some equations have no solutions in a given number system, but have a solution in a larger system. For example, the solution of x + 1 = 0 is an integer, not a whole number; the solution of 2 x + 1 = 0 is a rational number, not an integer; the solutions of x 2 - 2 = 0 are real numbers, not rational numbers; and the solutions of x 2 + 2 = 0 are complex numbers, not real numbers.

The same solution techniques used to solve equations can be used to rearrange formulas. For example, the formula for the area of a trapezoid, A = (( b 1 + b 2 )/2) h , can be solved for h using the same deductive process. Inequalities can be solved by reasoning about the properties of inequality. Many, but not all, of the properties of equality continue to hold for inequalities and can be useful in solving them.

Connections to Functions and Modeling. Expressions can define functions, and equivalent expressions define the same function. Asking when two functions have the same value for the same input leads to an equation; graphing the two functions allows for finding approximate solutions of the equation. Converting a verbal description to an equation, inequality, or system of these is an essential skill in modeling.

Algebra Overview

• Seeing Structure in Expressions
• Interpret the structure of expressions
• Write expressions in equivalent forms to solve problems

Arithmetic with Polynomials and Rational Functions

• Perform arithmetic operations on polynomials
• Understand the relationship between zeros and factors of polynomials
• Use polynomial identities to solve problems
• Rewrite rational functions

Creating Equations

• Create equations that describe numbers or relationships

Reasoning with Equations and Inequalities

• Understand solving equations as a process of reasoning and explain the reasoning
• Solve equations and inequalities in one variable
• Solve systems of equations
• Represent and solve equations and inequalities graphically

Mathematical Practices

• Make sense of problems and persevere in solving them.
• Reason abstractly and quantitatively.
• Construct viable arguments and critique the reasoning of others.
• Model with mathematics.
• Use appropriate tools strategically.
• Attend to precision.
• Look for and make use of structure.
• Look for and express regularity in repeated reasoning.

Arithmetic with Polynomials & Rational Expressions

Creating equations ✭, reasoning with equations & inequalities.

• Standards for Mathematical Practice
• How to read the grade level standards
• Introduction
• Counting & Cardinality
• Operations & Algebraic Thinking
• Number & Operations in Base Ten
• Measurement & Data
• Number & Operations—Fractions¹
• Number & Operations in Base Ten¹
• Number & Operations—Fractions
• Ratios & Proportional Relationships
• The Number System
• Expressions & Equations
• Statistics & Probability
• The Real Number System
• Quantities*
• The Complex Number System
• Vector & Matrix Quantities
• Arithmetic with Polynomials & Rational Expressions
• Creating Equations*
• Reasoning with Equations & Inequalities
• Interpreting Functions
• Building Functions
• Linear, Quadratic, & Exponential Models*
• Trigonometric Functions
• High School: Modeling
• Similarity, Right Triangles, & Trigonometry
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• Interpreting Categorical & Quantitative Data
• Making Inferences & Justifying Conclusions
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