Precalculus Used in Calculus

A.5 Sequences; Summation Notation; the Binomial Theorem
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A-38

OBJECTIVES

When you finish this section you should be able to:

Write the first several terms of a sequence (p. A-38)
Write the terms of a recursively defined sequence (p. A-39)
Use summation notation (p. A-40)
Find the sum of the first $n$ terms of a sequence (p. A-41)
Use the Binomial Theorem (p. A-42)

1 Write the First Several Terms of a Sequence
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DEFINITION Sequence

A sequence is a rule that assigns a real number to each positive integer.

A sequence is often represented by listing its values in order. For example, the sequence whose rule is to assign to each positive integer its reciprocal may be represented as $s_{1}=1, \quad s_{2}=\dfrac{1}{2}, \quad s_{3}=\dfrac{1}{3}\hbox{,} \quad s_{4}= \dfrac{1}{4},\ldots, \quad s_{n}=\dfrac{1}{n},\ldots$ or merely as the list $1,\quad \dfrac{1}{2}, \quad \dfrac{1}{3},\quad \dfrac{1}{4},\quad \ldots,\quad \dfrac{1}{n}\, , \quad \ldots$

The list never ends, as the dots (ellipsis) indicate. The real numbers in this ordered list are called the terms of the sequence.

Notice that when we use subscripted letters to represent the terms of a sequence, $s_{1}$ is the 1st term, $s_{2}$ is the 2nd term $,\ldots, s_{n}$ is the $n{\rm th}$ term, and so on. If we have a rule for the $n$ th term, as in the sequence above, it is used to identify the sequence. We represent such a sequence by enclosing the $n$ th term in braces. The notation $\{ s_{n}\} =\left\{ \dfrac{1}{n}\right\}$ or $\{ s_{n}\} _{n=1}^{\infty }=\left\{ \dfrac{1}{n}\right\} _{n=1}^{\infty }$ both represent the sequence $1$ , $\dfrac{1}{2}$ , $\dfrac{1}{3}$ , $\dfrac{1}{4 }$ , $\dfrac{1}{5}, \ldots\,$ .

For simplicity, we use the notation $\{ s_{n}\}$ to represent a sequence that has a rule for the $n$ th term. For example, the sequence whose $n$ th term is $b_{n}=\left( \dfrac{1}{2}\right) ^{n}$ can be represented either as $b_{1}=\dfrac{1}{2}, \quad b_{2}=\dfrac{1}{4}, \quad b_{3}=\dfrac{1}{8}, \ldots ,\quad b_{n}=\dfrac{1}{2^{n}},\ldots ,$ or as $\{ b_{n}\} =\left\{ \left( \dfrac{1}{2}\right) ^{n}\right\}$

EXAMPLE 1Writing the Terms of a Sequence

Write the first six terms of the sequence: $\{ b_{n}\} = \left\{ ( -1) ^{n-1} \, \dfrac{2}{n} \right\}$

$\begin{array}[t]{lllll} b_{1}= ( -1) ^{0}\,\dfrac{2}{1}=2 & &\qquad b_{2}= ( -1) ^{1}\, \dfrac{2}{2}=-1 & &\qquad b_{3}= ( -1) ^{2}\,\dfrac{2}{3}=\dfrac{2}{3} \\[7pt] b_{4}=-\dfrac{1}{2} & &\qquad b_{5}=\dfrac{2}{5} & &\qquad b_{6}=-\dfrac{1}{3} \end{array}$

The first six terms of the sequence are $2,-1,\,\dfrac{2}{3},-\dfrac{1}{2},\, \dfrac{2}{5},-\dfrac{1}{3}$ .

A-39

EXAMPLE 2Determining a Sequence from a Pattern

The pattern on the left suggests the $n{\rm th}$ term of the sequence on the right.

Pattern	$n$ th	Sequence
$e, \dfrac{e^{2}}{2}, \dfrac{e^{3}}{3}, \dfrac{e^{4}}{4},\ldots$	$a_{n}=\dfrac{e^{n}}{n}$	$\{ a_{n}\} =\left\{ \dfrac{e^{n}}{n}\right\}$
$1,~\dfrac{1}{3},~\dfrac{1}{9},~\dfrac{1}{27},\ldots$	$b_{n}=\dfrac{1}{3^{n-1}}$	$\{ b_{n}\} =\left\{ \dfrac{1}{3^{n-1}}\right\}$
$1,~3,~5,~7,\ldots$	$c_{n}=2n-1$	$\{ c_{n}\} = \{2n-1\}$
$1,~4,~9,~16,~25,\ldots$	$d_{n}=n^{2}$	$\{ d_{n}\} = \{n^{2}\}$
$1,~{-}\dfrac{1}{2},~\dfrac{1}{3},~{-}\dfrac{1}{4},~\dfrac{1}{5},\ldots$	$e_{n}= ( -1) ^{n+1}\left( \dfrac{1}{n}\right)$	$\{ e_{n}\} =\left\{( -1) ^{n+1}\left( \dfrac{1}{n}\right) \right\}$

2 Write the Terms of a Recursively Defined Sequence
Printed Page A-39

A second way of defining a sequence is to assign a value to the first (or the first few) term(s) and specify the $n$ th term by a formula or an equation that involves one or more of the terms preceding it. Such a sequence is defined recursively, and the rule or formula is called a recursive formula.

EXAMPLE 3Writing the Terms of a Recursively Defined Sequence

Write the first five terms of the recursively defined sequence. $s_{1}=1\qquad s_{n}=4s_{n-1}\qquad n\geq 2$

The first term is given as $s_{1}=1$ . To get the second term, we use $n=2$ in the formula to obtain $s_{2}=4s_{1}=4\cdot 1=4$ . To obtain the third term, we use $n=3$ in the formula, getting $s_{3}=4s_{2}=4\cdot 4=16.$ Each new term requires that we know the value of the preceding term. The first five terms are $\begin{eqnarray*} s_{1}&=&1 \\ s_{2}&=&4s_{2-1}=4s_{1}=4\cdot 1=4 \\ s_{3}&=&4s_{3-1}=4s_{2}=4\cdot 4=16 \\ s_{4}&=&4s_{4-1}=4s_{3}=4\cdot 16=64 \\ s_{5}&=&4s_{5-1}=4s_{4}=4\cdot 64=256 \end{eqnarray*}$

EXAMPLE 4Writing the Terms of a Recursively Defined Sequence

Write the first five terms of the recursively defined sequence. $u_{1}=1 \qquad u_{2}=1 \qquad u_{n}=u_{n-2}+u_{n-1}\quad n\geq 3$

We are given the first two terms. To obtain the third term requires that we know each of the previous two terms. $\begin{eqnarray*} u_{1}&=&1 \\[3pt] u_{2}&=&1 \\[3pt] u_{3}&=&u_{1}+u_{2}=1+1=2 \\[3pt] u_{4}&=&u_{2}+u_{3}=1+2=3 \\[3pt] u_{5}&=&u_{3}+u_{4}=2+3=5 \end{eqnarray*}$

The sequence defined in Example 4 is called a Fibonacci sequence; its terms are called Fibonacci numbers. These numbers appear in a wide variety of applications.

3 Use Summation Notation
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A-40

It is often important to be able to find the sum of the first $n$ terms of a sequence $\{ a_{n}\}$ , namely, $a_{1}+a_{2}+a_{3}+\cdots+a_{n}$

NOTE

Summation notation is called sigma notation in some books.

Rather than write down all these terms, we introduce summation notation. Using summation notation, the sum is written $a_{1}+a_{2}+a_{3}+\cdots+a_{n}=\sum\limits_{k=1}^{n}a_{k}$

The symbol $\sum$ (the uppercase Greek letter sigma, which is an $S$ in our alphabet) is simply an instruction to sum, or add up, the terms. The integer $k$ is called the index of summation; it tells us where to start the sum $( k=1)$ and where to end it $(k=n)$ . The expression $\sum\limits_{k=1}^{n}a_{k}$ is read, “The sum of the terms $a_{k}$ from $k=1$ to $k=n$ .”

NOTE

If ${n\geq 0}$ is an integer, the factorial symbol $n!$ means $0!=1$ , $1!=1$ , and $n!=1\cdot 2\cdot 3\cdot\ \cdots\ \cdot (n-1) \cdot n$ , where $n>1$ .

EXAMPLE 5Expanding Summation Notation

Write out each sum:

(a) $\displaystyle\sum\limits_{k=1}^{n}\dfrac{1}{k}$

(b) $\displaystyle\sum\limits_{k=1}^{n}k!$

(a) $\displaystyle\sum\limits_{k=1}^{n}\dfrac{1}{k}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$

(b) $\displaystyle\sum\limits_{k=1}^{n}k!=1!+2!+\cdots+n!$

EXAMPLE 6Writing a Sum in Summation Notation

Express each sum using summation notation:

(a) $1^{2}+2^{2}+3^{2}+\cdots+9^{2}$

(b) $1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots+\dfrac{1}{2^{n-1}}$

(a) The sum $1^{2}+2^{2}+3^{2}+\cdots+9^{2}$ has $9$ terms, each of the form $k^{2}$ ; it starts at $k=1$ and ends at $k=9$ : $1^{2}+2^{2}+3^{2}+\cdots+9^{2}=\sum\limits_{k=1}^{9}k^{2}$

(b) The sum $1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots+\dfrac{1}{2^{n-1}}$ has $n$ terms, each of the form $\dfrac{1}{2^{k-1}}$ . It starts at $k=1$ and ends at $k=n$ . $1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots+\dfrac{1}{2^{n-1}} =\sum\limits_{k=1}^{n}\dfrac{1}{2^{k-1}}$

The index of summation does not need to begin at $1$ or end at $n$ . For example, the sum in Example 6(b) could be expressed as $\sum\limits_{k=0}^{n-1}\dfrac{1}{2^{k}}=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1 }{8}+\cdots+\dfrac{1}{2^{n-1}}$ Letters other than $k$ can be used as the index. For example, $\sum\limits_{j=1}^{n}j! \qquad \hbox{and} \qquad \sum\limits_{i=1}^{n}i!$ each represent the same sum as Example 5(b).

4 Find the Sum of the First $n$ Terms of a Sequence
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A-41

When working with summation notation, the following properties are useful for finding the sum of the first $n$ terms of a sequence.

THEOREM Properties Involving Summation Notation

If $\{ a_{n}\}$ and $\{ b_{n}\}$ are two sequences and $c$ is a real number, then: $\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \begin{array}{l} {\color{#0066A7}•}\enspace \sum\limits_{k=1}^{n} (ca_{k}) =c\sum\limits_{k=1}^{n}a_{k} \hphantom{00000000} (1)\\ {\color{#0066A7}•}\enspace \sum\limits_{k=1}^{n} ( a_{k}+b_{k}) =\sum\limits_{k=1}^{n}a_{k}+\sum\limits_{k=1}^{n}b_{k}\hphantom{00000000} (2)\\ {\color{#0066A7}•}\enspace \sum\limits_{k=1}^{n} ( a_{k}-b_{k}) =\sum\limits_{k=1}^{n}a_{k}-\sum\limits_{k=1}^{n}b_{k}\hphantom{00000000} (3)\\ {\color{#0066A7}•}\enspace \sum\limits_{k=j+1}^{n}a_{k}=\sum\limits_{k=1}^{n}a_{k}-\sum\limits_{k=1}^{j}a_{k},\quad \hbox{where } 0<j<n\hphantom{00000000} (4) \end{array} }}$

The proof of property (1) follows from the distributive property of real numbers. The proofs of properties (2) and (3) are based on the commutative and associative properties of real numbers. Property (4) states that the sum from $j+1$ to $n$ equals the sum from $1$ to $n$ minus the sum from $1$ to $j$ . It can be helpful to use this property when the index of summation begins at a number larger than $1.$

Formulas for Sums of the First $n$ Terms of a Sequence

$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \begin{array}{l} {\color{#0066A7}•}\enspace \sum\limits_{k=1}^{n}1= {\hbox{$\stackrel{\underbrace{1+1+1+\cdots+1}}{n{\ \rm terms}}$}} =n\hphantom{00000000} (5)\\ {\color{#0066A7}•}\enspace \sum\limits_{k=1}^{n}k=1+2+3+\cdots+n=\dfrac{n ( n+1) }{2}\hphantom{00000000} (6)\\ {\color{#0066A7}•}\enspace \sum\limits_{k=1}^{n}k^{2}=1^{2}+2^{2}+3^{2}+\cdots +n^{2}=\dfrac{n (n+1) ( 2n+1) }{6}\hphantom{00000} (7) \\ {\color{#0066A7}•}\enspace \sum\limits_{k=1}^{n}k^{3}=1^{3}+2^{3}+3^{3}+\cdots+n^{3}=\left[ \dfrac{n (n+1) }{2}\right] ^{2}\hphantom{00000000} (8) \end{array} }}$

Notice the difference between formulas (5) and (6). In formula (5), the constant $1$ is being summed from $1$ to $n$ , while in (6) the index of summation $k$ is being summed from $1$ to $n$ .

EXAMPLE 7Finding Sums

Find each sum:

(a) $\displaystyle\sum\limits_{k=1}^{100}\dfrac{1}{2}$

(b) $\displaystyle\sum\limits_{k=1}^{5} ( 3k)$

(c) $\displaystyle\sum\limits_{k=1}^{10}( k^{3}+1)$

(d) $\displaystyle\sum\limits_{k=6}^{20} ( 4k^{2})$

(a) $\displaystyle\sum\limits_{k=1}^{100}\dfrac{1}{2}\underset{\underset{\color{#0066A7}{\hbox{(1)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=} \frac{1}{2}\textstyle\displaystyle\sum\limits_{k=1}^{100}1\underset{\underset{\color{#0066A7}{\hbox{(5)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=} \dfrac{1}{2}\cdot 100=50$

A-42

(b) $\displaystyle\sum\limits_{k=1}^{5} ( 3k) \underset{\underset{\color{#0066A7}{\hbox{(1)}}}{\color{#0066A7}{\displaystyle\uparrow}}}{=} 3\displaystyle\sum\limits_{k=1}^{5}k\underset{\underset{\color{#0066A7}{\hbox{(6)}}}{\color{#0066A7}{\displaystyle\uparrow}}}{=} 3\left( \dfrac{5 ( 5+1) }{2}\right)=3 ( 15) =45$

(c) $\displaystyle\sum\limits_{k=1}^{10} ( k^{3}+1) \underset{\underset{\color{#0066A7}{\hbox{(2)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=} \displaystyle\sum \limits_{k=1}^{10}k^{3}+\displaystyle\sum\limits_{k=1}^{10}1 \underset{\underset{\color{#0066A7}{\hbox{(8) and (5)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=} \left[ \dfrac{10( 10+1) }{2}\right] ^{2} +10 =3025+10=3035$

(d)Notice that the index of summation starts at 6. $\begin{eqnarray*} \sum\limits_{k=6}^{20} ( 4k^{2}) &\underset{\underset{\color{#0066A7}{\hbox{(1)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=}& 4\sum\limits_{k=6}^{20}k^{2}\underset{\underset{\color{#0066A7}{\hbox{(4)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=} 4\left[ \sum\limits_{k=1}^{20}k^{2}-\sum \limits_{k=1}^{5}k^{2}\right] \nonumber \\ & \underset{\underset{\color{#0066A7}{\hbox{(7)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=} & 4\left[ \dfrac{20(21) ( 41) }{6}-\dfrac{5( 6) ( 11) }{6}\right] =4[ 2870-55] =11,260 \\[-10.5pt] \end{eqnarray*}$

5 Use the Binomial Theorem
Printed Page A-42

You already know formulas for expanding $( x+a) ^{n}$ for $n=2$ and $n=3$ . The Binomial Theorem is a formula for the expansion of $( x+a) ^{n}$ for any positive integer $n$ . If $n=1$ , 2, 3, and 4, the expansion of $( x+a)^{n}$ is straightforward.

• $(x+a)^{1}=x+a$	Two terms, beginning with $x^{1}$ and ending with $a^{1}$
• $(x+a)^{2}=x^{2}+2ax+a^{2}$	Three terms, beginning with $x^{2}$ and ending with $a^{2}$
• $(x+a)^{3}=x^{3}+3ax^{2}+3a^{2}x+a^{3}$	Four terms, beginning with $x^{3}$ and ending with $a^{3}$
• $(x+a)^{4}=x^{4}+4ax^{3}+6a^{2}x^{2}+4a^{3}x+a^{4}$	Five terms, beginning with $x^{4}$ and ending with $a^{4}$

Notice that each expansion of $( x+a) ^{n}$ begins with $x^{n}$ , ends with $a^{n}$ , and has $n+1$ terms. As you look at the terms from left to right, the powers of $x$ are decreasing by one, while the powers of $a$ are increasing by one. As a result, we might conjecture that the expansion of $( x+a) ^{n}$ would look like this: $( x+a) ^{n}=x^{n}+ \_\_\_\_\_\_ax^{n-1}+ \_\_\_\_\_\_a^{2}x^{n-2}+\cdots+ \_\_\_\_\_\_a^{n-1}x+a^{n}$ where the blanks are numbers to be found. This, in fact, is the case.

Before filling in the blanks, we introduce the symbol $\displaystyle{ n\choose{j} }$ .

DEFINITION

If $j$ and $n$ are integers with $0\leq j\leq n$ , the symbol $\displaystyle{ n\choose{j} }$ is defined as $\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \displaystyle{ n\choose{j} } =\dfrac{n!}{j! ( n-j) !}}}$

A-43

Evaluating $\displaystyle{ n\choose{j}}$

$\displaystyle{{3}\choose{1}} =\dfrac{3!}{1! ( 3-1) !}=\dfrac{3!}{1!2!}=\dfrac{3\cdot 2\cdot 1}{1 ( 2\cdot 1) }=\dfrac{6}{2}=3$
$\displaystyle{{4}\choose{2}} =\dfrac{4!}{2! ( 4-2) !}=\dfrac{4!}{2!2!}=\dfrac{4\cdot 3\cdot 2\cdot 1}{ ( 2\cdot 1) ( 2\cdot 1) }=\dfrac{24}{4}=6$
$\displaystyle{{8}\choose{7}} =\dfrac{8!}{7! ( 8-7) !}=\dfrac{8!}{7!1!} \underset{\underset{\color{#0066A7}{\hbox{$8!=8\cdot 7!$}}}{\color{#0066A7}{\displaystyle\uparrow}}}{=} \dfrac{8\cdot 7!}{7!\cdot 1!}=\dfrac{8}{1}=8$
$\displaystyle{{n}\choose{0}} =\dfrac{n!}{0! ( n-0) !}=\dfrac{n!}{0!n!}=\dfrac{n!}{1\cdot n!}=1$
$\displaystyle{{n}\choose{n}} =\dfrac{n!}{n! ( n-n) !}=\dfrac{n!}{n!0!}=\dfrac{n!}{n!\cdot 1}=1$

RECALL

$0!=1$

THEOREM Binomial Theorem

Let $x$ and $a$ be real numbers. For any positive integer $n$ , $\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \begin{array} ( x+a) ^{n}&=& \displaystyle{{n}\choose{0}} x^{n}+ \displaystyle{{n}\choose{1}} ax^{n-1}+ \displaystyle{{n}\choose{2}} a^{2}x^{n-2}+\cdots+\displaystyle{ n\choose{j} } a^{j}x^{n-j}\\[11pt] &+&\cdots + \displaystyle{{n}\choose{n}} a^{n}=\sum\limits_{j=0}^{n}\displaystyle{ n\choose{j} } a^{j}x^{n-j} \end{array} }}$

Because of its appearance in the Binomial Theorem, the symbol $\displaystyle{ n\choose{j} }$ is called a binomial coefficient.

EXAMPLE 9Expanding a Binomial

Use the Binomial Theorem to expand $( x+2) ^{5}$ .

We use the Binomial Theorem with $a=2$ and $n=5$ . Then $\begin{eqnarray*} ( x+2) ^{5} &=& {{5}\choose{0}} x^{5}+ {{5}\choose{1}} 2x^{4}+ {{5}\choose{2}} 2^{2}x^{3}+ {{5}\choose{3}} 2^{3}x^{2}+ \displaystyle{{5}\choose{4}} 2^{4}x+ {{5}\choose{5}} 2^{5} \nonumber \\[4pt] &=&1\cdot x^{5}+5\cdot 2 x^{4}+10\cdot 4 x^{3}+10\cdot 8 x^{2}+5\cdot 16 x+1\cdot 32 \nonumber \\[4pt] &=&x^{5}+10 x^{4}+40 x^{3}+80 x^{2}+80 x+32 \end{eqnarray*}$

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A.5 Sequences; Summation Notation; the Binomial TheoremPrinted Page A-38

OBJECTIVES

1 Write the First Several Terms of a SequencePrinted Page A-38

DEFINITION Sequence

EXAMPLE 1Writing the Terms of a Sequence

EXAMPLE 2Determining a Sequence from a Pattern

2 Write the Terms of a Recursively Defined SequencePrinted Page A-39

EXAMPLE 3Writing the Terms of a Recursively Defined Sequence

EXAMPLE 4Writing the Terms of a Recursively Defined Sequence

3 Use Summation NotationPrinted Page A-40

NOTE

NOTE

EXAMPLE 5Expanding Summation Notation

EXAMPLE 6Writing a Sum in Summation Notation

4 Find the Sum of the First nn Terms of a SequencePrinted Page A-41

THEOREM Properties Involving Summation Notation

Formulas for Sums of the First nn Terms of a Sequence

EXAMPLE 7Finding Sums

5 Use the Binomial TheoremPrinted Page A-42

DEFINITION

EXAMPLE 8Evaluating \displaystyle{ n\choose{j}}\displaystyle{ n\choose{j}}

RECALL

THEOREM Binomial Theorem

EXAMPLE 9Expanding a Binomial

A.5 Sequences; Summation Notation; the Binomial Theorem
Printed Page A-38

1 Write the First Several Terms of a Sequence
Printed Page A-38

2 Write the Terms of a Recursively Defined Sequence
Printed Page A-39

3 Use Summation Notation
Printed Page A-40

4 Find the Sum of the First $n$ Terms of a Sequence
Printed Page A-41

Formulas for Sums of the First $n$ Terms of a Sequence

5 Use the Binomial Theorem
Printed Page A-42

Evaluating $\displaystyle{ n\choose{j}}$