A-38
When you finish this section you should be able to:
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 \[ \{ b_{n}\} =\left\{ \left( \dfrac{1}{2}\right) ^{n}\right\} \]
Write the first six terms of the sequence: \( \{ b_{n}\} = \left\{ ( -1) ^{n-1} \, \dfrac{2}{n} \right\} \)
Solution \( \begin{array}[t]{lllll} b_{1}= ( -1) ^{0}\,\dfrac{2}{1}=2 & &\qquad b_{2}= ( -1) ^{1}\, \dfrac{2}{2}=-1 & &\qquad b_{3}= ( -1) ^{2}\,\dfrac{2}{3}=\dfrac{2}{3} \\[7pt] b_{4}=-\dfrac{1}{2} & &\qquad b_{5}=\dfrac{2}{5} & &\qquad b_{6}=-\dfrac{1}{3} \end{array} \)
The first six terms of the sequence are \(2,-1,\,\dfrac{2}{3},-\dfrac{1}{2},\, \dfrac{2}{5},-\dfrac{1}{3}\).
A-39
The pattern on the left suggests the \(n{\rm th}\) term of the sequence on the right.
| Pattern | \( n\)th | Sequence |
|---|---|---|
| \(e, \dfrac{e^{2}}{2}, \dfrac{e^{3}}{3}, \dfrac{e^{4}}{4},\ldots\) | \(a_{n}=\dfrac{e^{n}}{n}\) | \( \{ a_{n}\} =\left\{ \dfrac{e^{n}}{n}\right\}\) |
| \(1,~\dfrac{1}{3},~\dfrac{1}{9},~\dfrac{1}{27},\ldots\) | \(b_{n}=\dfrac{1}{3^{n-1}}\) | \( \{ b_{n}\} =\left\{ \dfrac{1}{3^{n-1}}\right\}\) |
| \(1,~3,~5,~7,\ldots\) | \(c_{n}=2n-1\) | \( \{ c_{n}\} = \{2n-1\}\) |
| \(1,~4,~9,~16,~25,\ldots\) | \(d_{n}=n^{2}\) | \( \{ d_{n}\} = \{n^{2}\}\) |
| \(1,~{-}\dfrac{1}{2},~\dfrac{1}{3},~{-}\dfrac{1}{4},~\dfrac{1}{5},\ldots\) | \(e_{n}= ( -1) ^{n+1}\left( \dfrac{1}{n}\right)\) | \( \{ e_{n}\} =\left\{( -1) ^{n+1}\left( \dfrac{1}{n}\right) \right\}\) |
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.
Write the first five terms of the recursively defined sequence. \[ s_{1}=1\qquad s_{n}=4s_{n-1}\qquad n\geq 2 \]
Solution 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*}
Write the first five terms of the recursively defined sequence. \[ u_{1}=1 \qquad u_{2}=1 \qquad u_{n}=u_{n-2}+u_{n-1}\quad n\geq 3 \]
Solution 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.
A-40
It is often important to be able to find the sum of the first \(n\) terms of a sequence \(\{ a_{n}\} \), namely, \[ a_{1}+a_{2}+a_{3}+\cdots+a_{n} \]
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 \(k=n\).”
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 \(n>1\).
Write out each sum:
(a) \(\displaystyle\sum\limits_{k=1}^{n}\dfrac{1}{k}\)
(b) \(\displaystyle\sum\limits_{k=1}^{n}k!\)
Solution (a) \(\displaystyle\sum\limits_{k=1}^{n}\dfrac{1}{k}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}\)
(b) \(\displaystyle\sum\limits_{k=1}^{n}k!=1!+2!+\cdots+n!\)
Express each sum using summation notation:
(a) \(1^{2}+2^{2}+3^{2}+\cdots+9^{2}\)
(b) \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots+\dfrac{1}{2^{n-1}}\)
Solution (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).
A-41
When working with summation notation, the following properties are useful for finding the sum of the first \(n\) terms of a sequence.
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 \(1.\)
\[\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 \(n\).
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) \(\displaystyle\sum\limits_{k=6}^{20} ( 4k^{2})\)
Solution (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*}
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 \(\displaystyle{ n\choose{j} } \).
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
\(0!=1\)
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.
Use the Binomial Theorem to expand \(( x+2) ^{5}\).
Solution 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*}