Learning Objectives

By the end of this section, you will be able to:

- Write the first few terms of a sequence
- Find a formula for the general term (nth term) of a sequence
- Use factorial notation
- Find the partial sum
- Use summation notation to write a sum

Before you get started, take this readiness quiz.

- Evaluate (2n+3) for the integers (1, 2, 3), and (4).

If you missed this problem, review Example 1.6. - Evaluate ((−1)^{n}) for the integers (1, 2, 3), and (4).

If you missed this problem, review Example 1.19. - If (f(n)=n^{2}+2), find (f(1)+f(2)+f(3)).

If you missed this problem, review Example 3.49.

## Write the First Few Terms of a Sequence

Let’s look at the function (f(x)=2x) and evaluate it for just the counting numbers.

(f(x)=2x) | |

(x) | (2x) |

(1) | (2) |

(2) | (4) |

(3) | (6) |

(4) | (8) |

(5) | (10) |

(...) | (...) |

If we list the function values in order as (2, 4, 6, 8), and (10), … we have a sequence. A **sequence** is a function whose domain is the counting numbers.

Definition (PageIndex{1})

A **sequence** is a function whose domain is the counting numbers.

A sequence can also be seen as an ordered list of numbers and each number in the list is a *term*. A sequence may have an infinite number of terms or a finite number of terms. Our sequence has three dots (ellipsis) at the end which indicates the list never ends. If the domain is the set of all counting numbers, then the sequence is an **infinite sequence**. Its domain is all counting numbers and there is an infinite number of counting numbers.

(2,4,6,8,10, dots)

If we limit the domain to a finite number of counting numbers, then the sequence is a **finite sequence**. If we use only the first four counting numbers, (1, 2, 3, 4) our sequence would be the finite sequence,

(2,4,6,8)

Often when working with sequences we do not want to write out all the terms. We want more compact way to show how each term is defined. When we worked with functions, we wrote (f(x)=2x) and we said the expression (2x) was the rule that defined values in the range. While a sequence is a function, we do not use the usual function notation. Instead of writing the function as (f(x)=2x), we would write it as (a_{n}=2n). The (a_{n}) is the (n)th term of the sequence, the term in the (n)th position where (n) is a value in the domain. The formula for writing the (n)th term of the sequence is called the **general term** or formula of the sequence.

Definition (PageIndex{2})

The **general term** of the sequence is found from the formula for writing the (n)th term of the sequence. The (n)th term of the sequence, (a_{n}), is the term in the (n)th position where (n) is a value in the domain.

When we are given the general term of the sequence, we can find the terms by replacing (n) with the counting numbers in order. For (a_{n}=2 n),

(n) | (1) | (2) | (3) | (4) | (5) | (6) |

(a_{n}) | 2(cdot 1) | 2(cdot 2) | 2(cdot 3) | 2(cdot 4) | 2(cdot 5) | 2(cdot 6) |

(2) | (4) | (6) | (8) | (10) |

(a_{1}, quad a_{2}, quad a_{3}, quad a_{4}, quad a_{5}, ldots, quad a_{n}, dots)

(2, quad 4, quad 6, quad 8, quad10, dots)

To find the values of a sequence, we substitute in the counting numbers in order into the general term of the sequence.

Example (PageIndex{1})

Write the first five terms of the sequence whose general term is (a_{n}=4 n-3).

**Solution**:

We substitute the values (1, 2, 3, 4), and (5) into the formula, (a_{n}=4n−3), in order.

**Answer**:

The first five terms of the sequence are (1, 5, 9, 13), and (17).

Exercise (PageIndex{1})

Write the first five terms of the sequence whose general term is (a_{n}=3n-4).

**Answer**(-1,2,5,8,11)

Exercise (PageIndex{2})

Write the first five terms of the sequence whose general term is (a_{n}=2n-5).

**Answer**(-3,-1,1,3,5)

For some sequences, the variable is an exponent.

Example (PageIndex{2})

Write the first five terms of the sequence whose general term is (a_{n}=2^{n}+1).

**Solution**:

We substitute the values (1, 2, 3, 4), and (5) into the formula, (a_{n}=2^{n}+1), in order.

**Answer**:

The first five terms of the sequence are (3, 5, 9, 17), and (33).

Exercise (PageIndex{3})

Write the first five terms of the sequence whose general term is (a_{n}=3^{n}+4).

**Answer**(7,13,31,85,247)

Exercise (PageIndex{4})

Write the first five terms of the sequence whose general term is (a_{n}=2^{n}-5).

**Answer**(-3,-1,3,11,27)

It is not uncommon to see the expressions ((−1)^{n}) or ((−1)^{n+1}) in the general term for a sequence. If we evaluate each of these expressions for a few values, we see that this expression alternates the sign for the terms.

(n) | (1) | (2) | (3) | (4) | (5) |
---|---|---|---|---|---|

((-1)^{n}) | ((-1)^{1}) (-1) | ((-1)^{2}) 1 | ((-1)^{3}) (-1) | ((-1)^{4}) (1) | ((-1)^{5}) (-1) |

((-1)^{n+1}) | ((-1)^{1+1}) 1 | ((-1)^{2+1}) (-1) | ((-1)^{3+1}) 1 | ((-1)^{4+1}) (-1) | ((-1)^{5+1}) 1 |

(a_{1}, quad a_{2}, quad a_{3}, quad a_{4}, quad a_{5}, dots, quad a_{n}, dots)

Example (PageIndex{3})

Write the first five terms of the sequence whose general term is (a_{n}=(-1)^{n} n^{3}).

**Solution**:

We substitute the values (1, 2, 3, 4), and (5) into the formula, (a_{n}=(-1)^{n} n^{3}), in order.

**Answer**:

The first five terms of the sequence are (−1, 8, −27, 64, −1, 8, −27, 64), and (−125).

Exercise (PageIndex{5})

Write the first five terms of the sequence whose general term is (a_{n}=(-1)^{n} n^{2}).

**Answer**(-1,4,-9,16,-25)

Exercise (PageIndex{6})

Write the first five terms of the sequence whose general term is (a_{n}=(-1)^{n+1} n^{3}).

**Answer**(1,-8,27,-64,125)

## Find a Formula for the General Term ((n)th Term) of a Sequence

Sometimes we have a few terms of a sequence and it would be helpful to know the general term or (n)th term. To find the general term, we look for patterns in the terms. Often the patterns involve multiples or powers. We also look for a pattern in the signs of the terms.

Example (PageIndex{4})

Find a general term for the sequence whose first five terms are shown. (4,8,12,16,20, dots)

**Solution**:

We look for a pattern in the terms. The numbers are all multiples of (4). The general term of the sequence is (a_{n}=4n). Table 12.1.4 **Answer**:The general term of the sequence is (a_{n}=4n).

Exercise (PageIndex{7})

Find a general term for the sequence whose first five terms are shown.

(3,6,9,12,15, dots)

**Answer**(a_{n}=3 n)

Exercise (PageIndex{8})

Find a general term for the sequence whose first five terms are shown.

(5,10,15,20,25, dots)

**Answer**(a_{n}=5 n)

Example (PageIndex{5})

Find a general term for the sequence whose first five terms are shown. (2,-4,8,-16,32, dots)

**Solution**:

We look for a pattern in the terms. The numbers are powers of (2). The signs are alternating, with even (n) negative. The general term of the sequence is (a_{n}=(-1)^{n+1} 2^{n}) Table 12.1.5 **Answer**:The general term of the sequence is (a_{n}=(-1)^{n+1}2^{n}).

Exercise (PageIndex{9})

Find a general term for the sequence whose first five terms are shown.

(-3,9,-27,81,-243, dots)

**Answer**(a_{n}=(-1)^{n} 3^{n})

Exercise (PageIndex{10})

Find a general term for the sequence whose first five terms are shown

(1,-4,9,-16,25, dots)

**Answer**(a_{n}=(-1)^{n+1} n^{2})

Example (PageIndex{6})

Find a general term for the sequence whose first five terms are shown. (frac{1}{3}, frac{1}{9}, frac{1}{27}, frac{1}{81}, frac{1}{243}, dots)

**Solution**:

We look for a pattern in the terms. The numerators are all (1). The denominators are powers of (3). The general term of the sequence is (a_{n}=frac{1}{3^{n}}). Table 12.1.6 **Answer**:The general term of the sequence is (a_{n}=frac{1}{3^{n}}).

Exercise (PageIndex{11})

Find a general term for the sequence whose first five terms are shown.

(frac{1}{2}, frac{1}{4}, frac{1}{8}, frac{1}{16}, frac{1}{32}, dots)

**Answer**(a_{n}=frac{1}{2^{n}})

Exercise (PageIndex{12})

Find a general term for the sequence whose first five terms are shown.

(frac{1}{1}, frac{1}{4}, frac{1}{9}, frac{1}{16}, frac{1}{25}, dots)

**Answer**(a_{n}=frac{1}{n^{2}})

## Use Factorial Notation

Sequences often have terms that are products of consecutive integers. We indicate these products with a special notation called *factorial notation*. For example, (5!), read (5) factorial, means (5⋅4⋅3⋅2⋅1). The exclamation point is not punctuation here; it indicates the **factorial notation**.

Definition (PageIndex{3})

If (n) is a positive integer, then (n!) is

(n !=n(n-1)(n-2) dots)

We define (0!) as (1), so (0!=1).

The values of (n!) for the first (5) positive integers are shown.

(egin{array}{ccccc}{1 !} & {2 !} & {3 !} & {4 !} & {5 !} {1} & quad{2 cdot 1} & quad {3 cdot 2 cdot 1} & quad{4 cdot 3 cdot 2 cdot 1} & quad {5 cdot 4 cdot 3 cdot 2 cdot 1} {1} & {2} & {6} & {24} & {120}end{array})

Example (PageIndex{7})

Write the first five terms of the sequence whose general term is (a_{n}=frac{1}{n !}).

**Solution**:

We substitute the values (1, 2, 3, 4, 5) into the formula, (a_{n}=frac{1}{n !}), in order.

**Answer**:

The first five terms of the sequence are (1, frac{1}{2}, frac{1}{6}, frac{1}{24}, frac{1}{120}).

Exercise (PageIndex{13})

Write the first five terms of the sequence whose general term is (a_{n}=frac{2}{n !}).

**Answer**(2,1, frac{1}{3}, frac{1}{12}, frac{1}{60})

Exercise (PageIndex{14})

Write the first five terms of the sequence whose general term is (a_{n}=frac{3}{n !}).

**Answer**(3, frac{3}{2}, frac{1}{2}, frac{1}{8}, frac{1}{40})

When there is a fraction with factorials in the numerator and denominator, we line up the factors vertically to make our calculations easier.

Example (PageIndex{8})

Write the first five terms of the sequence whose general term is (a_{n}=frac{(n+1) !}{(n-1) !}).

**Solution**:

We substitute the values (1, 2, 3, 4, 5) into the formula, (a_{n}=frac{(n+1) !}{(n-1) !}), in order.

**Answer**:

The first five terms of the sequence are (2, 6, 12, 20), and (30).

Exercise (PageIndex{15})

Write the first five terms of the sequence whose general term is (a_{n}=frac{(n-1) !}{(n+1) !})

**Answer**(frac{1}{2}, frac{1}{6}, frac{1}{12}, frac{1}{20}, frac{1}{30})

Exercise (PageIndex{16})

Write the first five terms of the sequence whose general term is (a_{n}=frac{n !}{(n+1) !}).

**Answer**(frac{1}{2}, frac{1}{3}, frac{1}{4}, frac{1}{5}, frac{1}{6})

## Find the Partial Sum

Sometimes in applications, rather than just list the terms, it is important for us to add the terms of a sequence. Rather than just connect the terms with plus signs, we can use **summation notation**.

For example, (a_{1}+a_{2}+a_{3}+a_{4}+a_{5}) can be written as (sum_{i=1}^{5} a_{i}). We read this as “the sum of (a) sub (i) from (i) equals one to five.” The symbol (∑) means to add and the (i) is the index of summation. The (1) tells us where to start (initial value) and the (5) tells us where to end (terminal value).

Definition (PageIndex{4})

The sum of the first (n) terms of a sequence whose (n)th term is (a_{n}) is written in summation notation as:

(sum_{i=1}^{n} a_{i}=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+ldots+a_{n})

The (i) is the index of summation and the (1) tells us where to start and the (n) tells us where to end.

When we add a finite number of terms, we call the sum a **partial sum**.

Example (PageIndex{9})

Expand the partial sum and find its value: (sum_{i=1}^{5} 2 i).

**Solution**:

(sum_{i=1}^{5} 2 i) We substitute the values (1, 2, 3, 4, 5) in order. (2 cdot 1+2 cdot 2+2 cdot 3+2 cdot 4 + 2 cdot 5) Simplify. (2+4+6+8+10) Add. (egin{array} {c} 30 sum_{i=1}^{5} 2 i=30 end{array}) Table 12.1.7

(egin{array} {c} 30 sum_{i=1}^{5} 2 i=30 end{array})**Answer**:

Exercise (PageIndex{17})

Expand the partial sum and find its value: (sum_{i=1}^{5} 3 i).

**Answer**(45)

Exercise (PageIndex{18})

Expand the partial sum and find its value: (sum_{i=1}^{5} 4 i).

**Answer**(60)

The index does not always have to be (i) we can use any letter, but (i) and (k) are commonly used. The index does not have to start with (1) either—it can start and end with any positive integer.

Example (PageIndex{10})

Expand the partial sum and find its value: (sum_{k=0}^{3} frac{1}{k !}).

**Solution**:

(egin{array}{c c} {}&{sum_{k=0}^{3} frac{1}{k !}} {We:substitute:the:values:0,1,2,3:in:order.}&{frac{1}{1}+frac{1}{1 !}+frac{1}{2 !}+frac{1}{3 !}} {Evaluate:the:factorials.}& {frac{1}{1}+frac{1}{1}+frac{1}{2 !}+frac{1}{6}} {Simplify.}&{1+1+frac{3}{6}+frac{1}{6}} {Simplify.}& {frac{16}{6}} {Simplify.}&{frac{8}{3}} {}& {sum_{k=0}^{3} frac{1}{k !}=frac{8}{3}}end{array})

Exercise (PageIndex{19})

Expand the partial sum and find its value: (sum_{k=0}^{3} frac{2}{k !}).

**Answer**(frac{16}{3})

Exercise (PageIndex{20})

Expand the partial sum and find its value: (sum_{k=0}^{3} frac{3}{k !}).

**Answer**(8)

## Use Summation Notation to Write a Sum

In the last two examples, we went from summation notation to writing out the sum. Now we will start with a sum and change it to summation notation. This is very similar to finding the general term of a sequence. We will need to look at the terms and find a pattern. Often the patterns involve multiples or powers.

Example (PageIndex{11})

Write the sum using summation notation: (1+frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}).

**Solution**:

(egin{array} {}&{ 1+frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}} {}&{n : 1,2,3,4,5} { ext{We look for a pattern in the terms.}}&{ ext { Terms: } 1, frac{1}{2}, frac{1}{3}, frac{1}{4}, frac{1}{5}} { ext{The numerators are all one.}}&{ ext { Pattern: } frac{1}{1}, frac{1}{2}, frac{1}{3}, frac{1}{4}, frac{1}{5}, ldots frac{1}{n}} { ext{The denominators are the counting numbers from one to five.}}&{ ext{The sum written in summation notation}} {}&{1 + frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}=sum^{5}_{n=1} frac{1}{n}.} end{array})

Exercise (PageIndex{21})

Write the sum using summation notation: (frac{1}{2}+frac{1}{4}+frac{1}{8}+frac{1}{16}+frac{1}{32}).

**Answer**(sum_{n=1}^{5} frac{1}{2^{n}})

Exercise (PageIndex{22})

Write the sum using summation notation: (1+frac{1}{4}+frac{1}{9}+frac{1}{16}+frac{1}{25})

**Answer**(sum_{n=1}^{5} frac{1}{n^{2}})

When the terms of a sum have negative coefficients, we must carefully analyze the pattern of the signs.

Example (PageIndex{12})

Write the sum using summation notation: (-1+8-27+64-125).

**Solution**:

We look for a pattern in the terms. The signs of the terms alternate,

and the odd terms are negative.The numbers are the cubes of the

counting numbers from one to five.The sum written in summation notation is (-1+8-27+64-125=sum_{n=1}^{5}(-1)^{n} cdot n^{3}) Table 12.1.8

Exercise (PageIndex{23})

Write each sum using summation notation: (1-4+9-16+25).

**Answer**(sum_{n=1}^{5}(-1)^{n+1} n^{2})

Exercise (PageIndex{24})

Write each sum using summation notation: (-2+4-6+8-10).

**Answer**(sum_{n=1}^{5}(-1)^{n} 2 n)

Access this online resource for additional instruction and practice with sequences.

## Key Concepts

**Factorial Notation**

If (n) is a positive integer, then (n!) is

(n !=n(n-1)(n-2) ldots(3)(2)(1))

We define (0!) as (1), so (0!=1)

**Summation Notation**

The sum of the first (n) terms of a sequence whose (n)th term (a_{n}) is written in summation notation as:

(sum_{i=1}^{n} a_{i}=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+ldots+a_{n})

The (i) is the index of summation and the (1) tells us where to start and the (n) tells us where to end.

### Glossary

**finite sequence**- A sequence with a domain that is limited to a finite number of counting numbers.

**general term of a sequence**- The general term of the sequence is the formula for writing the (n)th term of the sequence. The (n)th term of the sequence, (a_{n}), is the term in the (n)th position where (n) is a value in the domain.

**infinite sequence**- A sequence whose domain is all counting numbers and there is an infinite number of counting numbers.

**partial sum**- When we add a finite number of terms of a sequence, we call the sum a partial sum.

**sequence**- A sequence is a function whose domain is the counting numbers.

## 9.1: Sequences

Many of the sequence functions take keyword arguments see Argument Lists. All keyword arguments are optional and, if specified, may appear in any order.

The :key argument should be passed either nil , or a function of one argument. This key function is used as a filter through which the elements of the sequence are seen for example, (cl-find x y :key 'car) is similar to (cl-assoc x y) . It searches for an element of the list whose CAR equals x , rather than for an element which equals x itself. If :key is omitted or nil , the filter is effectively the identity function.

The :test and :test-not arguments should be either nil , or functions of two arguments. The test function is used to compare two sequence elements, or to compare a search value with sequence elements. (The two values are passed to the test function in the same order as the original sequence function arguments from which they are derived, or, if they both come from the same sequence, in the same order as they appear in that sequence.) The :test argument specifies a function which must return true (non- nil ) to indicate a match instead, you may use :test-not to give a function which returns *false* to indicate a match. The default test function is eql .

Many functions that take item and :test or :test-not arguments also come in -if and -if-not varieties, where a predicate function is passed instead of item , and sequence elements match if the predicate returns true on them (or false in the case of -if-not ). For example:

to remove all zeros from sequence seq .

Some operations can work on a subsequence of the argument sequence these function take :start and :end arguments, which default to zero and the length of the sequence, respectively. Only elements between start (inclusive) and end (exclusive) are affected by the operation. The end argument may be passed nil to signify the length of the sequence otherwise, both start and end must be integers, with 0 <= start <= end <= (length seq ) . If the function takes two sequence arguments, the limits are defined by keywords :start1 and :end1 for the first, and :start2 and :end2 for the second.

A few functions accept a :from-end argument, which, if non- nil , causes the operation to go from right-to-left through the sequence instead of left-to-right, and a :count argument, which specifies an integer maximum number of elements to be removed or otherwise processed.

The sequence functions make no guarantees about the order in which the :test , :test-not , and :key functions are called on various elements. Therefore, it is a bad idea to depend on side effects of these functions. For example, :from-end may cause the sequence to be scanned actually in reverse, or it may be scanned forwards but computing a result &ldquoas if&rdquo it were scanned backwards. (Some functions, like cl-mapcar and cl-every , *do* specify exactly the order in which the function is called so side effects are perfectly acceptable in those cases.)

Strings may contain &ldquotext properties&rdquo as well as character data. Except as noted, it is undefined whether or not text properties are preserved by sequence functions. For example, (cl-remove ?A str ) may or may not preserve the properties of the characters copied from str into the result.

## NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series

Topics and Sub Topics in Class 11 Maths Chapter 9 Sequences and Series:

Section Name | Topic Name |

9 | Sequences and Series |

9.1 | Introduction |

9.2 | Sequences |

9.3 | Series |

9.4 | Arithmetic Progression (A.P.) |

9.5 | Geometric Progression (G.P.) |

9.6 | Relationship Between A.M. and G.M. |

9.7 | Sum to n terms of Special Series |

### NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1

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### NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series (अनुक्रम तथा श्रेणी) Hindi Medium Ex 9.1

### NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2

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### NCERT Solutions for class 11 Maths Chapter 9 Exercise 9.3

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### NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.4

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### Class 11 Maths NCERT Miscellaneous Solutions

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## PDI transformation counter generated sequence

The following table contains options for generating a sequence from a PDI transformation counter:

Select this check box if you want the sequence to be generated by PDI. This option is set by default

Use DB to generate the sequence? is automatically selected if this check box is cleared.

For example, if you set Start at value to 1 , Increment by to 1 , and Maximum value to 3 , then the resulting sequence will be 1, 2, 3, 1, 2, 3, 1, 2 . If you set Start at value to 0 , Increment by to -1 , and Maximum value to -2 , then the resulting sequence will be 0, -1, -2, 0, -1, -2, 0 .

### Precalculus 9.1 Sequences and Series

A sequence is a function whose domain is the set of positive integers.

Definition of Sequence :

An infinite sequence is a function whose domain is the set of positive integers. The function values

a_{1}, a_{2}, a_{3}, a_{4}, . , a_{n}, .

are the terms of the sequence. If the domain of the function consists of the first n positive integers only, the sequence is a finite sequence .

A. Finding the Terms of a Sequence :

Example 1 : find the first five terms of a_{n} = 4n - 7

a_{1} = 4(1) - 7 = -3

a_{2} = 4(2) - 7 = 8 - 7 = 1

a_{3},= 4(3) - 7 = 12 - 7 = 5

a_{4} = 4(4) - 7 = 16 - 7 = 9

a_{5} = 4(5) - 7 = 20 - 7 = 13

Example 2 : Find the 16th term of the sequence a_{n} = (-1) n-1 (n(n-1))

a_{16} = (-1) 16-1 (16(16-1)) = (-1) 15 (16(15)) = (-1)(240) = -240

Example 3 : Write the first five terms of the sequence defined recursively:

a_{1} = 15, a_{k+1} = a_{k}+ 3

Let k = 1 so we have:

a_{1+1} = a_{2} = a_{1} + 3 = 15 + 3 = 18

Let k = 2 so we have:

a_{2+1} = a_{3} = a_{2} + 3 = 18 + 3 = 21

a_{3+1} = a_{4} = a_{3} + 3 = 21 + 3 = 24

a_{4+1} = a_{5} = a_{4} + 3 = 24 + 3 = 27

B. Finding the nth term of a Sequence

Example 4 : Write an expression for the apparent nth term of the sequence

(assume n begins with 1): 3, 7, 11, 15, 19, .

As you can see, the terms are going up by 4 and the first term is one less than 4 so

a_{n} = 4n - 1

Example 5 : 1, 1/4, 1/9, 1/16, 1/25, .

As you can see, the terms denominators are perfect squares so

a_{n} = 1/(n 2 ) = n -2

C. The Fibonacci Sequence: A Recursive Sequence

The Fibonacci sequence is defined recursively as follows.

a_{0} = 1, a_{1} = 1, a_{k} = a_{k-2} + a_{k-1} , where k is greater than or equal to 2

Example 6: Write the first five terms of the sequence defined recursively. Use this pattern to write the nth term of the sequence as a function of n.

a_{1} = 25, a_{k+1} = a_{k} - 5

a_{2} = a_{1+1} = a_{1} -5 = 25 -5 = 20

a_{3} = a_{2+1} = a_{2} -5 = 20 - 5 = 15

a_{4} = a_{3+1} = a_{3} -5 = 15 - 5 = 10

a_{5} = a_{4+1} = a_{4} -5 = 10 - 5 = 5

Therefore a_{n} = 25 - 5n

C. Definition of Factorial :

In n is a positive integer, n factorial is defined by

n! = 1 x 2 x 3 x 4 x . (n - 1) x n

As a special case, zero factorial is defined as 0! = 1

Example 7 : Simplify the ratio of factorials.

(4!)/(7!) = (4!) / (7 x 6 x 5 x 4!) = 1/ (7 x 6 x 5 ) = 1 / 210

Example 8: Simplify the ratio of factorials .

(n + 2)! / (n!) = (n + 2)(n + 1)(n!) / (n!) = (n + 2)(n + 1) = n 2 + 3n + 2

D. Definition of Summation Notation:

The sum of the first n terms of a sequence is represented by

The summation of a_{i} when i = 1 to i = n is equal to

where *i* is called the index of summation, *n* is the upper limit of summation, and 1 is the lower limit of summation. (check out http://www.cs.fsu.edu/

Example 9: Find the sum of 3i - 1 when i = 1 to i = 6

3 - 1 + 6 - 1 + 9 - 1 + 12 - 1 + 15 - 1 + 18 - 1 = 57

To use your TI - 83, TI - 83 plus calculators:

Go to mode - change from function to sequential mode, then

2nd Stat arrow to math #5 Sum, 2nd Stat arrow to ops #5 seq then put in the sequence, then n to show the calculator the variable, then the lower bound number, then the upper bound number, then close the parenthesis twice, then press enter.

Your screen should look like this:

sum(seq(3n-1,n,1,6)) = 57

Definition of a Series:

Consider the infinite sequence a_{1}, a_{2}, a_{3}, a_{4}, . , a_{i}, .

1. The sum of all terms of the infinite sequence is called an **infinite series** and is denoted by

a_{1} + a_{2} + a_{3} + a_{4} + . + a_{i} + . = The summation of a_{i} when i = 1 to i = infinity.

2. The sum of the first n terms of the sequence is called a finite series or the nth partial sum of the sequence and is denoted by

a_{1} + a_{2} + a_{3} + a_{4} + . + a_{n} = The summation of a_{i} when i = 1 to i = n.

Example 10 : Find the sum of the partial sum of the series:

The summation of 8(-1/2) n when n = 1 to n = infinity, the 4th partial sum

= 8(-½ ) 1 + 8(-½ ) 2 + 8(-½ ) 3 + 8(-½ ) 4 = -4 + 2 + -1 + .5 = -5/2

(Calculator check: you can check your individual answers by putting you calculator in function mode, put the series into y1 = 8 (-.5) x and looking in your table to get your values.)

Example 11 : Find the sum of the infinite series:

## Chapter 9 Ex.9.1 Question 12

Write the first five terms of the following sequence and obtain the corresponding series:(

### Solution

**Video Solution**

Hence, the first five terms of the sequence are ( - 1,frac<< - 1>><2>,frac<< - 1>><6>,frac<< - 1>><<24>>) and (frac<< - 1>><<120>>).

The corresponding series is (left( < - 1>
ight) + left(

## 9.1: PN Sequence

A *pseudo-noise* (PN) sequence is extensively used in the LTE system for various purposes such as scrambling of reference signals, scrambling of downlink and uplink data transmissions as well as in generating various hopping sequences.

### 9.1.1 Maximal length sequence

A PN sequence can be generated by using linear feedback shift registers (LFSR). The shiftregister sequences with the maximum possible period for an *l*-stage shift register are called *maximal length sequences* or *m-sequences*. A necessary and sufficient condition for a sequence generated by an LFSR to be of maximal length ( *m*-sequence) is that its corresponding polynomial be primitive. The periodic autocorrelation function for an *m*-sequence *x*( *n*) is defined as:

The periodic autocorrelation function *R* ( *k*) equals:

We note that the autocorrelation of an *m*-sequence is 1 for zero-lag, and nearly zero ( ? 1 */M*, where *M* is the sequence length) for all other lags. In other words, the autocorrelation of the *m*-sequence can be said to approach unit impulse function as *m*-sequence length increases.

### 9.1.2 Gold sequence

The Gold sequences have been proposed by Gold in 1967 [1]. These sequences are constructed by EXOR-ing two *m*-sequences of the same length as shown in Figure 9.1. Thus, for a Gold sequence of length *n* = 2 *l* ? 1 we need to use two LFSR sequences, each of length *n* = 2 *l* ? 1. If.

## Many Rules

One of the troubles with finding "the next number" in a sequence is that mathematics is so powerful we can find more than one Rule that works.

### What is the next number in the sequence 1, 2, 4, 7, ?

Here are three solutions (there can be more!):

Solution 1: Add 1, then add 2, 3, 4, .

So, 1+**1**=2, 2+**2**=4, 4+**3**=7, 7+**4**=11, etc.

Sequence: 1, 2, 4, 7, **11, 16, 22, .**

(That rule looks a bit complicated, but it works)

Solution 2: After 1 and 2, add the two previous numbers, plus 1:

Sequence: 1, 2, 4, 7, **12, 20, 33, .**

Solution 3: After 1, 2 and 4, add the three previous numbers

Sequence: 1, 2, 4, 7, **13, 24, 44, .**

So, we have three perfectly reasonable solutions, and they create totally different sequences.

Which is right? **They are all right.**

. it may be a list of the winners' numbers . so the next number could be . anything!

## 9.1: Sequences

A string is a sequence of bytes or characters, enclosed within either single quote ( ' ) or double quote ( " ) characters. Examples:

Quoted strings placed next to each other are concatenated to a single string. The following lines are equivalent:

If the ANSI_QUOTES SQL mode is enabled, string literals can be quoted only within single quotation marks because a string quoted within double quotation marks is interpreted as an identifier.

A binary string is a string of bytes. Every binary string has a character set and collation named binary . A nonbinary string is a string of characters. It has a character set other than binary and a collation that is compatible with the character set.

For both types of strings, comparisons are based on the numeric values of the string unit. For binary strings, the unit is the byte comparisons use numeric byte values. For nonbinary strings, the unit is the character and some character sets support multibyte characters comparisons use numeric character code values. Character code ordering is a function of the string collation. (For more information, see Section 10.8.5, “The binary Collation Compared to _bin Collations”.)

Within the **mysql** client, binary strings display using hexadecimal notation, depending on the value of the --binary-as-hex . For more information about that option, see Section 4.5.1, “mysql — The MySQL Command-Line Client”.

A character string literal may have an optional character set introducer and COLLATE clause, to designate it as a string that uses a particular character set and collation:

You can use N' *literal* ' (or n' *literal* ' ) to create a string in the national character set. These statements are equivalent:

Within a string, certain sequences have special meaning unless the NO_BACKSLASH_ESCAPES SQL mode is enabled. Each of these sequences begins with a backslash ( ), known as the *escape character* . MySQL recognizes the escape sequences shown in Table 9.1, “Special Character Escape Sequences”. For all other escape sequences, backslash is ignored. That is, the escaped character is interpreted as if it was not escaped. For example, x is just x . These sequences are case-sensitive. For example, is interpreted as a backspace, but B is interpreted as B . Escape processing is done according to the character set indicated by the character_set_connection system variable. This is true even for strings that are preceded by an introducer that indicates a different character set, as discussed in Section 10.3.6, “Character String Literal Character Set and Collation”.

**Table 9.1 Special Character Escape Sequences**

Escape Sequence | Character Represented by Sequence |
---|---|

An ASCII NUL ( X'00' ) character | |

' | A single quote ( ' ) character |

" | A double quote ( " ) character |

A backspace character | |

A newline (linefeed) character | |

A carriage return character | |

A tab character | |

ASCII 26 (Control+Z) see note following the table | |

\ | A backslash ( ) character |

\% | A % character see note following the table |

\_ | A _ character see note following the table |

The ASCII 26 character can be encoded as to enable you to work around the problem that ASCII 26 stands for END-OF-FILE on Windows. ASCII 26 within a file causes problems if you try to use mysql *db_name* < *file_name* .

The \% and \_ sequences are used to search for literal instances of % and _ in pattern-matching contexts where they would otherwise be interpreted as wildcard characters. See the description of the LIKE operator in Section 12.8.1, “String Comparison Functions and Operators”. If you use \% or \_ outside of pattern-matching contexts, they evaluate to the strings \% and \_ , not to % and _ .

There are several ways to include quote characters within a string:

A ' inside a string quoted with ' may be written as '' .

A " inside a string quoted with " may be written as "" .

Precede the quote character by an escape character ( ).

A ' inside a string quoted with " needs no special treatment and need not be doubled or escaped. In the same way, " inside a string quoted with ' needs no special treatment.

The following SELECT statements demonstrate how quoting and escaping work:

To insert binary data into a string column (such as a BLOB column), you should represent certain characters by escape sequences. Backslash ( ) and the quote character used to quote the string must be escaped. In certain client environments, it may also be necessary to escape NUL or Control+Z. The **mysql** client truncates quoted strings containing NUL characters if they are not escaped, and Control+Z may be taken for END-OF-FILE on Windows if not escaped. For the escape sequences that represent each of these characters, see Table 9.1, “Special Character Escape Sequences”.

When writing application programs, any string that might contain any of these special characters must be properly escaped before the string is used as a data value in an SQL statement that is sent to the MySQL server. You can do this in two ways:

Process the string with a function that escapes the special characters. In a C program, you can use the mysql_real_escape_string_quote() C API function to escape characters. See mysql_real_escape_string_quote(). Within SQL statements that construct other SQL statements, you can use the QUOTE() function. The Perl DBI interface provides a quote method to convert special characters to the proper escape sequences. See Section 29.9, “MySQL Perl API”. Other language interfaces may provide a similar capability.

As an alternative to explicitly escaping special characters, many MySQL APIs provide a placeholder capability that enables you to insert special markers into a statement string, and then bind data values to them when you issue the statement. In this case, the API takes care of escaping special characters in the values for you.