**THE GENERAL TERM OF A NUMBER SEQUENCE**

Sequences may be defined in one of the following ways:

● listing all terms (of a finite sequence)

● listing the first few terms and assuming that the pattern represented continues indefinitely

● giving a description in words

● using a formula which represents the general term or

*n*th term.

Consider the illustrated tower of bricks. The first row has three bricks, the second row has four bricks, and the third row has five bricks.

If *u _{n}* represents the number of bricks in row

*n*(from the top) then

*u*

_{1}=3,

*u*

_{2}=4,

*u*

_{3}=5,

*u*

_{4}=6, … .

This sequence can be specified by:

●

**listing terms**

●

**using words**

“The top row has three bricks and each successive row under it has one more brick than the previous row.”

●

**using an explicit formula**

*u*=

_{n}*n*+2 is the general term or

*n*th term formula for

*n*=1, 2, 3, 4, … .

*Check*:

*u*

_{1}=1+2=3,

*u*

_{2}=2+2=4,

*u*

_{3}=3+2=5,

*u*

_{4}=4+2=6

●

**a pictorial or graphical representation**

Early members of a sequence can be graphed with each represented by a dot.

The dots

*must not*be joined because

*n*must be an integer.

**The General Term**

The **general term** or ** nth term** of a sequence is represented by a symbol with a subscript, for example

*a*,

_{n}*t*, or

_{n}*u*. The general term is defined for

_{n}*n*=1, 2, 3, 4, … .

{

*u*} represents the sequence that can be generated by using

_{n}*u*as the

_{n}*n*th term.

The general term

*u*is a function where

_{n}*n*↦

*u*. The domain is

_{n}*n*∈Z

^{+}for an infinite sequence, or a subset of Z

^{+}for a finite sequence.

For example, {2

*n*+1} generates the sequence 3, 5, 7, 9, 11, … .

A ball is dropped from height of 16 feet. Each time that it bounces, it reaches a height that is half of its previous height. We can list the height to which the ball bounces in order until it finally comes to rest.

After Bounce | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
---|---|---|---|---|---|---|---|

Height (ft) | 8 | 4 | 2 | 1 | ½ | ¼ | ⅛ |

The numbers 8, 4, 2, 1, ½, ¼, ⅛ form a *sequence*.

A

sequenceis a set of numbers written in a given order.

We can list these heights as ordered pairs of num- bers in which each height is paired with the number that indicates its position in the list. The set of ordered pairs would be:

We associate each term of the sequence with the positive integer that specifies its position in the ordered set. Therefore, a sequence is a special type of function.

The function that lists the height of the ball after 7 bounces is shown on the graph at the right.

Often the sequence can continue without end. In this case, the domain is the set of positive integers.

The terms of a sequence are often designated as

*a*

_{1},

*a*

_{2},

*a*

_{3},

*a*

_{4},

*a*

_{5}, … If the sequence is designated as the function f, then f(1)=

*a*

_{1}, f(2)=

*a*

_{2}, or in general:

*n*)=

*a*

_{n}Most sequences are sets of numbers that are related by some pattern that can be expressed as a formula. The formula that allows any term of a sequence, except the first, to be computed from the previous term is called a

**recursive definition**.

For example, the sequence that lists the heights to which a ball bounces when dropped from a height of 16 feet is 8, 4, 2, 1, 0.5, 0.25, 0.125, … . In this sequence, each term after the first is ½ the previous term. Therefore, for each term after the first,

For

*n*>1, we can write the recursive definition:

*a*=½

_{n}*a*

_{(n-1)}

Alternatively, we can write the recursive definition as:

*a*

_{(n+1)}=½

*a*for

_{n}*n*≥1

A rule that designates any term of a sequence can often be determined from the first few terms of the sequence.

Example 1

a. List the next three terms of the sequence 2, 4, 8, 16, … .

b. Write a general expression for

*a*.

_{n}c. Write a recursive definition for the sequence.

solution:

a. It appears that each term of the sequence is a power of 2: 2

^{1}, 2

^{2}, 2

^{3}, 2

^{4}, … .

Therefore, the next three terms should be 2

^{5}, 2

^{6}, and 2

^{7}or 32, 64, and 128.

b. Each term is a power of 2 with the exponent equal to the number of the term. Therefore,

*a*=2

_{n}^{n}

c. Each term is twice the previous term. Therefore, for

*n*>1,

*a*=2

_{n}*a*

_{(n-1)}.

Alternatively, for

*n*≥1,

*a*

_{(n+1)}=2

*a*.

_{n}answers:

a. 32, 64, 128

b.

*a*=2

_{n}^{n}

c. For

*n*>1,

*a*=2

_{n}*a*

_{(n-1)}or for

*n*≥1,

*a*

_{(n+1)}=2

*a*.

_{n}Example 2

Write the first three terms of the sequence *a _{n}*=3

*n*-1.

solution:

*a*

_{1}=3(1)-1=2;

*a*

_{2}=3(2)-1=5;

*a*

_{3}=3(3)-1=8

answer:

The first three terms of the sequence are 2, 5, and 8.

**RECURSIVE RULES**

What You Will Learn

► Evaluate recursive rules for sequences.

► Write recursive rules for sequences.

► Translate between recursive and explicit rules for sequences.

► Use recursive rules to solve real-life problems.

**Evaluating Recursive Rules**

So far in this chapter, you have worked with explicit rules for the *n*th term of a sequence, such as *a _{n}*=3

*n*-2 and

*a*=7⋅(0.5)

_{n}^{n}. An

**explicit rule**gives

*a*as a function of the term’s position number

_{n}*n*in the sequence.

In this section, you will learn another way to define a sequence—by a recursive rule. A **recursive rule** gives the beginning term(s) of a sequence and a recursive equation that tells how an is related to one or more preceding terms.

Example 3: **Evaluating Recursive Rules**

Write the first six terms of each sequence.

a. *a*_{0}=1, *a _{n}*=

*a*

_{(n-1)}+4 b. f(1)=1, f(

*n*)=3⋅f(

*n*-1)

Solution:

a. a_{0}=1a_{1}=a_{0}+4=1+4=5a_{2}=a_{1}+4=5+4=9a_{3}=a_{2}+4=9+4= 13a_{4}=a_{3}+4 =13+4=17a_{5}=a_{4}+4= 17+4=21 |
1st term 2 nd term3rd term 4th term 5th term 6th term |
b. f(1)=1 f(2)=3-f(1)=3(1)=3 f(3)=3-f(2)=3(3)=9 f(4)=3-f(3)=3(9)=27 f(5)=3-f(4)=3(27)=81 f(6)=3-f(5)=3(81)=243 |

**Writing Recursive Rules**

In part (a) of Example 3, the differences of consecutive terms of the sequence are constant, so the sequence is arithmetic. In part (b), the ratios of consecutive terms are constant, so the sequence is geometric. In general, rules for arithmetic and geometric sequences can be written recursively as follows.

Core Concept

Recursive Equations for Arithmetic and Geometric Sequences

Arithmetic Sequence

a=_{n}a_{(n-1)}+d, wheredis the common difference

Geometric Sequence

a=_{n}r⋅a_{(n-1)}, whereris the common ratio

Example 4: **Writing Recursive Rules**

Write a recursive rule for (a) 3, 13, 23, 33, 43, … and (b) 16, 40, 100, 250, 625, … .

Solution:

Use a table to organize the terms and find the pattern.

COMMON ERROR

A recursive equation for a sequence does not include the initial term. To write a recursive rule for a sequence, the initial term(s) must be included.

a.

The sequence is arithmetic with first term

*a*

_{1}=3 and common difference

*d*=10.

Recursive equation for arithmetic sequence

*a*=

_{n}*a*

_{(n-1)}+d.

Substitute 10 for

*d*.

*a*=

_{n}*a*

_{(n-1)}+10.

► A recursive rule for the sequence is

*a*

_{1}=3,

*a*=

_{n}*a*

_{(n-1)}+10.

b.

The sequence is geometric with first term

*a*

_{1}=16 and common ratio

*r*=5/2.

Recursive equation for geometric sequence

*a*=

_{n}*r*⋅

*a*

_{(n-1)}.

Substitute 5/2 for

*r*.

*a*=5/2⋅

_{n}*a*

_{(n-1)}.

► A recursive rule for the sequence is

*a*

_{1}=16,

*a*=5/2⋅

_{n}*a*

_{(n-1)}.

Let’s read the post Determine whether each sequence is arithmetic, geometric, or neither.

STUDY TIP

The sequence in part (a) of Example 5 is called thefibonacci sequence. The Sequence in part (b) listsfactorial numbers. You will learn more aboutfactorialsin the category of Probability.

Example 5: **Writing Recursive Rules**

Write a recursive rule for each sequence.

a. 1, 1, 2, 3, 5, …

b. 1, 1, 2, 6, 24, …

Solution:

a. The terms have neither a common difference nor a common ratio. Beginning with the third term in the sequence, each term is the sum of the two previous terms.

► A recursive rule for the sequence is *a*_{1}=1, *a*_{2}=1, *a _{n}*=

*a*

_{(n-2)}+

*a*

_{(n-1)}.

b. The terms have neither a common difference nor a common ratio. Denote the first term by

*a*

_{0}=1. Note that

*a*

_{1}=1=1⋅

*a*

_{0},

*a*

_{2}=2=2⋅

*a*

_{1},

*a*

_{3}=6=3⋅

*a*

_{2}, and so on.

► A recursive rule for the sequence is

*a*

_{0}=1,

*a*=n⋅

_{n}*a*

_{(n-1)}.

**Translating Between Recursive and Explicit Rules**

Example 6: **Translating from Explicit Rules to Recursive Rules**

Write a recursive rule for (a) *a _{n}*=-6+8n and (b)

*a*=-3⋅½

_{n}^{(n-1)}.

Solution:

a. The explicit rule represents an arithmetic sequence with first term

*a*

_{1}=-6+8(1)=2 and common difference

*d*=8.

Recursive equation for arithmetic sequence

*a*=

_{n}*a*

_{(n-1)}+d.

Substitute 8 for

*d*.

*a*=

_{n}*a*

_{(n-1)}+8.

► A recursive rule for the sequence is

*a*

_{1}=2,

*a*=

_{n}*a*

_{(n-1)}+8.

b. The explicit rule represents a geometric sequence with first term

*a*

_{1}=-3⋅½

^{0}=-3 and common ratio

*r*=½.

Recursive equation for geometric sequence

*a*=

_{n}*r*⋅

*a*

_{(n-1)}.

Substitute ½ for

*r*.

*a*=½⋅

_{n}*a*

_{(n-1)}

► A recursive rule for the sequence is

*a*

_{1}=-3,

*a*=½⋅

_{n}*a*

_{(n-1)}.

Example 7: **Translating from Recursive Rules to Explicit Rules**

Write an explicit rule for each sequence.

a. *a*_{1}=-5, *a _{n}*=

*a*

_{(n-1)}-2

b.

*a*

_{1}=10,

*a*=2⋅

_{n}*a*

_{(n-1)}

Solution:

a. The recursive rule represents an arithmetic sequence with first term

*a*

_{1}=-5 and common difference

*d*=-2.

Explicit rule for arithmetic sequence

*a*=

_{n}*a*

_{1}+(

*n*-1)

*d*

Substitute -5 for

*a*

_{1}and -2 for

*d*.

*a*=-5+(

_{n}*n*-1)(-2)

Simplify.

*a*=-3-2

_{n}*n*

► An explicit rule for the sequence is

*a*=-3-2

_{n}*n*.

b. The recursive rule represents a geometric sequence with first term *a*_{1}=10 and common ratio *r*=2.

Explicit rule for geometric sequence *a _{n}*=

*a*

_{1}⋅r

^{(n-1)}.

Substitute 10 for

*a*

_{1}and 2 for

*r*.

*a*=10⋅2

_{n}^{(n-1)}

► An explicit rule for the sequence is

*a*=10⋅2

_{n}^{(n-1)}.

**Solving Real-Life Problems**

Example 8: **Solving a Real-Life Problem**

A lake initially contains 5200 fish. Each year, the population declines 30% due to fishing and other causes, so the lake is restocked with 400 fish.

a. Write a recursive rule for the number

*a*of fish at the start of the

_{n}*n*th year.

b. Find the number of fish at the start of the fifth year.

c. Describe what happens to the population of fish over time.

Solution:

a. Write a recursive rule. The initial value is 5200. Because the population declines 30% each year, 70% of the fish remain in the lake from one year to the next. Also, 400 fish are added each year. Here is a Verbal model for the recursive equation.

► A recursive rule is

*a*

_{1}=5200,

*a*=(0.7)

_{n}*a*

_{(n-1)}+400.

b. Find the number of fish at the start of the fifth year. Enter 5200 (the value of *a*_{1}) in a graphing calculator. Then enter the rule

to find

*a*

_{2}. Press the enter button three more times to find

*a*

_{5}≈2262.

► There are about 2262 fish in the lake at the start of the fifth year.

c. Describe what happens to the population of fish over time. Continue pressing enter on the calculator. The screen at the right shows the fish populations for years 44 to 50. Observe that the population of fish approaches 1333.

► Over time, the population of fish in the lake stabilizes at about 1333 fish.

**Recursive Formulae For Sequences**

When discussing arithmetic and quadratic sequences, we noticed that the difference between two consecutive terms in the sequence could be written in a general way.

For an arithmetic sequence, where a new term is calculated by taking the previous term and adding a constant value, *d*:

*a*=

_{n}*a*

_{(n-1)}+d…(i)

The above equation is an example of a recursive equation since we can calculate the nth-term only by considering the previous term in the sequence. Compare this with equation (i),

*a*=

_{n}*a*

_{1}+

*d*⋅(

*n*-1)

where one can directly calculate the nth-term of an arithmetic sequence without knowing previous terms.

The recursive equation for a geometric sequence is:

*a*=

_{n}*r*⋅

*a*

_{(n-1)}

Recursive equations are extremely powerful: you can work out every term in the series just by knowing previous terms. As you can see from the examples above, working out

*a*using the previous term

_{n}*a*

_{(n-1)}can be a much simpler computation than working out

*a*from scratch using a general formula. This means that using a recursive formula when using a computer to work out a sequence would mean the computer would finish its calculations significantly quicker.

_{n} *Extension*: **The Fibonacci Sequence**

*Consider the following sequence*:

The above sequence is called the Fibonacci sequence. Each new term is calculated by adding the previous two terms. Hence, we can write down the recursive equation:

*a*=

_{n}*a*

_{(n-1)}+

*a*

_{(n-2)}

Let’s read the post The General Term of Fibonacci Sequence.

Example 9: Let the sequence an be defined as follows:

*a*

_{1}=1,

*a*=

_{n}*a*

_{(n-1)}+2 for

*n*≥2.

Find first five terms and write corresponding series.

Solution: We have

*a*

_{1}=1,

*a*

_{2}=

*a*

_{1}+2=1+2=3,

*a*

_{3}=

*a*

_{2}+2=3+2=5,

*a*

_{4}=

*a*

_{3}+2=5+2=7,

*a*

_{5}=

*a*

_{4}+2=7+2=9.

Hence, the first five terms of the sequence are 1, 3, 5, 7 and 9. The corresponding series is 1+3+5+7+9+⋯

Example 10: **Write the first 3 terms of the recursively defined sequence**.

(a) *a*_{1}=6, *a*_{(k+1)}=*a*_{(k+3)}

*a*

_{2}=

*a*

_{1}+3

*a*

_{2}=6+3

*a*

_{2}=9

*a*

_{3}=

*a*

_{2}+3

*a*

_{3}=9+3

*a*

_{3}=12

Answer: first 3 terms 6, 9, 12

(b) *a*_{1}=6, *a*_{(k+1)}=2(*a _{k}*-1)

*a*

_{2}=2(

*a*

_{1}-1)

*a*

_{2}=2(6-1)

*a*

_{2}=10

*a*

_{3}=2(

*a*

_{2}-1)

*a*

_{3}=2(10-1)

*a*

_{3}=18

Answer: first three terms are 6, 10, 18

(c) *a*_{1}=-3, *a*_{(k+1)}=2*a _{k}*+3

*a*

_{2}=2

*a*

_{1}+3

*a*

_{2}=2⋅(-3)+3

*a*

_{2}=-3

*a*

_{3}=2

*a*

_{2}+3

*a*

_{3}=2(-3)+3

*a*

_{3}=-3

Answer: first three terms -3, -3, -3

(d) *a*_{1}=6, *a*_{(k+1)}=½*a _{k}*+4

*a*_{3}=½*a*_{2}+4

*a*_{3}=½⋅7+4

*a*_{3}=3½+4=7½

Answer: First three terms 6, 7, 7½

Example 11. Find the next five terms of each of the following sequences given by:

Solution:

We have to find next five terms of following sequences.

(i) *a*_{1}=1, *a _{n}*=

*a*

_{(n-1)}+2,

*n*≥2

Given, first term

*a*

_{1}=1,

*n*th term

*a*=

_{n}*a*

_{(n-1)}+2,

*n*≥2.

To find 2

*n*d, 3rd, 4th, 5th, 6th terms, we use given condition

*n*≥2 for

*n*th term

_{n}=

*a*

_{(n-1)}+2.

*a*_{2}=*a*_{(2-1)}+2=*a*_{1}+2=1+2=3*a*_{3}=*a*_{(3-1)}+2=*a*_{2}+2=3+2=5*a*_{4}=*a*_{(4-1)}+2=*a*_{3}+2=5+2=7*a*_{5}=*a*_{(5-1)}+2=*a*_{4}+2=7+2=9*a*_{6}=*a*_{(6-1)}+2=*a*_{5}+2=9+2=11

∴ The next five terms are,

∴ The next five terms are,

Given, first term

To find next five terms i.e.,

∴ The next five terms are

The required next five terms are,

∴ The next five terms are,

*a*_{2}=3,*a*_{3}=5,*a*_{4}=7,*a*_{5}=9,*a*_{6}=11.(ii) *a*_{1}=*a*_{2}=2, *a _{n}*=

*a*

_{(n-1)}-3, n>2

Given,

First term

*a*

_{1}=2,

Second term

*a*

_{2}=2,

nth term

*a*=

_{n}*a*

_{(n-1)}-3,

To find next five terms i.e.,

*a*

_{3},

*a*

_{4},

*a*

_{5},

*a*

_{6},

*a*

_{7}we put

*n*=3, 4, 5, 6, 7 in

*a*.

_{n}*a*

_{3}=

*a*

_{(3-1)}-3=2-3=-1

*a*

_{4}=

*a*

_{(4-1)}-3=

*a*

_{3}-3=-1-3=-4

*a*

_{5}=

*a*

_{(5-1)}-3=

*a*

_{4}-3=-4-3=-7

*a*

_{6}=

*a*

_{(6-1)}-3=

*a*

_{5}-3=-7-3=-10

*a*

_{7}=

*a*

_{(7-1)}-3=

*a*

_{6}-3=-10-3=-13

∴ The next five terms are,

*a*_{3}=-1,*a*_{4}=-4,*a*_{5}=-7,*a*_{6}=-10,*a*_{7}=-13.Given, first term

*a*_{1}=2.To find next five terms i.e.,

*a*_{2},*a*_{3},*a*_{4},*a*_{5},*a*_{6}, we put*n*=2, 3, 4, 5, 6 in*a*._{n}∴ The next five terms are

*a*

_{2}=-½,

*a*

_{3}=-⅙,

*a*

_{4}=-1/24,

*a*

_{5}=-1/120,

*a*

_{6}=-1/720.

(iv) *a*_{1}=4, *a _{n}*=4

*a*

_{(n-1)}+3,

*n*>1

Given first term

*a*

_{1}=4,

*n*th term

*a*=4

_{n}*a*

_{(n-1)}+3,

*n*>1

To find next five terms i.e.,

*a*

_{2},

*a*

_{3},

*a*

_{4},

*a*

_{5},

*a*

_{6}we put

*n*=2, 3, 4, 5, 6 in

*a*. Then, we get

_{n}*a*

_{2}=4

*a*

_{(2-1)}+3=4

*a*

_{1}+3=4⋅4+3=19

*a*

_{3}=4

*a*

_{(3-1)}+3=4

*a*

_{2}+3=4⋅19+3=79

*a*

_{4}=4

*a*

_{(4-1)}+3 =4

*a*

_{3}+3=4⋅79+3=319

*a*

_{5}=4

*a*

_{(5-1)}+3=4

*a*

_{4}+3=4⋅319+3=1279

*a*

_{6}=4

*a*

_{(6-1)}+3=4

*a*

_{5}+3=4⋅1279+3=5119

The required next five terms are,

*a*

_{2}=19,

*a*

_{3}=79,

*a*

_{4}=319,

*a*

_{5}=1279,

*a*

_{6}=5119.

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