# Translating from Recursive Rules to Explicit Rules (General Terms)

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 nth 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 un represents the number of bricks in row n (from the top) then u1=3, u2=4, u3=5, u4=6, … .
This sequence can be specified by:
listing terms

3, 4, 5, 6, … .

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
un=n+2 is the general term or nth term formula for n=1, 2, 3, 4, … .
Check: u1=1+2=3, u2=2+2=4, u3=3+2=5, u4=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 an, tn, or un. The general term is defined for n=1, 2, 3, 4, … .
{un} represents the sequence that can be generated by using un as the nth term.
The general term un is a function where nun. The domain is n∈Z+ for an infinite sequence, or a subset of Z+ for a finite sequence.
For example, {2n+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 Height (ft) 1 2 3 4 5 6 7 8 4 2 1 ½ ¼ ⅛

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

A sequence is 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:

{(1, 8), (2, 4), (3, 2), (4, 1), (5, 0.5), (6, 0.25), (7, 0.125)}

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 a1, a2, a3, a4, a5, … If the sequence is designated as the function f, then f(1)=a1, f(2)=a2, or in general:
f(n)=an

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,
4=½(8), 2=½(4), 1=½(2), 0.5=½(1), 0.25=½(0.5), 0.125=½(0.25), …

For n>1, we can write the recursive definition:
ana(n-1)

Alternatively, we can write the recursive definition as:
a(n+1)an for 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 an.
c. Write a recursive definition for the sequence.
solution:
a. It appears that each term of the sequence is a power of 2: 21, 22, 23, 24, … .
Therefore, the next three terms should be 25, 26, and 27 or 32, 64, and 128.
b. Each term is a power of 2 with the exponent equal to the number of the term. Therefore, an=2n
c. Each term is twice the previous term. Therefore, for n>1, an=2a(n-1).
Alternatively, for n≥1, a(n+1)=2an.
a. 32, 64, 128
b. an=2n
c. For n>1, an=2a(n-1) or for n≥1, a(n+1)=2an.

Example 2
Write the first three terms of the sequence an=3n-1.
solution:

a1=3(1)-1=2;a2=3(2)-1=5;a3=3(3)-1=8

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 nth term of a sequence, such as an=3n-2 and an=7⋅(0.5)n. An explicit rule gives an as a function of the term’s position number 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. a0=1, an=a(n-1)+4 b. f(1)=1, f(n)=3⋅f(n-1)
Solution:

 a. a0=1 a1=a0+4=1+4=5 a2=a1+4=5+4=9 a3=a2+4=9+4= 13 a4=a3+4 =13+4=17 a5=a4+4= 17+4=21 1st term 2nd term 3rd 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
an=a(n-1)+d, where d is the common difference
Geometric Sequence
an=ra(n-1), where r is 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 a1=3 and common difference d=10.
Recursive equation for arithmetic sequence an=a(n-1)+d.
Substitute 10 for d. an=a(n-1)+10.
► A recursive rule for the sequence is a1=3, an=a(n-1)+10.

b. The sequence is geometric with first term a1=16 and common ratio r=5/2.
Recursive equation for geometric sequence an=ra(n-1).
Substitute 5/2 for r. an=5/2⋅a(n-1).
► A recursive rule for the sequence is a1=16, an=5/2⋅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 the fibonacci sequence. The Sequence in part (b) lists factorial numbers. You will learn more about factorials in 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 a1=1, a2=1, an=a(n-2)+a(n-1).
b. The terms have neither a common difference nor a common ratio. Denote the first term by a0=1. Note that a1=1=1⋅a0, a2=2=2⋅a1, a3=6=3⋅a2, and so on.
► A recursive rule for the sequence is a0=1, an=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) an=-6+8n and (b) an=-3⋅½(n-1).
Solution:
a. The explicit rule represents an arithmetic sequence with first term
a1=-6+8(1)=2 and common difference d=8.
Recursive equation for arithmetic sequence an=a(n-1)+d.
Substitute 8 for d. an=a(n-1)+8.
► A recursive rule for the sequence is a1=2, an=a(n-1)+8.
b. The explicit rule represents a geometric sequence with first term a1=-3⋅½0=-3 and common ratio r=½.
Recursive equation for geometric sequence an=ra(n-1).
Substitute ½ for r. an=½⋅a(n-1)
► A recursive rule for the sequence is a1=-3, an=½⋅a(n-1).

Example 7: Translating from Recursive Rules to Explicit Rules
Write an explicit rule for each sequence.
a. a1=-5, an=a(n-1)-2
b. a1=10, an=2⋅a(n-1)
Solution:
a. The recursive rule represents an arithmetic sequence with first term a1=-5 and common difference d=-2.
Explicit rule for arithmetic sequence an=a1+(n-1)d
Substitute -5 for a1 and -2 for d. an=-5+(n-1)(-2)
Simplify. an=-3-2n
► An explicit rule for the sequence is an=-3-2n.

b. The recursive rule represents a geometric sequence with first term a1=10 and common ratio r=2.
Explicit rule for geometric sequence an=a1⋅r(n-1).
Substitute 10 for a1 and 2 for r. an=10⋅2(n-1)
► An explicit rule for the sequence is an=10⋅2(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 an of fish at the start of the nth 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 a1=5200, an=(0.7) a(n-1)+400.

b. Find the number of fish at the start of the fifth year. Enter 5200 (the value of a1) in a graphing calculator. Then enter the rule

.7×Ans+400

to find a2. Press the enter button three more times to find a5≈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:

an=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),
an=a1+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:

an=ra(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 an using the previous term a(n-1) can be a much simpler computation than working out an 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.

Extension: The Fibonacci Sequence
Consider the following sequence:

0; 1; 1; 2; 3; 5; 8; 13; 21; 34; …

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:
an=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:
a1=1, an=a(n-1)+2 for n≥2.

Find first five terms and write corresponding series.
Solution: We have
a1=1, a2=a1+2=1+2=3, a3=a2+2=3+2=5,
a4=a3+2=5+2=7, a5=a4+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) a1=6, a(k+1)=a(k+3)

a2=a1+3
a2=6+3
a2=9
a3=a2+3
a3=9+3
a3=12

Answer: first 3 terms 6, 9, 12

(b) a1=6, a(k+1)=2(ak-1)

a2=2(a1-1)
a2=2(6-1)
a2=10
a3=2(a2-1)
a3=2(10-1)
a3=18

Answer: first three terms are 6, 10, 18

(c) a1=-3, a(k+1)=2ak+3

a2=2a1+3
a2=2⋅(-3)+3
a2=-3
a3=2a2+3
a3=2(-3)+3
a3=-3

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

(d) a1=6, a(k+1)ak+4

a3a2+4
a3=½⋅7+4
a3=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) a1=1, an=a(n-1)+2, n≥2
Given, first term a1=1, nth term an=a(n-1)+2, n≥2.
To find 2nd, 3rd, 4th, 5th, 6th terms, we use given condition n≥2 for nth term

an=a(n-1)+2.
a2=a(2-1)+2=a1+2=1+2=3
a3=a(3-1)+2=a2+2=3+2=5
a4=a(4-1)+2=a3+2=5+2=7
a5=a(5-1)+2=a4+2=7+2=9
a6=a(6-1)+2=a5+2=9+2=11

∴ The next five terms are, a2=3, a3=5, a4=7, a5=9, a6=11.

(ii) a1=a2=2, an=a(n-1)-3, n>2
Given,
First term a1=2,
Second term a2=2,
nth term an=a(n-1)-3,
To find next five terms i.e., a3, a4, a5, a6, a7 we put n=3, 4, 5, 6, 7 in an.

a3=a(3-1)-3=2-3=-1
a4=a(4-1)-3=a3-3=-1-3=-4
a5=a(5-1)-3=a4-3=-4-3=-7
a6=a(6-1)-3=a5-3=-7-3=-10
a7=a(7-1)-3=a6-3=-10-3=-13

∴ The next five terms are, a3=-1, a4=-4, a5=-7, a6=-10, a7=-13. Given, first term a1=2.
To find next five terms i.e., a2, a3, a4, a5, a6, we put n=2, 3, 4, 5, 6 in an. ∴ The next five terms are
a2=-½, a3=-⅙, a4=-1/24, a5=-1/120, a6=-1/720.

(iv) a1=4, an=4a(n-1)+3, n>1
Given first term a1=4, nth term an=4a(n-1)+3, n>1
To find next five terms i.e., a2, a3, a4, a5, a6 we put n=2, 3, 4, 5, 6 in an. Then, we get

a2=4a(2-1)+3=4a1+3=4⋅4+3=19
a3=4a(3-1)+3=4a2+3=4⋅19+3=79
a4 =4a(4-1)+3 =4a3+3=4⋅79+3=319
a5=4a(5-1)+3=4a4+3=4⋅319+3=1279
a6=4a(6-1)+3=4a5+3=4⋅1279+3=5119

The required next five terms are,
a2=19, a3=79, a4=319, a5=1279, a6=5119.