Unit 10 Section 4 : Special Sequences

Before going on to look at harder examples, we list some of the important sequences that you should know:

1, 3, 5, 7, 9, 11, 13, ...

the odd numbers u_n = 2n – 1

2, 4, 6, 8, 10, 12, 14, ...

the even numbers u_n = 2n

1, 4, 9, 15, 25, 36, 49, ...

the square numbers u_n = n^2

1, 8, 27, 64, 125, 216, 343, ...

the cube numbers u_n = n^3

1, 3, 6, 10, 15, 21, 28, ...

the triangular numbers un = n(n + 1)

There is one other important sequence, namely the prime numbers

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, ...

Note: there is no formula for calculating the nth prime number.

We now look at other, harder sequences generated by algebraic rules.

Example 1

(a)

Write down the next 3 terms of the sequence,

1, 1, 2, 3, 5, 8, 13, ...

Use the first differences to extend the sequence:

11235813...
011235

Note that the first differences, ignoring the first 0, are in fact the actual sequence itself. These can then be used to extend the sequence:

11235813213455...
01123581321
(b)

Determine a formula for calculating the nth term.

Each term is the sum of the two previous terms; for example,

u_3 = u_2 + u_1

u_4 = u_3 + u_2

We can express this mathematically as,

un = un-1 + un-2

This formula connects un to the two previous terms, rather than n which we used in the earlier sections. This sequence is actually a special sequence and is called the Fibonacci sequence.

Example 2

The first two terms of a sequence are 1 and 2. The sequence is defined as,

un = 2un-1 + un-2

Calculate the next 3 terms of the sequence.

Note that u1 = 1 and u2 = 2.

u3 = 2u_2 + u_1
= 2 × 2 + 1
= 5
u4 = 2u_3 + u_2
= 2 × 5 + 2
= 12
u5 = 2u_4 + u_3
= 2 × 12 + 5
= 29

So the first 5 terms of the sequence are,

1, 2, 5, 12, 29

Example 3

For the sequence,

6, 12, 24, 48, 96, ...

(a)

calculate the next 2 terms of the sequence,

Note that, in this sequence, each term is twice the previous term.

612244896192384...
×2×2×2×2×2×2
(b)

determine a general formula for the nth term.

Consider how each term is formed:

u_1 = 6 = 3 × 2
u_2 = 12 = 3 × 2 × 2 = 3 × 22
u_3 = 24 = 3 × 2 × 2 × 2 = 3 × 23
u_4 = 48 = 3 × 2 × 2 × 2 × 2 = 3 × 24

Hence the general term will be u_n = 3 × 2^n.

This sequence is an example of an exponential sequence.

Example 4

Consider the sequence,

, , , , , , ...

Determine the general formula for the nth term of the sequence.

It is best to consider the numerators and the denominators separately.

First consider the sequence of numerators,

3711151923...
44444

As the difference between the terms is 4, we have

u_n = 4n + a

and using the first term,

3 = 4 × 1 + a

a = –1

Hence

u_n = 4n – 1

Now consider the sequence of denominators,

5811141720...
33333

As the difference between the terms is 3, we have

u_n = 3n + b

and using the first term,

5 = 3 × 1 + b

b = 2

Hence

u_n = 3n + 2

So for the given sequence of fractions we have,

un =

Example 5

What happens to the sequence defined by,

un =

as n becomes larger and larger?

The following table lists n and un for several values of n.

From the table it can be seen that the values of un = get larger and larger as n increases.

However, the numerator is always smaller than the denominator, so each value un must be smaller than 1.

It follows that, as n gets larger and larger, the values of un must get closer and closer to 1.

nun = un to 3 decimal places
1 00
2 0.333
3 0.5
4 0.6
5 0.667
10 0.818
20 0.905
50 0.961
1000.980
5000.996
10000.998
20000.999

Exercises

Note: To type indeces on this page use ^ sign. e.g.  n^2 :  
Use braces to write formulas as indexes. e.g.  2^(n+2) :  

To use multiplication sign ×  write * instead.

Question 1

Calculate the next three terms in each of the following sequences:

(a)1, 3, 4, 7, 11, 18, ..., ,
(b)4, 9, 13, 22, 35, ..., ,
(c) , , , , , ... , ,
(d)5, 15, 45, 135, 405, ..., ,
Question 2

Calculate the first 6 terms of each of the following sequences:

(a)

u1 = 0, u2 = 3, un = un-1 + un-2

, , , , ,
(b)

u1 = 3, u2 = 4, un = 2un-1 + un-2

, , , , ,
(c)

u1 = 6, u2 = 10, un = 3un-1un-2

, , , , ,
(d)

u1 = 1, u2 = 2, un = un-1 × un-2

, , , , ,
Question 3

(a)

Calculate the next 3 terms of the sequence,

1, 4, 16, 64, 256, ...

, ,
(b)

Determine a formula for the nth term of the sequence.

u_n =
Question 4

Determine the formula for the general term of each of the following sequences:

(a) 15, 75, 375, 1875, 9375, ... un =
(b) 1, 3, 9, 27, 81, ... un =
(c) 20, 200, 2000, 20 000, 200 000, ... un =
(d) 4, 28, 196, 1372, 9604, ... un =
Question 5

(a)

Determine the general formula for the terms of the sequence,

1, 7, 13, 19, 25, 31, ...

un =
(b)

Determine the general formula for the terms of the sequence,

2, 10, 18, 26, 34, 42, ...

un =
(c)

Determine the general formula for the terms of the sequence,

, , , , , , ...

un =
Question 6

Determine the general formula for the terms of each of the following sequences:

(a)

, , , , , ...

un =
(b)

, , , , , ...

un =
(c)

, , , , , ...

un =
(d)

, , , , , ...

un =
Question 7

Determine the formula for the general term of each of the following sequences, and also calculate the 10th term of each sequence.

(a)

1, , , , , ...

un = 10th term:  
(b)

, , 1, , , ...

un = 10th term:  
Question 8

(a)

Complete the following table for un = .

nunun to 3 decimal places
1
5
10
50
100
500
1000
2000
(b)

Determine the formula for the general term of the sequence.

As n becomes , un gets closer and closer to
Question 9

Complete tables similar to the one in question 8, for each of the following sequences:

(a)

un =

nun to 3 decimal places
1
5
10
50
100
500
1000
2000
As n becomes larger and larger, un gets closer and closer to
(b)

un =

nun to 3 decimal places
1
5
10
50
100
500
1000
2000
As n becomes larger and larger, un gets closer and closer to
(c)

un =

nun to 3 decimal places
1
5
10
50
100
500
1000
2000
As n becomes larger and larger, un gets closer and closer to
(d)

un =

nun to 3 decimal places
1
5
10
50
100
500
1000
2000
As n becomes larger and larger, un gets closer and closer to
Question 10

What do you think will happen to each of the sequences below as n becomes large?

(a)

un =

un as n incerases
nun to 3 d.p.
1 2.000
10 3.636
100 3.960
1000 3.996
(b)

un =

un as n incerases
nun to 3 d.p.
1 8.000
10 7.100
100 7.010
1000 7.001
(c)

un =

un as n incerases
nun to 3 d.p.
1 0.333
10 0.476
100 0.497
1000 0.499
(d)

un =

un as n incerases
nun to 3 d.p.
1 4.000
10 2.105
100 2.010
1000 2.001
Question 11
(a)

Here is a number chain:

2 → 4 → 6 → 8 → 10 → 12 →

What could the rule be?

each time.
(b)

A different number chain is:

2 → 4 → 8 → 16 → 32 → 64 →

What could the rule be?

each time.
Question 12

Each term of a number sequence is made by adding 1 to the numerator and 2 to the denominator of the previous term.

Here is the beginning of the number sequence:

, , , , , ...

(a)

Write an expression for the nth term of the sequence.

un =
(b)

The first five terms of the sequence are shown on the graph.

The sequence goes on and on for ever.

Which of the following four graphs shows how the sequence continues?

Graph shows how the sequence continues.
(c)

The nth term of a different sequence is .

The first term of the sequence is .

Write down the next three terms.

, ,
(d)

This new sequence also goes on and on for ever.

Which of the four graphs below shows how the sequence continues?

Graph shows how the sequence continues.