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 *u*_{n} = *n*(*n* + 1)

There is one other important sequence, namely the prime numbers

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

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

(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:

1 | 1 | 2 | 3 | 5 | 8 | 13 | ... |

0 | 1 | 1 | 2 | 3 | 5 |

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:

1 | 1 | 2 | 3 | 5 | 8 | 13 | 21 | 34 | 55 | ... |

0 | 1 | 1 | 2 | 3 | 5 | 8 | 13 | 21 |

(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,

*u*_{n} = *u*_{n-1} + *u*_{n-2}

This formula connects *u _{n}* to the two previous terms, rather than

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

*u*_{n} = 2*u*_{n-1} + *u*_{n-2}

Calculate the next 3 terms of the sequence.

Note that *u*_{1} = 1 and *u*_{2} = 2.

u_{3} | = 2u_2 + u_1 |

= 2 × 2 + 1 | |

= 5 |

u_{4} | = 2u_3 + u_2 |

= 2 × 5 + 2 | |

= 12 |

u_{5} | = 2u_4 + u_3 |

= 2 × 12 + 5 | |

= 29 |

So the first 5 terms of the sequence are,

1, 2, 5, 12, 29

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.

6 | 12 | 24 | 48 | 96 | 192 | 384 | ... |

×2 | ×2 | ×2 | ×2 | ×2 | ×2 |

(b)

determine a general formula for the *n*th term.

Consider how each term is formed:

u_1 | = 6 | = 3 × 2 |

u_2 | = 12 | = 3 × 2 × 2 = 3 × 2^{2} |

u_3 | = 24 | = 3 × 2 × 2 × 2 = 3 × 2^{3} |

u_4 | = 48 | = 3 × 2 × 2 × 2 × 2 = 3 × 2^{4} |

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

This sequence is an example of an *exponential* sequence.

Consider the sequence,

, , , , , , ...

Determine the general formula for the *n*th term of the sequence.

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

First consider the sequence of numerators,

3 | 7 | 11 | 15 | 19 | 23 | ... |

4 | 4 | 4 | 4 | 4 |

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,

5 | 8 | 11 | 14 | 17 | 20 | ... |

3 | 3 | 3 | 3 | 3 |

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,

*u _{n}* =

What happens to the sequence defined by,

*u _{n}* =

as *n* becomes larger and larger?

The following table lists *n* and *u _{n}* for several values of

From the table it can be seen that the values of *u _{n}* = get larger and larger as

However, the numerator is always smaller than the denominator, so each value *u _{n}* must be smaller than 1.

It follows that, as *n* gets larger and larger, the values of *u _{n}* must get closer and closer to 1.

n | u = _{n} | u_{n} to 3 decimal places |

1 | 0 | 0 |

2 | 0.333 | |

3 | 0.5 | |

4 | 0.6 | |

5 | 0.667 | |

10 | 0.818 | |

20 | 0.905 | |

50 | 0.961 | |

100 | 0.980 | |

500 | 0.996 | |

1000 | 0.998 | |

2000 | 0.999 |

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.