Transcript Introduction

ARITHMETIC SEQUENCES AND SERIES
This chapter focuses on:
o
Recognising, generating and analysing sequences
o
Formulating rules for obtaining a sequence
o
Working with arithmetic sequences and series:
general terms & sum
o
Using Σ notation to find the sum
ARITHMETIC SEQUENCES AND SERIES
Contents:
1.
Pre-knowledge
2.
Introduction
3.
Finding terms of a sequence
4.
Recurrence Relationships
5.
Assignment 1
6.
Pre-knowledge Revision
ARITHMETIC SEQUENCES AND SERIES
PRE-KNOWLEDGE
Before starting this chapter you should be able to:
oSolve simple equations
oFactorise and solve quadratic equations
oFinding the algebraic rule for a sequence
Note:
The underlined words provide links to additional information –
further slides, video tutorials etc. Whenever you see these
underlined words a hyperlink is attached to further
information.
ARITHMETIC SEQUENCES AND SERIES
Week Commencing Monday 21st September
Learning Intention:
To be able to find the nth term of a sequence.
ARITHMETIC SEQUENCES AND SERIES
INTRODUCTION
2 3 4
, , ,...
3 4 5
The above list is called a sequence of numbers. Each
term in the sequence follows a common rule.
Can you find the next 3 numbers in the sequence?
ARITHMETIC SEQUENCES AND SERIES
FINDING TERMS OF A SEQUENCE
If we know the formula for a sequence we can find any term
in the sequence.
The formula for a sequence is usually expressed in terms of
n. The formula is for the nth term of a sequence, often
called the general term.
To work out a term of a sequence we replace the n in the
formula with the number of the term we want.
ARITHMETIC SEQUENCES AND SERIES
FINDING TERMS OF A SEQUENCE
Example:
The nth term of a sequence is given by Un = 10 – 5n.
Find: (i) the first term
(ii) the fifth term
(iii) the tenth term
Solutions:
(i)
term
we
replace
nn in
the
formula
with
1.
(ii)
(iii)For
Forthe
thefirst
fifth
tenth
term
term
wewe
replace
replace
nin in
the
the
formula
formula
with
with
5.10.
U1 U=U510
10 – =5(1)
= 10
10 –– 5(5)
5(10)
= 10 – =5= 10
=105–– 25
50 == -15
-40
ARITHMETIC SEQUENCES AND SERIES
FINDING TERMS OF A SEQUENCE
Example:
Find the value of n for which Un has the given value.
Un = n2 – 2
Un = 398
Solution:
If Un = n2 – 2 and Un = 398 then n2 – 2 = 398
Solving n2 – 2 = 398 we get
n2 = 398 + 2 = 400
n = √400 = 20
Note: we only take the positive value of √400 as we cannot have a
negative number of terms in a sequence.
ARITHMETIC SEQUENCES AND SERIES
FINDING TERMS OF A SEQUENCE
Example:
Prove that the terms of the sequence Un = n2 – 10n + 27 are all
positive. For what value of n is Un smallest?
Solution:
U
n2 –that
10n all
+ 27
= (nof
– 5)
+ 2 positive we need to complete the
Ton =show
terms
Un2 are
square for n2 – 10n + 27.
The smallest value for Un occurs when (n – 5)2 = 0.
n2 – 10n
+ 27 = (n – 10/2)2 – (10/2)2 + 27
2
(n – 5) = 0 when n = 5.
= (n - 5)2 – (5)2 + 27
Furthermore, when n is 5, Un equals 2.
= (n – 5)2 – 25 + 27 = (n -5)2 + 2
(n – 5)2 + 2 > 0 for all values of n. Therefore all terms of the
sequence Un = n2 – 10n + 27 are positive.
ARITHMETIC SEQUENCES AND SERIES
RECURRENCE RELATIONSHIPS
Sometimes the terms of a sequence are dependent upon the previous
term.
Look at this sequence of numbers:
5, 8, 11, 14, 17, ...
This sequence can be described by the rule add 3 to the previous
term. That is
U2 = U1 + 3
U3 = U2 + 3
U4 = U3 + 3 etc.
This type of rule which determines the next term is called a
recurrence formula or recurrence relationship.
ARITHMETIC SEQUENCES AND SERIES
RECURRENCE RELATIONSHIPS
Example:
Find the first four terms in the sequence Un+1 = Un +3, U1 = 2
Solution:
U1 = 2
U2 = U1 + 3 = 2 + 3 = 5
U3 = U2 + 3 = 5 + 3 = 8
U4 = U3 + 3 = 8 + 3 = 11
First four terms are: 2, 5, 8, 11
ARITHMETIC SEQUENCES AND SERIES
RECURRENCE RELATIONSHIPS
Example:
Find the first four terms in the sequence Un+2 = 2Un+1 – Un, U1 = 1
and U2 = 2
Solution:
U1 = 1
U2 = 2
U3 = 2U2 – U1 = 2(2) – 1 = 4 – 1 = 3
U4 = 2U3 - U3 = 2(3) – 2 = 6 – 2 = 4
ARITHMETIC SEQUENCES AND SERIES
Assignment 1
Follow the link for Assignment 1 on Recurrence
Relationships in the Moodle Course Area.
Completed assignments must be submitted by 5:00pm on
Monday 28th September.
ARITHMETIC SEQUENCES AND SERIES
PRE-KNOWLEDGE REVISION
ARITHMETIC SEQUENCES AND SERIES
BACK
SOLVING SIMPLE EQUATIONS
A simple equation is an equation of the type:
2a + 5 = 13.
To solve these equation we take all the numbers to one
side and the letters to the other.
2a = 13 – 5 = 8
a = 8 ÷ 2 = 4
For further tutorials on solving equations visit:
http://stream.port.ac.uk/streams/play/play.asp?id=472&stream=MediumBand
ARITHMETIC SEQUENCES AND SERIES
BACK
FACTORISING – COMMON FACTORS
We can factorise some expressions by taking out a common
factor.
Example:
Factorise 6x2 – 2xy
Solution:
2 is a common factor to both 6 and 2. Both
terms also contain an x.
So we factorise by taking 2x outside a bracket.
6x2 – 2xy = 2x(3x – y)
Full tutorial at:
http://stream.port.ac.uk/streams/play/play.asp?id=466&stream=MediumBand
ARITHMETIC SEQUENCES AND SERIES
BACK
FACTORISING QUADRATICS
Factorising quadratics of the form ax2 + bx + c
The method is:
Example:
Factorise
6x2 + xthat
– 12.
Step 1: Find
two numbers
multiply together to make
ac
and add to make b.
Solution: We need to find two numbers that multiply to
Step 2: Split up the bx term using the numbers found in
make 6 × -12 = -72 and add to make 1. These
step 1.
two numbers are -8 and 9.
2 + x – 12 = 6x2 - 8x + 9x – 12
Therefore,
Step 3: Factorise
the6x
front
and back pair of expressions
as fully as possible.
= 2x(3x – 4) + 3(3x – 4)
Step 4: There should be a common bracket. Take this
(the two brackets must be identical)
out as a common factor.
= (3x – 4)(2x + 3)