ARITHMETIC SEQUENCES AND SERIES This chapter focuses on: oRecognising, generating and analysing sequences oFormulating rules for obtaining a sequence oWorking with arithmetic sequences and series: general terms & sum oUsing Σ notation to find the sum

ARITHMETIC SEQUENCES AND SERIES Contents: 1.Pre-knowledgePre-knowledge 2.IntroductionIntroduction 3.Finding terms of a sequenceFinding terms of a sequence 4.Recurrence RelationshipsRecurrence Relationships 5.Assignment 1Assignment 1 6.Pre-knowledge RevisionPre-knowledge Revision

ARITHMETIC SEQUENCES AND SERIES PRE-KNOWLEDGE Before starting this chapter you should be able to: oSolve simple equationsSolve simple equations oFactorise and solve quadratic equationsFactorisequadratic oFinding the algebraic rule for a sequencealgebraic rule 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 21 st September Learning Intention: To be able to find the nth term of a sequence.

ARITHMETIC SEQUENCES AND SERIES INTRODUCTION 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 U n = 10 – 5n. Find: (i) the first term (ii) the fifth term (iii) the tenth term Solutions: (i)For the first term we replace n in the formula with 1. U 1 = 10 – 5(1) = 10 – 5 = 5 (ii) For the fifth term we replace n in the formula with 5. U 5 = 10 – 5(5) = 10 – 25 = -15 (iii) For the tenth term we replace n in the formula with 10. U 10 = 10 – 5(10) = 10 – 50 = -40

ARITHMETIC SEQUENCES AND SERIES FINDING TERMS OF A SEQUENCE Example: Find the value of n for which U n has the given value. U n = n 2 – 2U n = 398 Solution: If U n = n 2 – 2and U n = 398 then n 2 – 2 = 398 Solving n 2 – 2 = 398 we get n 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 U n = n 2 – 10n + 27 are all positive. For what value of n is U n smallest? Solution: To show that all terms of U n are positive we need to complete the square for n 2 – 10n n 2 – 10n + 27 = (n – 10/2) 2 – (10/2) = (n - 5) 2 – (5) = (n – 5) 2 – = (n -5) (n – 5) > 0 for all values of n. Therefore all terms of the sequence U n = n 2 – 10n + 27 are positive. U n = n 2 – 10n + 27 = (n – 5) The smallest value for U n occurs when (n – 5) 2 = 0. (n – 5) 2 = 0 when n = 5. Furthermore, when n is 5, U n equals 2.

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 U 2 = U U 3 = U U 4 = U 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 U n+1 = U n +3, U 1 = 2 Solution: U 1 = 2 U 2 = U = = 5 U 3 = U = = 8 U 4 = U = = 11 First four terms are: 2, 5, 8, 11

ARITHMETIC SEQUENCES AND SERIES RECURRENCE RELATIONSHIPS Example: Find the first four terms in the sequence U n+2 = 2U n+1 – U n, U 1 = 1 and U 2 = 2 Solution: U 1 = 1 U 2 = 2 U 3 = 2U 2 – U 1 = 2(2) – 1 = 4 – 1 = 3 U 4 = 2U 3 - U 3 = 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 28 th 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:

ARITHMETIC SEQUENCES AND SERIES BACK FACTORISING – COMMON FACTORS We can factorise some expressions by taking out a common factor. Example:Factorise 6x 2 – 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. 6x 2 – 2xy = 2x(3x – y) Full tutorial at:

ARITHMETIC SEQUENCES AND SERIES BACK FACTORISING QUADRATICS Factorising quadratics of the form ax 2 + bx + c The method is: Step 1: Find two numbers that multiply together to make ac and add to make b. Step 2: Split up the bx term using the numbers found in step 1. Step 3: Factorise the front and back pair of expressions as fully as possible. Step 4: There should be a common bracket. Take this out as a common factor. Example: Factorise 6x 2 + x – 12. Solution: We need to find two numbers that multiply to make 6 × -12 = -72 and add to make 1. These two numbers are -8 and 9. Therefore, 6x 2 + x – 12 = 6x 2 - 8x + 9x – 12 = 2x(3x – 4) + 3(3x – 4) (the two brackets must be identical) = (3x – 4)(2x + 3)