ARITHMETIC SEQUENCES AND SERIES This chapter focuses on:

1. ARITHMETIC SEQUENCES AND SERIES • This chapter focuses on: • Recognising, generating and analysing sequences • Formulating rules for obtaining a sequence • Working with arithmetic sequences and series: general terms & sum • Using Σ notation to find the sum

2. ARITHMETIC SEQUENCES AND SERIES • Contents: • Pre-knowledge • Introduction • Finding terms of a sequence • Recurrence Relationships • Assignment 1 • Pre-knowledge Revision

3. ARITHMETIC SEQUENCES AND SERIES • PRE-KNOWLEDGE • Before starting this chapter you should be able to: • Solve simple equations • Factorise and solve quadratic equations • Finding 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.

4. ARITHMETIC SEQUENCES AND SERIES Week Commencing Monday 21st September Learning Intention: To be able to find the nth term of a sequence.

5. 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?

6. 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.

7. 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: • For the first term we replace n in the formula with 1. • U1 = 10 – 5(1) • = 10 – 5 = 5 (ii) For the fifth term we replace n in the formula with 5. U5 = 10 – 5(5) = 10 – 25 = -15 (iii) For the tenth term we replace n in the formula with 10. U10 = 10 – 5(10) = 10 – 50 = -40

8. 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.

9. 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: Un = n2 – 10n + 27 = (n – 5)2 + 2 The smallest value for Un occurs when (n – 5)2 = 0. (n – 5)2 = 0 when n = 5. Furthermore, when n is 5, Un equals 2. To show that all terms of Un are positive we need to complete the square for n2 – 10n + 27. n2 – 10n + 27 = (n – 10/2)2 – (10/2)2 + 27 = (n - 5)2 – (5)2 + 27 = (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.

10. 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.

11. 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

12. 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

13. 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.

14. ARITHMETIC SEQUENCES AND SERIES PRE-KNOWLEDGE REVISION

15. 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

16. 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

17. ARITHMETIC SEQUENCES AND SERIES BACK FACTORISING QUADRATICS Factorising quadratics of the form ax2 + 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 6x2 + 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, 6x2 + x – 12 = 6x2 - 8x + 9x – 12 = 2x(3x – 4) + 3(3x – 4) (the two brackets must be identical) = (3x – 4)(2x + 3)