Topic 1 Number and Algebra

Topic 1: Number and Algebra PART A Number systems: R → Real Numbers The set of real numbers includes all the rational and irrational numbers. The real numbers are “all the numbers” on the number line. There are infinitely many real numbers just as there are infinitely many numbers in each of the other sets of numbers. Q → Rational Numbers Rational Numbers are numbers that can be expressed as a ratio of two integers, or by the fraction 1 p/q, where p and q are integers. For instance, the fractions 2 and −7 23 are rational numbers. Another very important thing to note here is that only terminating decimals are rational numbers. Decimals that follow a repeating pattern after some point are also rationals: for example, 1 0.333333 = 3is a rational number. Z → Integers Integers are the set of real numbers consisting of natural numbers or whole numbers, their additive inverses and zero. The sum, product and difference of two integers is also an integer. For example: {1,2,32,-4,20} are all integers. N → Natural Numbers The natural numbers are essentially counting numbers like 1,2,3,4,5, etc. There are infinitely many natural numbers. I → Irrational Numbers 1 An irrational number is a number that cannot be written as a ratio of two integers. In its decimal form, it never ends or repeats. For instance, √3 and π = 3.14159265358979.. , are both irrational numbers. The following venn diagram displays the conventional number systems that are involved in mathematics: Real Numbers, Rational Numbers, Integers, Natural Numbers and Irrational Numbers. It gives an understanding of how these number systems are related and identified. PART B Arithmetic Sequences and Series: Finding an equation of an arithmetic sequence is essentially like finding a pattern in the numbers. A sequence is a list of numbers that is written in a defined order which can be ascending or descending. Each of the numbers in a sequence is called a term, and an arithmetic sequence is often called a progression. A sequence is defined with a general formula shown below. Where 𝑢1 is the first term of the sequence and ‘d’ is the common difference between each term in the sequence. This is used to find the 𝑛𝑡ℎ term of a particular arithmetic sequence. 2 For example, 𝑢𝑛 = 3𝑛 − 5. Find the following terms of the sequence. i. 𝑢1 ii. 𝑢27 iii. 𝑢5 To solve the above, we need to substitute ‘n’ with the term that we wish to find. Answers: i. 𝑢1 = 3(1) − 5 = −2 ii. 𝑢27 = 3(27) − 1 = 80 iii. 𝑢5 = 3(5) − 1 = 14 Exercise A: Q1. For the following sequences find 𝑢7 , 𝑢100 , 𝑢2 . i. 𝑢𝑛 = 17𝑛 − 108 ii. 𝑢𝑛 = 2(𝑛 − 28)+ 57 Q2. An arithmetic sequence starts 23,36,49,62, ... Find the first term of the sequence to exceed 10,000. Note:- It is helpful to remember that in a arithmetic sequence 𝑢𝑛+1 − 𝑢𝑛 = 𝑑. When asked to prove that a sequence is arithmetic, this is one of the methods you are expected to use. Application of Arithmetic Sequences - Approximating using Arithmetic Sequences Q. Raina has put an empty egg carton on a weighing scale. Its mass is 32g. When the carton is filled with 12 eggs, the total mass of the eggs and cartons is 743g. Find, I. The average mass of the eggs that are in the carton II. Hence, find an arithmetic sequence for 𝑢𝑛 , the approximate total mass when ‘n’ eggs have been added to the carton. III. For what values of n is your model valid? To solve the above question it is important to go step by step. I. II. To find the average mass of the eggs in the carton we need to know the total mass of the eggs and the number of eggs (12) one of which has been given. To find the total mass of the eggs only we subtract 32g (mass of carton) from 743g (mass of carton + eggs). 743 - 32 = 711 (total mass of eggs only). Now we need to divide this by the number of eggs: 711/12 = 59.25g. Each egg weighs approximately 59.25g. To write a general formula for the above word problem we need to understand it in depth. When ‘n’ eggs have been added to the basket, the weight of the eggs would be n(59.25). However, we must not forget to include the weight of the basket as well. Therefore, the formula would be: 𝑢𝑛 = 59.25𝑛 + 32. 3 III. The value of ‘n’ for which this model would be valid only till we know the capacity of the basket. Since the basket remains constant and we have not been told otherwise, the amount of eggs that can be placed depends on the dimensions of the basket. Therefore, for 0 < 𝑛 ≤ 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑘𝑒𝑡, this model is valid. Exercise: Q. Parth joins a social media platform. After a week he has 34 friends, and after 9 weeks he has 80 friends on the platform. Find, I. The average number of friends Parth has made each week from week 1 to 9. II. Assuming that the total number of friends he has after ‘n’ weeks is an arithmetic sequence, find the model that models the number of Parth’s friends in ‘n’ weeks. Q. A party event venue advertises all-inclusive venue hire and catering costs of $ 6950for 50 guests and $11950 for 100 guests. Assume that the cost of venue hire and catering for ‘n’ guests forms an arithmetic sequence. I. Write a general formula for 𝑢𝑛 . II. Explain the significance of a) the common difference and b) the constant term III. Estimate the cost of venue hiring and catering for a party with 95 guests. Sum of an arithmetic sequence To find the sum of the n terms in an arithmetic sequence use the formula below. Q. Determine the sum of the first 8 terms of the arithmetic sequence 𝑢𝑛 = 5𝑛 + 27. A. In order to do this we simply need to go step by step. First, we shall find 𝑢1 . To find this, we substitute 1 in place of ‘n’ in the general formula of the sequence to get 5(1) + 27 = 32. Then we need to find the common difference which can be directly obtained; 27. Now we need to apply the formula and get the sum of the first 8 terms of the sequence. 𝑆8 = 4{ 2(32) + (8 − 1)27 } = 253(4) = 1012. Geometric Sequences and Series All sequences cannot have a linear relationship and the ones that do not, are called Geometric Sequences. The formula below helps you determine the 𝑛𝑡ℎ term of a geometric sequence. Where 𝑢1 is the first term of the sequence and 𝑟is the common ratio. 4 For example, 𝑢𝑛 = 5(2) 𝑛−1 . Find the following terms of the sequence. i. 𝑢2 ii. 𝑢5 iii. 𝑢14 To solve the above, we need to substitute ‘n’ with the term that we wish to find. Answers: i. 𝑢1 = 5(2) 1−1 = 5 ii. 𝑢5 = 5(2) 5−1 = 5(16) = 80 iii. 𝑢14 = 5(2) 5(8192) =40980 14−1 = Exercise: Q1. For the following sequences find 𝑢7 , 𝑢100 , 𝑢200 . i. 𝑢𝑛 = 17(5)𝑛−1 ii. 𝑢𝑛 = 2(3)𝑛−1 Q2. Show that the sequence is geometric : 5,10,20,40. Find 𝑢𝑛 and hence the 15𝑡ℎ term. Q3. A Geometric sequence has 𝑢4 = −70 and 𝑢7 = 8.75. Find the second term of the sequence. Q4. A geometric sequence with common ratio ‘r’ and an arithmetic sequence with common difference ‘d’ have the same first two terms. The third terms of the geometric and arithmetic sequences are in the ratio 2:1. Find, I. the possible values of ‘r’ II. For each value of ‘r’, find the ratio of the 4𝑡ℎ terms Note:- It is imperative to remember that in a geometric sequence 𝑢𝑛+1 ÷ 𝑢𝑛 = 𝑟. If you are asked to prove that a sequence is geometric, one of the methods you are expected to use is this. Sum of finite Geometric Sequences The general formula below gives you the sum of the n terms of a finite geometric sequence. Q. Find the sum of the first 8 terms of the arithmetic sequence 𝑢𝑛 = 5(2) 𝑛−1. A. In order to do this we simply need to go step by step. First, we shall find 𝑢1 . To find this, we substitute 1 in place of ‘n’ in the general formula of the sequence to get 5(1) = 5. Then we need to find the common ratio which is 2. Now we need to apply the formula and get the sum of the first 8 terms of the sequence. 𝑆8 = 5(28 −1) 2−1 = 640. 5 Sum of infinite convergent geometric sequences Q. The sum of the first three terms of a convergent infinite geometric series is 19. The sum of the series is 27. Find the first term and the common ratio. Q. The second term of a convergent infinite geometric series is 8/5. The sum of the series is 10. Show that there are two possible series, and find the first term and common ratio in each case. Sigma Notation and Arithmetic/ Geometric Series 𝑒𝑛𝑑 ∑ = 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑠𝑡𝑎𝑟𝑡 The sigma notation is used to find the sum of the series. Expand and evaluate: 1. ∑5𝑘=1 (11 − 2𝑘) 2. ∑3𝑘=1 (4𝑘) Applications of Geometric Sequences Geometric sequences have many applications. Growth and Decay, Financial math, Population growth and Annual Depreciation are only a few of them that the IBDP syllabus focuses on. 1. Compound Interest & Annual Depreciation Using the formula above we can calculate the amount or the future value of a deposit that is compounded annually. Another method to perform calculations of compound interest is through the Graphic Display Calculator, where you have to plug in values that are given in a word problem and you will receive your answer. 6 Example: Q. A certain sum amounts to $ 72900 in 2 years at 8% per annum compound interest, compounded annually. Find the sum. A. Let the sum be $ 100. Then, amount = $ {100 × (1 + 8/100)²} = $ (100 × 27/25 × 27/25) = $ (2916/25) If the amount is $ 2916/25 then the sum = $ 100. If the amount is $ 72900 then the sum = $ (100 × 25/2916 × 72900) = $ 62500. Hence, the required sum is $ 62500. Exercise: It is known that the number of fish in a certain lake will decrease by 7% each year unless the addition of new fish occurs. At the end of each year, 250 new fish are added to the lake. At the start of 2018, there are 2500 fish in the lake. A. Show that there will be approximately 2645 fish in the lake at the start of 2020. B. Find the approximate number of fish in the lake at the start of 2042. 2. Recurring Deposits One of the applications of Geometric sequences are recurring deposits. As the name suggests 7 in a recurring deposit, a fixed amount of money is invested at a fixed duration for a fixed period of time. These installments all mature on the same date. Essentially, a recurring deposit is like having multiple fixed deposit investments, all of which mature on the same day. Example: Q. Parth has started renting an apartment. He paid $5000 rent in the first year, and his rent increased by 5% each year. a. Find, to the nearest $10, the rent Parth pays in the 5th year b. Determine an expression for the total rent Parth pays during the first n years. c. How much rent did Parth pay during his 7 year stay? A. To solve this we must use our knowledge of sequences and their sums. a. We need to determine a general formula for the rent she pays. Since the principal amount is 5000 dollar, and the rate is 5% the general formula for the compounded rent is the following: 5 𝑢𝑛 = 5000(1 + ) 𝑛−1 100 To find the amount in the 4th year, simply substitute 5 in place of n. b. The total rent paid by Paula is essentially the sum of the finite geometric sequence that we derived in part (a). Thus using the formula we can say: 𝑆𝑛 = 5000(1.05𝑛 −1) 1.05−1 , which is equivalent to 𝑆𝑛 = 100000(1.05𝑛 − 1). c. To determine how much she paid in the first 7 years we can use the formula that we derived in subpart (b) since it gives us the total amount paid in the first ‘n’ years. 𝑆7 = 100000(1.057 − 1). Example: Q. Prashant initially puts $6000 in an account, which earns 5% interest paid annually. At the end of each year, he invests another $1000 in the same account. Find the amount in the account after 8 years. A. To solve this particular question we need to derive an equation that shows the amount he saves per year. Every year we are told Prashant deposits $6000 and at the end another $1000. If 𝐴𝑛 is the amount he has at the end of each year then we can say that : 𝐴1 = 6000(1.05) + 1000 The amount he has at the end of the second year would be the interest he earns on the amount he had in the first year, plus the $1000 he deposited. Thus 𝐴2 = 𝐴1 (1.05) + 1000 = (6000(1.05) + 1000)1.05 + 1000 = 6000(1.05)2 + 1000(1.05) + 1000. We can therefore say that 𝐴8 = 6000(1.05)8 + 1000(1.05)7 + 1000(1.05)6 + 1000(1.05)5 + 1000(1.05)4 + 1000(1.05)3 + 1000(1.05)2 + 1000(1.05)1 + 1000. 8 We see that the amount following 6000(1.05)8is a geometric series whose common ratio is 1.05. Thus 𝐴8 can be written as 𝐴8 = 6000(1.05)8 + 1000(1.058 −1) 1.05−1 = $18413.84 which is the value of the account after 8 years. Exercise: Q. On Mary’s birthday 1st January 1998, Mary’s grandparents put $x in a savings account. They kept on depositing $x on the first day of each month thereafter. The account paid a fixed rate of 0.4% interest every month. The interest was determined on the last day of every month. Then, the interest is added to the account. Let $𝐴𝑛 be the amount in Mary’s account on the last day of the nth month, immediately after the addition of the interest. a) Determine an expression for A1 and prove that A2 =1.0042 x + 1.004x b) (i) Find a similar expression for A3 and A4. (ii) Hence prove that the amount in Mary’s account the day before her 10th birthday is given by 251(1.004120 − 1)x. c) Write down an expression for 𝐴𝑛 in terms of x on the day before Mary’s 18th birthday, indicating clearly the value of n. d) Mary’s grandparents hoped that the amount in her account would be at least $20 000 the day before she was 18. Determine the minimum value of the monthly deposit $x required to achieve this and give your answer correct to the nearest dollar. e) As soon as Mary turned 18, she decided to invest $15 000 of the money from her grandparents in an account of the same type earning 0.4% interest per month. She withdraws $1000 every year on her birthday to get herself something. Determine how long she can do so until there is no money left in her account. Q. Raina takes a bank loan of $150 000 to buy a house, at an annual interest rate of 3.5%. The interest is calculated at the end of each year and is added to the amount outstanding. a) Find the amount Raina would owe the bank after 20 years. Give your answer to the nearest dollar. To pay off her loan, Raina makes annual deposits of $P at the end of every year in a savings account, paying an annual interest rate of 2% . She makes her first deposit at the end of the first year after taking out the loan. b) Show that Raina’s savings total to (1.0220 −1)P after 20 years . 9 c) Given that Raina’s aim is to own the house after 20 years, find P to the nearest dollar. Raina visits a different bank and makes a single deposit of $Q , the annual interest rate being 2.8%. d) Raina wants to withdraw $5000 at the end of each year for a period of n years. Prove that an 5000 expression for the minimum value of Q is 1.028+ 5000 2 1.028 5000 +...+ 1.028𝑛 . PART C Laws of Exponents: 1. 𝑎𝑚 × 𝑎𝑛 = 𝑎𝑚+𝑛 → Product Law 2. 𝑎𝑚 ÷ 𝑎𝑛 = 𝑎𝑚−𝑛 → Division Law 3. (𝑎𝑚 ) = 𝑎𝑚𝑛 → Power Law 𝑛 4. 𝑎0 = 1 1 5. 𝑎−𝑚 = 𝑎𝑚 → Negative Index Law 𝑚 𝑛 6. √𝑎𝑚 = 𝑎 𝑛 7. (𝑎 × 𝑏)𝑚 = 𝑎𝑚 × 𝑏 𝑚 8. ( 9. ( 𝑎 𝑏 𝑏 𝑎 ) 𝑛 ) −𝑛 = 𝑎𝑛 𝑏𝑛 =( 𝑎 𝑏 ) 𝑛 10.𝑏 = 𝑏 𝑐 → 𝑎 = 𝑐 PART D Laws of logarithms: 1. 𝑎 = 𝑏 𝑐 , then 𝑙𝑜𝑔𝑏 𝑎 = 𝑐 2. 𝑙𝑜𝑔𝑎 𝑥 + 𝑙𝑜𝑔𝑎 𝑦 = 𝑙𝑜𝑔𝑎 (𝑥𝑦) 𝑥 3. 𝑙𝑜𝑔𝑎 𝑥 − 𝑙𝑜𝑔𝑎 𝑦 = 𝑙𝑜𝑔𝑎 ( ) 𝑦 𝑛 4. 𝑙𝑜𝑔𝑎 𝑥 = 𝑛𝑙𝑜𝑔𝑎 𝑥 5. 𝑙𝑜𝑔𝑎 1 = 0 6. 𝑙𝑜𝑔𝑎 𝑎 = 1 1 7. −𝑙𝑜𝑔𝑎 𝑥 = 𝑙𝑜𝑔𝑎 ( ) 𝑥 10 8. 𝑙𝑜𝑔𝑎 𝑏 = 1 𝑙𝑜𝑔𝑏 𝑎 The same rules apply for logarithms with the base 𝑒: (ln x). Change of Base Theorem 𝑙𝑜𝑔𝑎 𝑥 = 𝑙𝑜𝑔𝑏 𝑥 𝑙𝑜𝑔𝑏 𝑎 Essentially, it is a formula that permits you to rewrite a logarithm in terms of logs with another base. This is very important when using a calculator to evaluate a log to any base other than 10 or e. Example: 𝑙𝑜𝑔16 32 = 𝑙𝑜𝑔2 32 5 = 𝑙𝑜𝑔2 16 4 Example: 𝑙𝑜𝑔2 3 = 𝑙𝑜𝑔10 3 0.47712 ≃ ≃ 1.585 𝑙𝑜𝑔10 2 0.30103 Example: 𝑙𝑜𝑔8 𝑥 = 𝑙𝑛 𝑥 1 = 𝑙𝑛 𝑥 𝑙𝑛 8 𝑙𝑛 8 Note:- Another very important type of question that often comes in the exam is the following. Q. Solve 4𝑥 + 2𝑥+2 = 3 using logs. A. To solve this question we must simplify the equation so that the terms with ‘x’ have the same base: 22𝑥 + 2𝑥 (4) − 3 = 0 B. Next, it is important to realise that this can be written as a quadratic equation where we let 2𝑥 = 𝑎. If we do that we get: 𝑎2 + 4𝑎 − 3 = 0. C. Using the quadratic formula we get 𝑎 = −2 ± √7. However, we do not have to find ‘a’ the questions asks us to find ‘x’, therefore we can rewrite the same as 2𝑥 = −2 ± √7. D. Using logarithms we can rewrite this as 𝑥 = 𝑙𝑜𝑔2 (−2 ± √7). However, we are not done yet. Since log of a negative number does not exist, 𝑙𝑜𝑔2 (−2 − √7) does not exist and should be omitted. The final answer is 𝑥 = 𝑙𝑜𝑔2 (−2 + √7). 11 Exercise: 1. Solve the simultaneous equations 𝑙𝑜𝑔2 6𝑥 = 1 + 2𝑙𝑜𝑔2 𝑦 1 + 𝑙𝑜𝑔6 𝑥 = 𝑙𝑜𝑔6 (15𝑦 − 25). 2 2. Solve (𝑙𝑛 𝑥) − (𝑙𝑛 2)(𝑙𝑛 𝑥) < 2(𝑙𝑛 𝑥) 2 3. Solve the equation 𝑙𝑜𝑔2 (𝑥 + 3) + 𝑙𝑜𝑔2 (𝑥 − 3) = 4 4. Find the integer values of m and n for which 𝑚 − 𝑛𝑙𝑜𝑔3 2 = 10𝑙𝑜𝑔9 6 5. The first terms of an arithmetic sequence are 1 , 1 , 1 , 1 𝑙𝑜𝑔2 𝑥 𝑙𝑜𝑔8 𝑥 𝑙𝑜𝑔32 𝑥 𝑙𝑜𝑔128 𝑥 ,…. Find x if the sum of the first 20 terms is equal to 100. Using the Graphic Display Calculator one can graph certain equations to solve questions as well. Example 1: Q. Solve the equation 4𝑥−1 − 2𝑥 = 8 A. To solve this by going to the Graphic function in the GDC we would type in Y1 = 4𝑥−1 Y2 = 2𝑥 + 8, and then locate the intersection. The ‘x’ value of the intersection would be the answer. 12 In this case the intersection point is (3,16). Therefore the answer is x = 3. PART E Binomial Expansion: The binomial theorem uses the concept of Pascal’s Triangle: And so on and so forth…. Binomial Expansion is the expansion of expressions into the sum of their individual terms. How to use Pascal’s triangle to perform binomial expansion? When we have to expand (𝑎 + 𝑏) 1 = 1𝑎 + 1𝑏 (𝑎 + 𝑏) 2 = 1𝑎2 + 2𝑎𝑏 + 1𝑏 2 (𝑎 + 𝑏) 3 = 1𝑎3 + 3𝑎2 𝑏 + 3𝑎𝑏 2 + 1𝑏 3 (𝑎 + 𝑏) 4 = 1𝑎4 + 4𝑎3 𝑏 + 6𝑎2 𝑏 2 + 4𝑎𝑏 3 + 1𝑏 4 ..And so on and so forth. We see that there is a pattern, which is the same as in Pascal’s triangle and therefore we use Pascal’s triangle in Binomial Expansion. For example, if we want to expand (𝑥 − 2𝑦 ) 3 , we can perform binomial expansion and do it. Since the power of the expansion is ‘3’ we use the 4th row of Pascal’s triangle to determine the coefficient of each term in the expression. One other important thing to note is that the number of terms ‘n’ in the expansion is always one more than the power. Similarly, this expansion would have 4 terms. First we must write down the coefficients of each of the terms. Note:- If the expansion has a negative sign involved like in the example shown here, the signs of the coefficients will alternate. +1 −3 +1 −3 13 Second, we write ‘x’ and ‘2y’ next to each of these coefficients. The powers for ‘x’ keep reducing from 3 to 0 and for 2y the powers keep increasing from 0 to 3. Then we simplify each of the terms. +1 × (𝑥)3 × (2𝑦)0 = 𝑥 3 −3 × (𝑥)2 × (2𝑦)1 = −6𝑥 2 𝑦 +1 × (𝑥)1 × (2𝑦)2 = 4𝑥𝑦 2 −3 × (𝑥)0 × (2𝑦)3 = −24𝑦 3 Thus, we can say that (𝑥 − 2𝑦 ) 3 = 𝑥 3 − 6𝑥 2 𝑦 + 4𝑥𝑦 2 − 24𝑦 3 . To determine the value of the coefficients according to Pascal’s triangle we use the concept of Combinations. 𝑛 𝐶𝑟 = 𝑛! 𝑟! (𝑛 − 𝑟)! For example, if we want to know the 3rd term in the Pascal’s Triangle for the 4th power. By looking at the triangle above we know it would be 4, however using this notation we can calculate it. 4 Similarly we can write (𝑎 + 𝑏)𝑛 = 𝐶3 = 𝑛 4! =4 3! (4 − 3)! 𝐶0 𝑎𝑛 𝑏 0 + 𝑛 𝐶1 𝑎𝑛−1 𝑏 1 + 𝐶2 𝑎𝑛−2 𝑏 2 . . . + 𝑛 𝐶𝑛 𝑎0 𝑏 𝑛 The above is equivalent to the formula given in the Math AA HL formula booklet : 𝑛 Note:- It is helpful to remember that 𝑛2 −𝑛 𝑛 𝐶0 = 1 ,𝑛 𝐶𝑛 = 1 , 𝑛 𝐶1 = 𝑛 , 𝑛 𝐶2 = . 2 To find a specific term in the expansion of an expression, rather than expanding the entire expression we can use a general formula that will help us find the term. General Formula : 𝑇𝑟+1 = appear in the exam. 𝑛 𝐶𝑟 𝑎𝑛−𝑟 𝑏 𝑟 . Below are a few examples of what kind of questions Example: Write down and simplify the expansion of (2 + 𝑥)4 in ascending powers of x. Hence find the exact value of (2.1)4 . A. To solve this we first follow the steps shown above. +1 × (2)4 × (𝑥)0 = 16 14 +4 × (2)3 × (𝑥)1 = 32𝑥 +6 × (2)2 × (𝑥)2 = 24𝑥 2 +4 × (2)1 × (𝑥)3 = 8𝑥 3 +1 × (2)4 × (𝑥)4 = 16𝑥 4 Thus, (2 + 𝑥)4 = 16 + 32𝑥 + 24𝑥 2 + 8𝑥 3 + 16𝑥 4 . In order to find the exact value of (2.1)4we must write it in a manner that is easier to compute. (2.1)4 = (2 + 0.1)4 𝑈𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 ∶ 16 + 32(0.1) + 24(0.1)2 + 8(0.1)3 + 16(0.1)4 = 19.4481 Example: Find the term in 𝑥 in the expansion of (3𝑥 + 𝐴)(2𝑥 + 𝐵)6 . 5 A. To solve this we need to first expand the expression (2𝑥 + 𝐵)6 and then multiply it with (3𝑥 + 𝐴). (2𝑥 + 𝐵)6 = +1 × (2𝑥)6 × (𝐵)0 = 64𝑥 6 +6 × (2𝑥)5 × (𝐵) = 192𝑥 5 𝐵 +15 × (2𝑥)4 × (𝐵)2 = 240𝑥 4 𝐵 2 +20 × (2𝑥)3 × (𝐵)3 = 160𝑥 3 𝐵 3 +15 × (2𝑥)2 × (𝐵)4 = 60𝑥 2 𝐵 4 +6 × (2𝑥)1 × (𝐵)5 = 12𝑥𝐵 5 +1 × (2𝑥)0 × (𝐵)6 = 𝐵 6 (3𝑥 + 𝐴)(2𝑥 + 𝐵)6 = (3𝑥 + 𝐴)(64𝑥 6 + 192𝑥 5 𝐵 + 240𝑥 4 𝐵 2 + 160𝑥 3 𝐵 3 + 60𝑥 2 𝐵 4 + 12𝑥𝐵 5 + 𝐵 6 ) To find the coefficient of 𝑥 5 we need to multiply only those terms in the two brackets that would give 𝑥 5 . So, (3𝑥)(240𝑥 4 𝐵2 ) + (𝐴)192𝑥 5 𝐵 = 𝑥 5 (720𝐵 2 + 192𝐴𝐵)and therefore the coefficient of 𝑥 5 is 720𝐵 2 + 192𝐴𝐵. Example: 2 Find the constant term in the expansion of (𝑥 − 𝑥)4 . A. To find the constant term in the expansion of the above expression we can use the general formula. 𝑇𝑟+1 = 4 𝐶𝑟 (𝑥)4−𝑟 (2𝑥 −1 )𝑟 ; To find the constant term we need the power of ‘x’ to equate to 0. Since 𝑥 0 = 1. Through this we can determine the value of ‘r’ and substitute it in the general formula to get the constant term. Equation the power of ‘x’ to 0: (𝑥)4−𝑟 (𝑥 −𝑟 ) = 𝑥 0 4−𝑟−𝑟 =0 4 − 2𝑟 = 0 15 4 = 2𝑟 ; 𝑟 = 2 Now, we can substitute this in the general form equation to get: 𝑇2+1 = 4 4 𝐶2 (𝑥)4−2 (2𝑥 −1 )2 = 6(𝑥)2 (𝑥 2 ) = 6(4) = 24. Thus, the constant term in the 2 expansion of (𝑥 − 𝑥)4 is 24. Exercise: Q. Expand the following expressions using binomial expansion:1. (2𝑥 + 5)5 2. (3𝑥 + 4)7 3. (2𝑥 2 + 5)3 4. (𝑥 + 9)8 5. (7𝑥 + 2)14 2 Q. Find the coefficient of 𝑥 8 in the expansion of (𝑥 2 − 𝑥)7 . 3 Q. Find the constant term in the expansion of (4𝑥 2 − 2𝑥)12 . Q. Consider the expansion of (1 + 𝑥)𝑛 where 𝑛 ≥ 3. a. Write down the first four terms of the expansion. The coefficients of the second, third and fourth terms of the expansion are consecutive terms of an arithmetic sequence. b. Show that 𝑛3 − 9𝑛2 + 14𝑛 = 0. c. Hence find the value of n. 𝑥 Q. When (1 + 2 )𝑛 , 𝑛 ∈ 𝑁, is expanded in ascending powers of x, the coefficient of 𝑥 3 is 70. a. Find the value of n. b. Hence, find the coefficient of 𝑥 2 . PART F Partial Fractions: Writing a fraction in the form of partial fractions makes it easier to integrate. For now, we will understand how to write fractions in the form of partial fractions. Example: Q. Write 2𝑥 −8 𝑥 2 −4 as partial fractions. A. To do this we first find simplify the denominator. 𝑥 2 − 4 = (𝑥 − 2)(𝑥 + 2) We can then write it in the form shown below- 16 2𝑥 − 8 = 𝐴 𝐵 + 𝑥−2 𝑥+2 (𝑥 − 2)(𝑥 + 2) 2𝑥 −8 + 𝐵(𝑥−2) = 𝐴(𝑥+2) ; We can now equate2𝑥 − 8 = 𝐴(𝑥 + 2) + 𝐵(𝑥 − 2). (𝑥−2)(𝑥+2) (𝑥−2)(𝑥+2) 2𝑥 − 8 = 𝐴𝑥 + 2𝐴 + 𝐵𝑥 − 2𝐵 2(𝑥) − 8 = 𝑥(𝐴 + 𝐵) + 2𝐴 − 2𝐵 ; Thus, we can say that 𝐴 + 𝐵 = 2; 2𝐴 − 2𝐵 = −8. Using the simultaneous equations we get A = -1 and B = 3. Thus we can write 1 2𝑥 −8 3 𝑥 2 −4 𝑥+2 = − . 𝑥−2 Exercise: 20 Q. Write (2𝑥−3)(𝑥+1)as the sum of partial fractions. 𝑥−9 Q. Write 𝑥 2−2𝑥−3as the sum of partial fractions. 6𝑥 2 +𝑥−19 Q. Write (𝑥+3)(𝑥−1)2 as the sum of partial fractions in the form 𝐴 𝑥+3 + 𝐵 𝑥−1 + 𝐶 𝑥−2 17