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1 1 Sequences 11 Overview A (numerical) sequence is a list of real numbers in which each entry is a function of its position in the list The entries in the list are called terms For example, 1, 1, 1 3, 1 4, is a sequence with first term 1, second term 1/, third term 1/3, etc A sequence is typically denoted {a n } n=1, where the subscript n, called the index, indicates the position of the term a n in the list That is, {a n } n=1 = a 1, a, a 3, 1 st term nd term 3 rd term The terms of a sequence { are } often given as a formula, which gives us the recipe for the sequence For 1 example, the sequence written out is n n=1 { } 1 = 1 n n=1 1, 1, 1 3, 1 4, Here, the general n th term is a n = 1/n, so a 1 = 1/1, a = 1/, and so on The index does not always start at n = 1 For that matter, the index need not be denoted with the letter n For example, Here, a k = k, k 0 Here s another example: { k } k=0 = 0, 1,, 3, = 1,, 4, 8, Example: Define a sequence by b k = k/(1 + k ), k = 1,, 3, Write down the first three terms of the sequence Solution: b 1, b, b 3 = , 1 +, = 1 3, 5, 1 3 We are interested in two specific types of sequences: (i) arithmetic and (ii) geometric Glen Pugh 010 p 1 of 14

2 1 Arithmetic Sequences Definition: A sequence a 1, a, a 3, with the property that a a 1 = d a 3 a = d a 4 a 3 = d a n a n 1 = d is called an arithmetic sequence with common difference d In simple terms, an arithmetic sequence is characterized by the property that the difference between consecutive terms is the same An arithmetic sequence is also called an arithmetic progression Example: Let a n = 5 n, n = 1,, 3, (i) List the first three terms of the sequence (ii) Is the sequence arithmetic? (iii) If yes to (ii), find the common difference Solution: (i) The first three terms are a 1, a, a 3 =5 (1), 5 (), 5 (3) =3, 1, 1 (ii) Suppose k 1 is any positive integer Then a k+1 a k =[5 (k + 1)] [5 (k)] =5 k 5 + k = Since k was arbitrary, we conclude that the difference between any two consecutive terms is, and so the sequence is arithmetic (iii) From (ii) we conclude that the common difference is d = Glen Pugh 010 p of 14

3 Example: Suppose {a n } n=1 is an arithmetic sequence with first term 3 and common difference 7/3 Find a formula for a n Solution: Since the common difference is 7/3 and the first term is 3, write out the first few terms to establish a pattern: a 1, a, a 3, a 4, =3, 3 + 7/3, 3 + 7/3 + 7/3, 3 + 7/3 + 7/3 + 7/3, =3, 3 + 7/3, 3 + (7/3), 3 + 3(7/3), By inspection, we see that term n has form 3 + (n 1)(7/3) That is, a n = 3 + (n 1)(7/3) This last example generalizes to give a standard form for arithmetic sequences: an arithmetic sequence {a n } n=1 with first term a and common difference d has nth term a n = a + (n 1)d Example: An arithmetic sequence has 4 th term 14/15 and 11 th term 7/3 Find a formula for a n Solution: Since a 4 = a + (4 1)d = a + 3d and a 11 = a + (11 1)d = a + 10d we have a + 10d = 7 3 a + 3d = Subtracting left hand sides and right hand sides respectively gives (a + 10d) (a + 3d) = d = 7 5 so d = 1 5 Now substitute this value for d into one of the expressions for the given terms of the series, for example a + 10d = 7 ( ) 3 1 a + 10 = a = 7 3 a = 1 3 Therefore, a n = (n 1)1 5 Glen Pugh 010 p 3 of 14

4 13 Geometric Sequences The general development of geometric sequences parallels that of arithmetic sequences, except that we consider division by a common value rather than addition: Definition: A sequence a 1, a, a 3, with the property that a a 1 = r a 3 a = r a 4 a 3 = r a n = r a n 1 is called a geometric sequence with common ratio r For a geometric sequence, the ratio of consecutive terms is the same A geometric sequence is also called a geometric progression { } 3 Example: Let {b n } n=1 = 7 n 1 n=1 (i) List the first three terms of the sequence (ii) Is the sequence geometric? (iii) If yes to (ii), find the common ratio Solution: (i) The first three terms are b 1, b, b 3 = , 3 7 1, =3, 3 7, 3 49 Glen Pugh 010 p 4 of 14

5 (ii) Suppose k 1 is any positive integer Then b k+1 b k = 3 7 k+1 1 / 3 7 k 1 = 3 7 k 7 k 1 3 = 1 7 Since k was arbitrary, we conclude that b k+1 /b k = 1/7 for all integers k 1, so that the sequence is geometric (iii) From (ii) we conclude that the common ratio is r = 1/7 Example: Suppose {a n } n=1 formula for a n is a geometric sequence with first term a and common ratio r Find a Solution: Since the common ratio is r and the first term is a, the sequence has the form a 1, a, a 3, a 4, =a, a r, a r r, a r r r, =a, ar, ar, ar 3, By inspection, we see that term n has form ar n 1 That is, a n = ar n 1 This last example leads us to conclude: a geometric sequence {a n } n=1 ratio r has n th term a n = ar n 1 with first term a and common Example: A geometric sequence has a 3 = 7/4 and a 6 = 7/3 Find a 8 Solution: We have a 3 = ar 3 1 = ar and a 6 = ar 6 1 = ar 5, so Dividing left and right hand sides respectively gives ar 5 = 7 3 ar = 7 4 ar 5 ar = ( 7 3 ( 7 4) ) r 3 = 1 8 so r = 1 Glen Pugh 010 p 5 of 14

6 Now substitute this value for r into one of the expressions for the given terms of the series, for example ar = 7 4 ( ) 1 a = 7 4 ( ) 1 a = so a = 7 Finally, a 8 = ar 7 ( 1 = 7 = 7 18 ) 7 14 Problems 1 Write the first four terms of the sequence defined by a n = n 1 n + n, n 1 ans: 1/3, 3/8, 1/3, 7/4 For the sequence {a n } n=1 with first five terms,, 6,, 10, give a possible expression for a n ans: n 3 Find an expression for a n for the arithmetic sequence 3/5, 1/10, /5, ans: 3/5 (n 1)5/10 4 Find the 14 th term of the arithmetic sequence 3, 7/3, 5/3, 5 An arithmetic sequence has a 17 = 5/3 and a 3 = 95/6 What is a 6? 6 If a 1 = 5, d = 14 and a n = 507 then what is n if the sequence is arithmetic? 7 Find the 3 rd term of the geometric sequence 7/65, 7/5, ans: 17/3 ans: 17/6 ans: 39 ans: Glen Pugh 010 p 6 of 14

7 8 Find an expression for a n for the geometric sequence /x, 4/x, 9 If a geometric sequence has a 4 = 8/3 and a 7 = 64/3, what is a 5? 10 A geometric sequence has the property that a n+3 = 7a n What then is r? ans: (/x) n ans: 3 ans: 16/3 Series Now that we have studied sequences in general, and arithmetic and geometric sequences in particular, we make use of these to define the notion of a series Given a sequence {a n } n=1 = a 1, a, a 3, we define the associated series to be sum of the terms of the sequence A series may be finite, in which case we use S n to denote the sum of the first n terms of the sequence: S n = a 1 + a + a a n For example, given the sequence defined by a n = 1/ n, n = 1,, 3,, we may form the series of the first 100 terms S 100 = a 1 + a + a a 100 = We will be concerned with finite series, and in particular how to add them up We will find convenient expressions for S n which allow us to easily determine the sum without having to add the terms one at a time Note that a series may also be infinite, meaning we continue to add terms without stopping For an infinite series there is no final term in the sum An example of an infinite series is S = To most people, the idea of adding up an infinite number of terms is difficult to grasp, as surely the resulting sum must be infinite This is not always the case however, and there is a well defined mathematical procedure for determining the sum of an infinite series when it exists For example, it turns out that S = = 1 Glen Pugh 010 p 7 of 14

8 Another well known infinite series is S = which, quite astoundingly, adds up to S = π /6 We will will not be looking at infinite series further in this course We will be concerned with two specific types of series related to our earlier work: arithmetic series and geometric series 1 Arithmetic Series Given an arithmetic sequence with first term a and common difference d, recall the n th term had the form a n = a + (n 1)d, n = 1,, 3, We define the n-term arithmetic series as the sum of the first n terms of the arithmetic sequence: S n = a 1 + a + a a n There is a very simple formula for S n which we will now derive Write S n = a 1 + a + a a n = [a + 0d] + [a + 1d] + [a + d] + + [a + (n )d] + [a + (n 1)d], where in the last line above we have written out the terms of the series in their standard form a n = a + (n 1)d Now reverse the order of the terms in the last line above and write them in the following convenient arrangement: S n = [a + 0d] +[a + 1d] + [a + d] + + [a + (n )d] +[a + (n 1)d] S n = [a + (n 1)d] +[a + (n )d] + [a + (n 3)d] + + [a + 1d] +[a + 0d] Now add both equations by adding down the columns as we ve written them: S n = [a + a + (n 1)d] +[a + a + (n 1)d] + +[a + a + (n 1)d] + [a + a + (n 1)d] The right hand side of this last equation is the sum of the n terms [a + a + (n 1)d], and so which gives S n = n[a + (n 1)d] S n = n[a + (n 1)d] Using this equation, another convenient form for S n is found by writing the term in square brackets as terms of the sum: n[a + (n 1)d] S n = n[a + a + (n 1)d] = Glen Pugh 010 p 8 of 14

9 So that S n = n (a 1 + a n ) This last expression for S n says that the sum of the first n terms of an arithmetic series is the average of the first and last terms, multiplied by the number of terms Example: What it sum of the first n positive integers? Solution: The question is to find S n = n The sequence 1,, 3,, n is arithmetic with a = 1 and d = 1, and we have a 1 = 1 and a n = n Thus S n = n (a 1 + a n ) = n 1 + n n(n + 1) = Example: What it sum of the first n odd positive integers? Solution: This time the sum is S n = a n The sequence of odd integers is arithmetic with a = 1 and d =, and so a n = 1 + (n 1) = n 1 Therefore S n = n (a 1 + a n ) (1 + n 1) = n = n n = n Example: A well drilling company charges \$16 to drill the first metre of a well, \$1630 for the next metre, \$1660 for the next, and so, the cost increasing by \$030 per metre How much will it cost to drill a 30 metre well? Solution: Let c n = cost of drilling the n th metre = 16 + (n 1)(030) Glen Pugh 010 p 9 of 14

10 Notice c 1, c, c 3, is an arithmetic sequence, and the cost to drill 30 metres is S 30 = c 1 +c + +c 30, an arithmetic series with 30 terms Therefore, S 30 = 30 (c 1 + c 30 ) [16 + (16 + (30 1)(030))] = 30 = \$61050 and so it will cost \$61050 to drill the 30 metre well Geometric Series Given a geometric sequence with first term a and common ratio r, recall the n th term had the form a n = ar n 1, n = 1,, 3, We define the n-term geometric series as the sum of the first n terms of the geometric sequence: S n = a 1 + a + a a n As with the arithmetic series, there is a very simple formula for S n First write S n = a 1 + a + a a n = a + ar + ar + + ar n + ar n 1 Now multiply both sides of this equation by r, giving and subtract the second equation from the first: rs n = ar + ar + + ar n 1 + ar n, S n rs n = (a + ar + ar + + ar n + ar n 1 ) (ar + ar + + ar n 1 + ar n ) Notice that if the bracketed terms on the right hand side are simplified, all terms cancel except a and ar n Also, the S n is common to both terms on the left Thus, this last equation simplifies to S n (1 r) = a ar n, and so the sum of the first n terms of the geometric series is S n = a(1 rn ) 1 r Sums of geometric series arise in many interesting examples; here are a few: Glen Pugh 010 p 10 of 14

11 Example: The game of chess is played on an 8 8 board of 64 squares There is a legend that the game was invented for a king, and the king was so pleased with the new game that he wanted to reward the inventor, so he asked the inventor what he would like The inventor replied that he would like one grain of rice for the first square of the chess board, two for the second square, four for the third, and so on, doubling the number of grains for each square of the board for all 64 squares The king saw that this was a fairly modest request and so ordered that the rice be delivered to the inventor If a single grain of rice has a mass of kilograms, how much rice (in kilograms) was required to deliver on the king s order? Solution: The chessboard with the grains of rice for the first few squares is shown below: Let N be the total number of grains required: N = Notice that the sequence 1,, 4, is geometric with a = 1 and r =, and we wish to sum the first n = 64 terms of this sequence Thus N = a(1 rn ) 1 r = 1(1 64 ) 1 = 64 1 The total mass of rice is the number of grains multiplied by the mass of a single grain: N = ( 64 1)(00000) = kilograms Glen Pugh 010 p 11 of 14

12 Needless to say, this is a rather large amount of rice Expressed in scientific notation, the enormity of this quantity may be difficult to appreciate Here is the number written out: N = 368,934,881,474, kilograms = 368,934,881, tonnes That is, about 369 billion metric tonnes of rice would be required, or about 369 trillion 1 kg bags of rice Compound Interest A very important application of geometric series is that of compound interest In general terms, when money is lent to a borrower by a lender, interest is paid by the borrower in exchange for the use of the lender s money For example, suppose you find \$5000 in your sock drawer, and you decide to take it to your bank and purchase a type of investment called a GIC (Guaranteed Investment Certificate) Under the terms of the GIC, you give your \$5000 to the bank, and at the end of ten years they agree to return the \$5000 to you along with earned interest In this case, you are the lender and the bank is the borrower In many examples, the lender is an investor depositing money into an investment, and the investment can be considered as the borrower, the expectation being that the investment will return the original amount invested plus interest Interest is normally quoted as a percentage For example, \$1000 invested for one year at an annual rate of interest of 7% accumulates to \$1070 after one year The \$1070 consists of the original \$1000 plus the interest of (7%)(\$1000) = (007)(\$1000) = \$70 Compound interest is the process of adding earned interest to the original amount of the investment, so that this new total (compound) amount then in turn earns interest That is, interest is then earned on previously earned interest The compounding frequency states how many times per year the earned interest is added to the interest bearing portion of the investment In our case we will stick to the simplest type of compound interest: interest compounded annually Let s return to our GIC example to see how this works Suppose you decide to invest \$5000 in a 10 year GIC which pays interest at the rate of 4% compounded annually Here is a table illustrating the accumulation of interest in the investment over the years: Year value at start of year ( ) { value at start of year } { interest + earned in year } value at = end of year ( ) (004)(5000) = 5000(104) 5000(104) 5000(104) + (004)5000(104) = 5000(104) (104) 5000(104) + (004)5000(104) = 5000(104) (104) (104) 9 + (004)5000(104) 9 = 5000(104) 10 Glen Pugh 010 p 1 of 14

13 Notice how, in each year, the value of the investment at the end of the year is the value at the beginning plus the interest earned in the year The accumulated value of the investment at the end of one year then becomes the amount subject to interest at the beginning of the next Also notice the form of the accumulated value of the investment at the end of each year This leads us to conclude: An amount \$x invested for n years at a rate of interest of i% compounded annually has an accumulated value at the end of n years of ( x 1 + i ) n 100 Example: If, on every birthday beginning with the 1 st and ending with the 65 th, a person deposits \$1000 into a retirement fund paying 10% interest compounded annually, what will the accumulated value of the fund be on the person s 65 th birthday? Solution: 1000 deposited on 1 st birthday accumulates to 1000(1 + 01) deposited on nd birthday accumulates to 1000(1 + 01) deposited on 3 rd birthday accumulates to 1000(1 + 01) deposited on 64 th birthday accumulates to 1000(1 + 01) deposited on 65 th birthday accumulates to 1000(1 + 01) 0 = 1000 The value of the fund at age 65 will be total of all of these accumulated amounts: (11) (11) (11) 44, a geometric series Here a = 1000, r = 11 and n = 45, and so the total value is given by a(1 r n ) 1 r = (11) = \$718,905 Notice here that only \$45,000 was paid into the retirement fund over the years, yet the fund is worth \$718,905 at age 65 3 Problems 1 Find the sum of the first 0 terms of the series a 0 Find the sum of the first 300 even positive integers ans: 670 ans: 90,300 Glen Pugh 010 p 13 of 14

14 3 Find the sum of the first 50 terms of the arithmetic sequence that has first term 8 and 50 th term Compute π 3 + π 3 + π + 4π π 3 5 Compute 1 e + 3 e + 5 e e ans: 375 ans: 91π/3 ans: 11/e 6 A theatre is designed with 8 seats in the front row, 3 in the second, 36 in the third, and so on, all the way back to the 0 th and final row How many seats are there in the theatre? 7 Find the sum of the first nine terms of the geometric series ans: 130 ans: 491/18 8 Find the sum of the geometric series ans: You convince your employer to pay you 1 for the first day of a month worked, for the next, 4 for the next, and so on, doubling with each day worked If you work each day of the month of March (31 days), how much will you earn? ans: \$1,474, To save for university, a parent makes a \$100 deposit on January 1 of each year to a savings fund which earns interest at 6% compounded annually If 18 deposits are made in total, what is the accumulated value of the fund at the time when the last deposit is made? ans: \$37, Parents of a newborn wish to accumulate \$150,000 in an education fund by making equal deposits starting on the day of the child s birth and ending with the 18 th birthday If the education fund pays 8% interest compounded annually, how large should the payments be in order to reach the goal? ans: \$ Glen Pugh 010 p 14 of 14

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