#### Transcript C2.3 Sequences and series

AS-Level Maths: Core 2 for Edexcel C2.3 Sequences and series This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 39 © Boardworks Ltd 2005 Geometric sequences Contents Geometric sequences Geometric series The sum to infinity of a geometric series Binomial expansions Examination-style questions 2 of 39 © Boardworks Ltd 2005 Geometric sequences In a geometric sequence (or geometric progression) each term is produced by multiplying the previous term by a constant value called the common ratio. For example, the sequence 3, 6, 12, 24, 48, 96, … is a geometric sequence that starts with 3 and has a common ratio of 2. We could write this sequence as 3, 3 × 2, 3 × 2 × 2, 3 × 2 × 2 × 2, 3 × 2 × 2 × 2 × 2, … or 3, 3 of 39 3 × 2, 3 × 22, 3 × 23, 3 × 24, 3 × 25, … © Boardworks Ltd 2005 Geometric sequences If we call the first term of a geometric sequence a and the common ratio r we can write a general geometric sequence as: a, ar, ar2, ar3, ar4, ar5, … The nth term of a geometric sequence with first term a and common ratio r is: un = arn–1 Also: The inductive definition of a geometric sequence with first term a and common ratio r is: u1 = a, un+1 = run 4 of 39 © Boardworks Ltd 2005 Geometric sequences What is the 7th term of the geometric sequence 8, 12, 18, 27, …? This is a geometric sequence with first term a = 7 and common ratio r = 12 ÷ 8 = 1.5. The nth term is given by arn–1 so the 7th term is: u7 = 8 × 1.56 = 91.125 The 3rd term in a geometric sequence is 36 and the 6th term is 972. What is the value of the 1st term and the common ratio? Using the 3rd term: Using the 6th term: 5 of 39 ar2 = 36 ar5 = 972 © Boardworks Ltd 2005 Geometric sequences Dividing these gives: ar 5 972 = 2 ar 36 r3 = 27 So: r=3 Substituting this into ar2 = 36 gives: a × 32 = 36 9a = 36 a=4 So the first term of the sequence is 4 and the common ratio is 3. The general term of this sequence is un = 4 × (3)n–1. 6 of 39 © Boardworks Ltd 2005 Find the missing terms 7 of 39 © Boardworks Ltd 2005 Convergent and divergent sequences Geometric sequences either converge or diverge depending on the value of the common ratio r. Suppose the first value of a geometric sequence is 8 and the common ratio is 41 . This gives the following sequence: 8, 2, 1 2 , 1 8 , 1 32 , 1 128 , ... This sequence converges to 0. If the common ratio is negative the signs of the terms will alternate. For example, if the first value of a geometric sequence is 6 and the common ratio is 21 , we have: 6, –3, 3 2 , 34 , 3 8 , 163 , ... This sequence also converges to 0. 8 of 39 © Boardworks Ltd 2005 Convergent and divergent sequences In general, if the common ratio r of a geometric sequence is between 0 and 1 or between –1 and 0, the terms of the sequence will converge to 0. We use |r| to represent the modulus of r, mod r. This is the numerical value of r, regardless of whether it is positive or negative. So, for example: |0.6| = 0.6 and |–0.6| = 0.6 In general: For a geometric sequence, if 0 < |r| < 1 the sequence will converge. 9 of 39 © Boardworks Ltd 2005 Convergent and divergent sequences For a geometric sequence, if |r| > 1 the sequence will diverge. This means that the terms will get larger and larger without limit. For example, if the first value of a geometric sequence is 0.2 and the common ratio is –5, the sequence will be: 0.2, –1, 5, –25, 125, –625, … This sequence is divergent. The only geometric sequences that neither converge nor diverge are those where the common ratio is –1. For example: 4, –4, 4, –4, 4, –4, … This sequence oscillates between two values. 10 of 39 © Boardworks Ltd 2005 Contents Geometric series Geometric sequences Geometric series The sum to infinity of a geometric series Binomial expansions Examination-style questions 11 of 39 © Boardworks Ltd 2005 Geometric series The sum of all the terms of a geometric sequence is called a geometric series. We can write the sum of the first n terms of a geometric series as: Sn = a + ar + ar2 + ar3 + … + arn–1 For example, the sum of the first 5 terms of the geometric series with first term 2 and common ratio 3 is: S4 = 2 + (2 × 3) + (2 × 32) + (2 × 33) + (2 × 34) = 2 + 6 + 18 + 54 + 162 = 242 When n is large, a more systematic approach for calculating the sum of a given number of terms is required. 12 of 39 © Boardworks Ltd 2005 The sum of a geometric series Start by writing the sum of the first n terms of a general geometric series with first term a and common ratio r as: Sn = a + ar + ar2 + ar3 + … + arn–1 Multiplying both sides by r gives: rSn = ar + ar2 + ar3 + … + arn–1 + arn Now if we subtract the first equation from the second we have: rSn – Sn= arn – a Sn(r – 1) = a(rn – 1) a( r n 1) Sn = r 1 13 of 39 © Boardworks Ltd 2005 The sum of a geometric series If we multiply the numerator and the denominator by –1 we can also write the sum of the first n terms as: a(1 r n ) Sn = 1 r This form is more useful when |r| < 1 since it avoids the use of negative numbers. For example: Find the sum of the first 8 terms of the geometric series that starts 4 – 3 + 2 41 –, … 14 of 39 © Boardworks Ltd 2005 Using Σ notation We can write the sum of a geometric series using Σ notation as: n n a ( r 1) a (1 r ) i 1 ar = = r 1 1 r i =1 n 7 Find i 1 5(2) i =1 7 i 1 6 5(2) = 5 +10 + 20 +... + 5(2) i =1 This is a geometric series with first term 5 and common ratio 2. There are 7 terms in this sequence so: 7 5(2 1) i 1 5(2) = 2 1 i =1 = 635 7 15 of 39 © Boardworks Ltd 2005 Contents The sum to infinity of a geometric series Geometric sequences Geometric series The sum to infinity of a geometric series Binomial expansions Examination-style questions 16 of 39 © Boardworks Ltd 2005 The sum to infinity of a geometric series When the common ratio of a geometric series is between –1 and 1, the sum of the series will tend to a particular value as more terms are added. For example, the geometric series 1 + 21 + 41 + 81 + 1 16 + ... tends to 2 as the number of terms increases. We can show this diagrammatically as follows: 2 1 2 1 1 1 16 1 8 17 of 39 1 4 1 2 © Boardworks Ltd 2005 The sum to infinity of a geometric series a(1 r n ) If we use the formula Sn = with a = 1 and r = 21 we have 1 r 1(1 ( 21 )n ) Sn = 1 21 = 1 ( 21 )n 1 2 = 2 ( 21 )n1 =2 1 As n , n1 0. 2 So: 18 of 39 1 2n1 S = 2 where S = lim Sn n © Boardworks Ltd 2005 The sum to infinity of a geometric series In general, the sum of the first n terms of a geometric series is: a(1 r n ) Sn = 1 r But if |r| < 1, r n 0 as n In this case, we can write the sum to infinity as: a S = 1 r For example: Find the sum to infinity of the geometric series with first term 6 and common ratio –0.2. 6 S = =5 1+ 0.2 19 of 39 © Boardworks Ltd 2005 The sum to infinity of a geometric series The first term of a geometric series is 20 and the sum to infinity is 15. What is the common ratio? a Using S = we have 1 r 20 15 = 1 r 15(1 r ) = 20 20 1 r = 15 20 r = 1 15 1 r= 3 20 of 39 © Boardworks Ltd 2005 Contents Binomial expansions Geometric sequences Geometric series The sum to infinity of a geometric series Binomial expansions Examination-style questions 21 of 39 © Boardworks Ltd 2005 Pascal’s Triangle 22 of 39 © Boardworks Ltd 2005 Binomial expansions An expression containing two terms, for example a + b, is called a binomial expression. When we find powers of binomial expressions an interesting pattern emerges. (a + b)0 = 1 (a + b)1 = 1a + 1b (a + b)2 = 1a2 + 2ab + 1b2 (a + b)3 = 1a3 + 3a2b + 3ab2 + 1b3 (a + b)4 = 1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4 (a + b)5 = 1a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + 1b5 What patterns do you notice? 23 of 39 © Boardworks Ltd 2005 Binomial expansions In general, in the expansion of (a + b)n: The coefficients are given by the (n + 1)th row of Pascal’s triangle. The sum of the powers of a and b is n for each term. Altogether, there are n + 1 terms in the expansion. As long as n is relatively small, we can expand a given binomial directly by comparing it to the equivalent expansion of (a + b)n. For example: Expand (x + 1)5. Using (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 and replacing a with x and b with 1 gives: (x + 1)5 = x5 + 5x4 + 10x3 + 10x2 + 5x + 1 24 of 39 © Boardworks Ltd 2005 Binomial expansions Expand (2x – y)4. (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 Using and replacing a with 2x and b with –y gives: (2x – y)4 = (2x)4 + 4(2x)3(–y) + 6(2x)2(–y)2 + 4(2x)(–y)3 + (–y)4 = 16x4 – 32x3y + 24x2y2 – 8xy3 + y4 Notice that when the second term in a binomial is negative the signs of the terms in the expansion will alternate. Suppose we wanted to expand (a + b)20. We could find the 21st row of Pascal’s triangle, but this would take a very long time. 25 of 39 © Boardworks Ltd 2005 Finding binomial coefficients When n is large we can find the binomial coefficients using combinations theory. Let’s look more closely at the expansion of (a + b)4 = (a + b)(a + b)(a + b)(a + b) Ways of getting a4 aaaa 1 way 26 of 39 Ways of Ways of Ways of getting a3b getting a2b2 getting ab3 aaab aabb abbb aaba abab babb abaa abba bbab baaa bbaa bbba baba baab 4 ways 6 ways 4 ways Ways of getting b4 bbbb 1 way © Boardworks Ltd 2005 Finding binomial coefficients The situation where no b’s (or four a’s) are chosen from any of the four brackets can be written as 4 4 4 This is the same as C4 or . or . 0 4 The situation where one b (or three a’s) can be chosen from any of the four brackets can be written as: 4C 0 4 4 4C or 1 . This is the same as 4C3 or 3 . 1 The situation where two b’s (or two a’s) can be chosen from any of the four brackets can be written as 4C 2 27 of 39 4 or . 2 © Boardworks Ltd 2005 Finding binomial coefficients The fifth row of Pascal’s triangle can be written as: 4 0 4 1 4 2 4 3 4 4 4 1 This corresponds to the values 1 4 6 The expansion of (a + b)4 can therefore be written as: 4 4 4 3 4 2 2 4 3 4 4 (a b) a a b a b ab b 0 1 2 3 4 4 (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 Or: 28 of 39 © Boardworks Ltd 2005 Finding binomial coefficients The number of ways to choose r objects from a group of n objects is written as nCr and is given by n n! r = r! n r ! n! is read as ‘n factorial’ and is the product of all the natural numbers from 1 to n. In general: n! = n × (n –1) × (n – 2) × (n – 3) … × 2 × 1 n can also be 0 and by definition 0! = 1. 29 of 39 © Boardworks Ltd 2005 Finding binomial coefficients The value of n! gets large very quickly as the value of n increases. For example: 5! = 5 × 4 × 3 × 2 × 1 = 120 12! = 12 × 11 × 10 × … × 2 × 1 = 479 001 600 20! = 20 × 19 × 18 × … × 2 × 1 = 2 432 902 008 176 640 000 Fortunately, when we use the formula n n! r = r! n r ! to calculate binomial coefficients, many of the numbers cancel out. For example, for 4C2 we have 2 4 × 3 × 2×1 4 4! 4×3 = = = =6 2 2! 2! (2×1)×(2×1) 2 ×1 30 of 39 © Boardworks Ltd 2005 Finding binomial coefficients Here are some more examples: 8 × 7 × 6 × 5 × 4 × 3 × 2×1 8 8! 8×7×6 = = = = 56 3 3! 5! (3 × 2×1)×(5 × 4 × 3 × 2×1) 3 × 2 ×1 This value corresponds to the number of ways of choosing 3 a’s from the 8 brackets in the expansion of (a + b)8. 56 is therefore the coefficient of a3b5 in the expansion of (a + b)8. 4 9× 8× 7× 6× 5× 4× 3× 2×1 9 9! 9×8 = = = = 36 7 7! 2! (7× 6× 5× 4× 3× 2×1)×(2×1) 2 ×1 This value corresponds to the number of ways of choosing 7 a’s from the 9 brackets in the expansion of (a + b)9. 36 is therefore the coefficient of a7b2 in the expansion of (a + b)9. 31 of 39 © Boardworks Ltd 2005 Finding binomial coefficients The effect of this cancelling gives an alternative form for nCr. n n ×( n 1)×( n 2)×...×( n r +1) r = r! In general, the expansion of (a + b)n can be written as: n n n n 1 n n 2 2 n n (a + b) = a + a b + a b + ...+ b 0 1 2 n n A special case is the expansion of (1 + x)n n n n 2 n 3 (1 + x ) = + x + x + x +... + x n 0 1 2 3 n( n 1) 2 n( n 1)( n 2) 3 = 1+ nx + x + x +... + x n 2! 3! n 32 of 39 © Boardworks Ltd 2005 Using the binomial theorem This method of finding the binomial coefficients is called the binomial theorem. Find the coefficient of a7b3 in the expansion of (a – 2b)10. The term in a7b3 is of the form: 3 4 10 7 10 × 9 × 8 7 3 3 a ( 2 b ) = a ( 8 b ) 3 1× 2× 3 = 120(–8a7b3) = –960a7b3 So the coefficient of a7b3 in the expansion of (a – 2b)10 is –960. 33 of 39 © Boardworks Ltd 2005 Using the binomial theorem Use the binomial theorem to write down the first four terms in the expansion of (1 + x)7 in ascending powers of x. 7 7 7 2 7 3 (1+ x ) = + x + x + x +... 0 1 2 3 7 3 = 1+ 7 x + 7×6 2 7×6×5 3 x + x +... 2×1 3 × 2×1 = 1+ 7 x + 21x2 + 35 x3 +... How could we use this expansion to find an approximate value for 1.17? 34 of 39 © Boardworks Ltd 2005 Using the binomial theorem To find an approximate value for 1.17 we can let x = 0.1 in the expansion (1 + x)7 = 1 + 7x + 21x2 + 35x3 + … This gives us 1.17 ≈ 1 + 7 × 0.1 + 21 × 0.12 + 35 × 0.13 As 0.1 is raised to ever higher powers it becomes much smaller and so less significant. We can therefore leave out higher powers of x and still have a reasonable approximation. 1.17 ≈ 1 + 0.7 + 0.21 + 0.035 ≈ 1.945 35 of 39 © Boardworks Ltd 2005 Contents Examination-style questions Geometric sequences Geometric series The sum to infinity of a geometric series Binomial expansions Examination-style questions 36 of 39 © Boardworks Ltd 2005 Examination-style question 1 The 2nd term of a geometric series is 40 and the 5th term is 20.48. a) Find the value of the first term and the common ratio. b) Calculate the sum of the first 10 terms of the series. c) Calculate the sum to infinity. a) Using the 2nd term: ar = 40 Using the 5th term: ar4 = 20.48 Dividing these gives: ar 4 20.48 = ar 40 r3 = 0.512 So: r = 0.8 37 of 39 © Boardworks Ltd 2005 Examination-style question 1 Substituting this into ar = 40 gives: a × 0.8 = 40 a = 50 a( r n 1) b) Using Sn = with n = 10, a = 50 and r = 0.8 gives: r 1 50(0.810 1) S10 = 0.8 1 = 223.16 (to 2 d.p.) a c) Using S = with a = 50 and r = 0.8 gives: 1 r 50 S = 1 0.8 = 250 38 of 39 © Boardworks Ltd 2005 Examination-style question 2 a) Write down the first four terms in the expansion of (1 + ax)13 in ascending powers of x, where a > 0. b) Given that in the expansion of (1 + ax)13 the coefficient of x is –b and the coefficient of x2 is 12b, find the value of a and b. a) (1 + 13 ×12 13 ×12×11 2 1+13ax + ( ax ) + ( ax )3 +... 2 3×2 = 1 + 13ax + 78a2x2 + 286a3x3 + … ax)13 = 13a = –b 1 78a2 = 12b 2 78a2 = 12 × –13a Substituting 1 into 2 : 78a = –156 a = –2 b = 26 b) 39 of 39 © Boardworks Ltd 2005