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Further Pure 1 Revision Topic 5: Sums of Series The OCR syllabus says that candidates should: Candidates should be able to: (a) use the standard results for Σr , Σr2 , Σr3 to find related sums; (b) use the method of differences to obtain the sum of a finite series; (c) recognise, by direct consideration of the sum to n terms, when a series is convergent, and find the sum to infinity in such cases. Section 1: Using standard formulae for r, r 2 , r 3 You need to learn the formula: n 1 r 2 n(n 1) r 1 The following formulae are given in the formula book: n 1 r 2 6 n(n 1)(2n 1) r 1 n 1 r 3 4 n2 (n 1)2 . r 1 n Note: 1 n (this should be obvious to you – if it isn’t then you need to remember it!). r 1 The above formulae can be used to find the sum of series. Worked examination question (Edexcel June 2004) n 1 (a) Show that (r 1)(r 5) = r 1 6 n(n+ 7)(2n + 7). (4) 40 (b) Hence calculate the value of (r 1)( r 5) r 10 (2) Solution: n n a) (r 1)(r 5) (r 2 6r 5) (expanding out brackets) r 1 r 1 We next split up the summation into several separate sums: n n n n (r 1)(r 5) r 2 6 r 51 r 1 r 1 r 1 r 1 So, using the results from above: n 1 1 1 (r 1)(r 5) 6 n(n 1)(2n 1) 6 2 n(n 1) 5n = 6 n(n 1)(2n 1) 3n(n 1) 5n r 1 1 We can try factorising this by taking out n as a factor: 6 n 1 (r 1)(r 5) 6 n (n 1)(2n 1) 18(n 1) 30 r 1 Expanding out the contents of the square brackets gives: n 1 1 (r 1)(r 5) 6 n 2n2 3n 1 18n 18 30 6 n 2n2 21n 49 r 1 The contents of the squared brackets can now be factorised. We get the result: n 1 (r 1)(r 5) = 6 n(n+ 7)(2n + 7). r 1 40 40 9 b) (r 1)(r 5) (r 1)(r 5) (r 1)(r 5) r 10 r 1 r 1 So: 40 1 1 (r 1)(r 5) 6 40(47)(87) 6 9 16 25 r 10 Therefore, 40 (r 1)(r 5) 27260 600 26660 . r 10 Examination question: Edexcel June 2002 Prove that n 6( r r 1 2 1) (n 1)n(2n + 5). Examination question (AQA June 2005) a) Use standard formulae to show that n 1 r 2 (r 1) 12n(n2 1)(3n 2) . r 1 b) Use the result from part (a) to find the value of 11 r 2 (r 1) . r 4 Examination question (AQA 2001) a) Find the sum of the integers from 1 to 300 inclusive. b) Evaluate: 300 (r 2 r ) r 1 Examination question (OCR May 2004) (i) Use the formula for r 3 to show that 13 23 33 ... (2n 1)3 n2 (2n 1)2 . (ii) Show also that 23 43 63 ... (2n 2)3 2n2 (n 1)2 . (iii) Hence find the sum of the series 13 23 33 43 ... (2n 2)3 (2n 1)3 , simplifying your answer. Section 2: Method of differences Some series can be summed using a difference method. Worked examination question (AQA 2006) a) Show that 1 1 2r 1 2 r 2 (r 1) 2 r (r 1) 2 b) Hence find the sum of the first n terms of the series 3 5 7 2 2 2 2 ... 1 2 2 2 2 3 3 4 Solution: a) Writing with a common denominator, we get 1 1 (r 1)2 r2 2 2 r 2 (r 1)2 r (r 1)2 r (r 1)2 Therefore, 1 1 r 2 2r 1 r 2 2r 1 2 r 2 (r 1) 2 r (r 1) 2 2 r (r 1)2 b) We wish to find n 2r 1 r 2 (r 1)2 . r 1 Using the result from (a), this is equivalent to finding n 1 1 r 2 (r 1)2 . r 1 Substituting r = 1, 2, 3, … into this summation gives: 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 ... 2 2 1 2 2 3 3 4 (n 1) 2 n n (n 1)2 Therefore, n 1 1 1 r 2 (r 1)2 = 1 (n 1)2 . r 1 Note: The series given in the above question is convergent. As the value of n increases (i.e. as the number of terms gets bigger and bigger), the sum of the series approaches 1 (since the value of 1 will tend to zero). (n 1) 2 We write 1 1 r 2 (r 1)2 = 1. r 1 Worked examination question (Edexcel Jan 2004) (a) Show that (r + 1)3 – (r – 1)3 Ar2 + B, where A and B are constants to be found. n (b) Prove by the method of differences that r r 1 2 = 1 6 n(n + 1)(2n + 1), n > 1. Solution a) On removing the brackets we find that (r + 1)3 – (r – 1)3 = r 3 3r 2 3r 1 r 3 3r 2 3r 1 = 6r 2 2 So, A = 6 and B = 2. n n b) So, (6r 2 2) (r 1)3 (r 1)3 . r 1 r 1 Working out the right hand side of this equation (by substituting r = 1, 2, 3, etc) gives: n (6r 2 2) ( 23 03 ) ( 33 13 ) ( 43 23 ) ( 53 33 ) r1 1st term 2nd term 3rd term 4th term ( 63 43 ) ... (n3 (n 2)3 ) ((n 1)3 (n 1)3 ) 5th term n1th term nth term i.e. n (6r 2 2) (n 1)3 n3 1 r 1 But the left-hand side of this equation is: n n n n (6r 2 2) 6 r 2 2 6 r 2 2n . r 1 r 1 r 1 r 1 Therefore: n 6 r 2 2n = (n 1)3 n3 1 r 1 n So: 6 r 2 = (n 1)3 n3 1 2n (n3 3n2 3n 1) n3 1 2n r 1 n i.e. 6 r 2 = 2n3 3n2 n n(2n2 3n 1) n(2n 1)(n 1) . r 1 So we get the result: n r r 1 2 = 1 n(n + 1)(2n + 1). 6 Examination question (Adapted from Edexcel June 2005) 2 1 1 (a) Show that = 4 r 1 2r 1 2r 1 2 n 2 1 (b) Hence, or otherwise, prove that =1– . r 1 4r 1 2 2n 1 20 2 (c) Hence find the exact value of r 11 4r 1 2 . Examination question (adapted from Edexcel Jan 2005) 1 1 1 (a) Show that . r (r 2) 2r 2(r 2) (b) Hence prove, by the method of differences, that n 4 n(3n 5) r (r 2) = (n 1)( n 2) . r 1 100 4 (c) Find the value of r 50 r (r 2) , to 4 decimal places. Examination question (AQA Jan 2005) a) Show that r 2 (r 1)2 (r 1)2 r 2 4r 3 . 100 b) Use the method of differences to find the value of r3 . r 50 Examination question (AQA June 2004) a) Show that 1 1 2 . r (r 1) (r 1)(r 2) r (r 1)(r 2) b) Hence find the sum of the series 1 1 1 1 ... , 1 2 3 2 3 4 3 4 5 30 31 32 giving your answer as a rational number. Examination Question (OCR January 2005) 4 You are given that f (r ) . (r 1)(r 3) 2 2 (i) Show that f (r ) . r 1 r 3 n (ii) Hence find f (r ) . (You need not express your answer as a single fraction.) r 1 (iii) Show that the series in part (ii) is convergent, and state its sum to infinity. Worked examination question (OCR May 2005) n 1 a) Show that (3r 1)2 n(6n2 3n 1) . r 1 2 1 b) (i) Given that f ( r ) , show that r (r 1) 2 f (r 1) f (r ) . r (r 1)(r 2) n 1 (ii) Hence find r (r 1)(r 2) . r 1 Solution: n n n n n a) (3r 1)2 (9r 2 6r 1) 9 r 2 6 r 1 r 1 r 1 r 1 r 1 r 1 Using the standard formulae given in section 1 of these revision notes, we see that n 1 1 (3r 1)2 9 6 n(n 1)(2n 1) 6 2 n(n 1) n r 1 3 = n(n 1)(2n 1) 3n( n 1) n . 2 Taking out ½n as a factor, we get: n 1 (3r 1)2 2 n 3(n 1)(2n 1) 6(n 1) 2 r 1 1 2 n 6n 2 9n 3 6n 6 2 1 2 n 6n 2 3n 1 1 1 b) If f ( r ) , then f (r 1) . r (r 1) (r 1)(r 2) 1 1 So, f (r 1) f (r ) = - . (r 1)( r 2) r (r 1) r r2 2 Therefore, f (r 1) f (r ) = . r (r 1)(r 2) r (r 1)(r 2) r (r 1)(r 2) n 1 1 n 2 1 n (ii) r (r 1)(r 2) 2 r (r 1)(r 2) 2 = ( f (r 1) f (r )) . r 1 r 1 r 1 Therefore, n 1 1 r (r 1)(r 2) = 2 ( f (2) f (1)) ( f (3) f (2) ) ( f (4) f (3) ) ... ( f (n 1) f (n) ) r 1 n 1 1 i.e. r (r 1)(r 2) = 2 f (n 1) f (1) . r 1 But, 1 1 1 f (1) and f (n 1) . 1 2 2 (n 1)(n 2) Therefore n 1 1 1 1 1 1 1 r (r 1)(r 2) = 2 f (1) f (n 1) 4 2( n 1)( n 2) . 2 2 (n 1)( n 2) r 1 Examination question (AQA June 2003) a) Show that 1 1 r r ! (r 1)! (r 1)! b) Hence find n r (r 1)! r 1 Hint: In part (a), use the result that (r + 1)! = (r + 1)r!

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