Revision Library

revisionlibrary.co.uk

A-Level · Further Pure Mathematics - Core Pure

FP.CP10 Series and Summation

EDEXCEL 9FM0 · Calculator allowed · about 110 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

The standard results for sums of powers of the first n positive integers are used throughout this worksheet.

(a)State the standard results for sum_{r=1}^{n} r, sum_{r=1}^{n} r² and sum_{r=1}^{n} r^3.(3)

(b)Calculate sum_{r=1}^{20} r.(2)

(c)Calculate sum_{r=1}^{12} r^2.(2)

(d)Hence calculate sum_{r=1}^{12} (2r² - 3r + 1).(3)

(Total for Question 1 is 10 marks)

Priya is investigating the series with general term u_r = 4r + 3, for r = 1, 2, 3, ...

(a)Show that sum_{r=1}^{n} (4r + 3) = n(2n + 5).(3)

(b)Hence find the value of n for which sum_{r=1}^{n} (4r + 3) = 900.(3)

(c)State why the other root of the quadratic in part (b) is rejected.(1)

(Total for Question 2 is 7 marks)

Consider the series with general term r(r+1), for r = 1, 2, 3, ...

(a)Show that sum_{r=1}^{n} r(r+1) = n(n+1)(n+2)/3.(4)

(b)Hence evaluate sum_{r=1}^{25} r(r+1).(2)

(c)Use your answer to part (a) to find sum_{r=10}^{25} r(r+1).(3)

(Total for Question 3 is 9 marks)

A quick check of understanding of σ notation.

(a)Which of the following is equal to sum_{r=1}^{n} (2r - 1), for all positive integers n?
A) n²
B) n(n+1)
C) n(2n+1)
D) [n(n+1)/2]²(1)

A) n²
B) n(n+1)
C) n(2n+1)
D) [n(n+1)/2]²

(b)Justify your answer to part (a) by expressing sum_{r=1}^{n} (2r - 1) in terms of n, showing your full working.(3)

(Total for Question 4 is 4 marks)

For a positive integer n, consider sum_{r=n+1}^{2n} r², the sum of the squares of the integers from n+1 to 2n.

(a)Show that sum_{r=n+1}^{2n} r² = n(2n+1)(7n+1)/6.(5)

(b)Hence find sum_{r=6}^{10} r² by using the result in part (a) with n = 5.(2)

(Total for Question 5 is 7 marks)

This question uses the method of differences on the series with general term 1/(r(r+1)).

(a)Show that 1/r - 1/(r+1) = 1/(r(r+1)).(2)

(b)Hence, using the method of differences, show that sum_{r=1}^{n} 1/(r(r+1)) = n/(n+1).(3)

(c)Hence evaluate sum_{r=1}^{99} 1/(r(r+1)) as an exact fraction.(2)

(d)State the value that sum_{r=1}^{n} 1/(r(r+1)) approaches as n tends to infinity, giving a reason.(2)

(Total for Question 6 is 9 marks)

This question uses the method of differences on the series with general term 1/((2r-1)(2r+1)).

(a)Express 1/((2r-1)(2r+1)) in the form A/(2r-1) + B/(2r+1), where A and B are constants to be found.(4)

(b)Hence, using the method of differences, show that sum_{r=1}^{n} 1/((2r-1)(2r+1)) = n/(2n+1).(4)

(c)Hence find sum_{r=1}^{50} 1/((2r-1)(2r+1)) as an exact fraction.(2)

(d)Find the smallest positive integer n for which sum_{r=1}^{n} 1/((2r-1)(2r+1)) > 0.499.(3)

(Total for Question 7 is 13 marks)

Prove, by mathematical induction, that for all positive integers n:
sum_{r=1}^{n} r(2r - 1) = n(n+1)(4n-1)/6

(a)Prove the given result by mathematical induction, showing the basis case, the inductive step and a full conclusion.(7)

(Total for Question 8 is 7 marks)

This question derives the standard result for sum_{r=1}^{n} r² from first principles, using the method of differences.

(a)Show that r³ - (r-1)³ = 3r² - 3r + 1.(2)

(b)By summing both sides of the identity in part (a) from r = 1 to n, and using the method of differences on the left-hand side, show that sum_{r=1}^{n} r² = n(n+1)(2n+1)/6. (You may use sum_{r=1}^{n} r = n(n+1)/2 without proof.)(5)

(Total for Question 9 is 7 marks)

Aisha and Daniel are exploring sums of the positive integers and their subsets.

(a)Find the smallest positive integer n such that sum_{r=1}^{n} r > 1000.(3)

(b)Calculate sum_{r=1}^{40} 2r, the sum of the first 40 positive even integers.(2)

(c)Hence, or otherwise, find the sum of the first 40 positive odd integers.(2)

(Total for Question 10 is 7 marks)

Tom and Freya are comparing partial sums of the series with general term r² - 2r.

(a)Show, using standard results, that sum_{r=5}^{15} (r² - 2r) = 990.(4)

(b)A second sum is defined by sum_{r=k}^{15} (r² - 2r) = 916, where k is a positive integer with 1 ≤ k ≤ 15. Find the value of k.(4)

(Total for Question 11 is 8 marks)

This question extends the method of differences to a series with a cubic general term, and then to a quartic general term.

(a)Show that r(r+1)(r+2)(r+3) - (r-1)r(r+1)(r+2) = 4r(r+1)(r+2).(2)

(b)Hence, using the method of differences, show that sum_{r=1}^{n} r(r+1)(r+2) = n(n+1)(n+2)(n+3)/4.(4)

(c)By defining g(r) = r(r+1)(r+2)(r+3)(r+4) and considering g(r) - g(r-1), show that sum_{r=1}^{n} r(r+1)(r+2)(r+3) = n(n+1)(n+2)(n+3)(n+4)/5.(4)

(d)Use the result in part (c) to evaluate sum_{r=1}^{10} r(r+1)(r+2)(r+3).(2)

(Total for Question 12 is 12 marks)

Mark scheme · FP.CP10 Series and Summation

Question 1

(a) B1 sum r = n(n+1)/2 oe
(a) B1 sum r² = n(n+1)(2n+1)/6 oe
(a) B1 sum r³ = [n(n+1)/2]² oe
(a) Answer: sum r = n(n+1)/2; sum r² = n(n+1)(2n+1)/6; sum r³ = [n(n+1)/2]²
(b) M1 substituting n = 20 into n(n+1)/2
(b) A1 210 cao
(b) Answer: 210
(c) M1 substituting n = 12 into n(n+1)(2n+1)/6
(c) A1 650 cao
(c) Answer: 650
(d) M1 splitting into 2*sum r² - 3*sum r + sum 1
(d) M1 substituting sum r² = 650 (or recomputed) and sum r = 78
(d) A1 1078 cao
(d) Answer: 1078

Question 2

(a) M1 splitting sum into 4*sum r + sum 3
(a) M1 substituting sum r = n(n+1)/2 to get 2n(n+1) + 3n
(a) A1 cso: simplifies to n(2n+5)
(a) Answer: n(2n + 5)
(b) M1 forming 2n² + 5n - 900 = 0
(b) M1 solving the quadratic (formula or factorisation) to get n = 20 or n = -22.5
(b) A1 n = 20 cao, rejecting the non-integer negative root
(b) Answer: n = 20
(c) B1 n must be a positive integer (number of terms), so the negative/non-integer root has no meaning here
(c) Answer: n represents a number of terms, so it must be a positive integer; the other root is negative.

Question 3

(a) M1 expanding r(r+1) = r² + r
(a) M1 substituting both standard results: sum r² + sum r = n(n+1)(2n+1)/6 + n(n+1)/2
(a) M1 combining over a common denominator of 6 and factorising out n(n+1)
(a) A1 cso: simplifies to n(n+1)(n+2)/3
(a) Answer: n(n+1)(n+2)/3
(b) M1 substituting n = 25 into the result of part (a)
(b) A1 5850 cao
(b) Answer: 5850
(c) M1 recognising sum_{r=10}^{25} = sum_{r=1}^{25} - sum_{r=1}^{9}
(c) M1 computing sum_{r=1}^{9} r(r+1) = 9*10*11/3 = 330
(c) A1 5520 cao
(c) Answer: 5520

Question 4

(a) B1 A cao
(a) Answer: A) n²
(b) M1 splitting into 2*sum r - sum 1
(b) M1 substituting sum r = n(n+1)/2 to get n(n+1) - n
(b) A1 cso: simplifies to n²
(b) Answer: n²

Question 5

(a) M1 recognising sum_{r=n+1}^{2n} r² = sum_{r=1}^{2n} r² - sum_{r=1}^{n} r²
(a) M1 substituting the standard result with 2n: (2n)(2n+1)(4n+1)/6
(a) M1 substituting the standard result with n: n(n+1)(2n+1)/6
(a) M1 combining over a common denominator and factorising out n(2n+1)
(a) A1 cso: simplifies to n(2n+1)(7n+1)/6
(a) Answer: n(2n+1)(7n+1)/6
(b) M1 substituting n = 5 into n(2n+1)(7n+1)/6
(b) A1 330 cao
(b) Answer: 330

Question 6

(a) M1 combining 1/r - 1/(r+1) over the common denominator r(r+1)
(a) A1 cso: numerator (r+1) - r = 1, giving 1/(r(r+1))
(a) Answer: 1/(r(r+1))
(b) M1 writing out enough leading and trailing terms to show the telescoping pattern, e.g. (1 - 1/2) + (1/2 - 1/3) + ... + (1/n - 1/(n+1))
(b) M1 cancelling all middle terms to leave 1 - 1/(n+1)
(b) A1 cso: simplifies to n/(n+1)
(b) Answer: n/(n+1)
(c) M1 substituting n = 99 into n/(n+1)
(c) A1 99/100 cao
(c) Answer: 99/100
(d) B1 1 cao
(d) B1 correct reason, e.g. n/(n+1) = 1/(1 + 1/n) -> 1 as n -> infinity, since 1/n -> 0
(d) Answer: 1

Question 7

(a) M1 forming 1 = A(2r+1) + B(2r-1)
(a) M1 substituting r = 1/2 and r = -1/2 (or comparing coefficients)
(a) A1 A = 1/2 cao
(a) A1 B = -1/2 cao
(a) Answer: A = 1/2, B = -1/2
(b) M1 writing sum = (1/2) sum [1/(2r-1) - 1/(2r+1)]
(b) M1 listing leading/trailing terms to show telescoping: (1/2)[(1/1 - 1/3) + (1/3 - 1/5) + ... + (1/(2n-1) - 1/(2n+1))]
(b) M1 cancelling middle terms to leave (1/2)[1 - 1/(2n+1)]
(b) A1 cso: simplifies to n/(2n+1)
(b) Answer: n/(2n+1)
(c) M1 substituting n = 50 into n/(2n+1)
(c) A1 50/101 cao
(c) Answer: 50/101
(d) M1 forming the inequality n/(2n+1) > 0.499
(d) M1 solving to obtain n > 249.5 (or equivalent correct rearrangement)
(d) A1 n = 250 cao, with a check that n=249 fails and n=250 succeeds
(d) Answer: n = 250

Question 8

(a) B1 basis: n=1 gives LHS = 1*1 = 1 and RHS = 1*2*3/6 = 1, so true for n=1
(a) M1 assuming true for n=k and adding the (k+1)th term (k+1)(2(k+1)-1) = (k+1)(2k+1) to both sides
(a) M1 combining over a common denominator of 6: (k+1)[k(4k-1) + 6(2k+1)]/6
(a) A1 correctly simplifying the bracket to 4k² + 11k + 6 oe
(a) M1 factorising 4k²+11k+6 = (k+2)(4k+3)
(a) A1 cso: obtaining (k+1)(k+2)(4k+3)/6, which is the given formula with n = k+1
(a) dB1 dependent on all previous marks: correct conclusion stating true for n=1, and if true for n=k then true for n=k+1, so by induction true for all positive integers n
(a) Answer: Proved by induction: true for n=1; if true for n=k then true for n=k+1; therefore true for all positive integers n.

Question 9

(a) M1 expanding (r-1)³ = r³ - 3r² + 3r - 1
(a) A1 cso: r³ - (r-1)³ = 3r² - 3r + 1
(a) Answer: 3r² - 3r + 1
(b) M1 recognising the left-hand side telescopes: sum_{r=1}^{n} [r³-(r-1)³] = n³ - 0³ = n³
(b) M1 writing the right-hand side as 3*sum r² - 3*sum r + n and substituting sum r = n(n+1)/2
(b) M1 rearranging to make sum r² the subject: 3*sum r² = n³ - n + 3n(n+1)/2
(b) M1 combining over a common denominator and factorising: 3*sum r² = n(n+1)(2n+1)/2
(b) A1 cso: dividing by 3 to obtain sum r² = n(n+1)(2n+1)/6
(b) Answer: n(n+1)(2n+1)/6

Question 10

(a) M1 forming the inequality n(n+1)/2 > 1000, i.e. n(n+1) > 2000
(a) M1 trial (or solving the quadratic) to compare n=44 and n=45
(a) A1 n = 45 cao, with 44*45=1980<2000 and 45*46=2070>2000 shown
(a) Answer: n = 45
(b) M1 using sum 2r = 2*sum r with n=40
(b) A1 1640 cao
(b) Answer: 1640
(c) M1 valid method, e.g. sum_{r=1}^{80} r - sum_{r=1}^{40} 2r, or using sum(2r-1)=n² with n=40
(c) A1 1600 cao
(c) Answer: 1600

Question 11

(a) M1 splitting into sum_{r=5}^{15} r² - 2*sum_{r=5}^{15} r
(a) M1 using sum_{r=5}^{15} r² = sum_{r=1}^{15} r² - sum_{r=1}^{4} r² = 1240 - 30 = 1210
(a) M1 using sum_{r=5}^{15} r = sum_{r=1}^{15} r - sum_{r=1}^{4} r = 120 - 10 = 110
(a) A1 cso: 1210 - 2(110) = 990
(a) Answer: 990
(b) M1 expressing sum_{r=k}^{15} (r²-2r) as sum_{r=1}^{15}(r²-2r) - sum_{r=1}^{k-1}(r²-2r)
(b) M1 computing sum_{r=1}^{15} (r²-2r) = 1240 - 2(120) = 1000
(b) M1 forming and solving 1000 - sum_{r=1}^{k-1}(r²-2r) = 916, i.e. sum_{r=1}^{k-1}(r²-2r) = 84, by trial or algebraically
(b) A1 k = 8 cao
(b) Answer: k = 8

Question 12

(a) M1 factorising out the common factor r(r+1)(r+2)
(a) A1 cso: r(r+1)(r+2)[(r+3)-(r-1)] = 4r(r+1)(r+2)
(a) Answer: 4r(r+1)(r+2)
(b) M1 writing sum_{r=1}^{n} 4r(r+1)(r+2) as sum_{r=1}^{n} [f(r)-f(r-1)] where f(r)=r(r+1)(r+2)(r+3)
(b) M1 recognising the sum telescopes to f(n) - f(0)
(b) M1 using f(0) = 0*1*2*3 = 0 so the sum equals n(n+1)(n+2)(n+3)
(b) A1 cso: dividing by 4 to obtain sum r(r+1)(r+2) = n(n+1)(n+2)(n+3)/4
(b) Answer: n(n+1)(n+2)(n+3)/4
(c) M1 showing g(r) - g(r-1) = r(r+1)(r+2)(r+3)[(r+4)-(r-1)] = 5r(r+1)(r+2)(r+3)
(c) M1 summing both sides from r=1 to n so the left-hand side telescopes to g(n) - g(0)
(c) A1 using g(0) = 0*1*2*3*4 = 0, so sum 5r(r+1)(r+2)(r+3) = n(n+1)(n+2)(n+3)(n+4)
(c) A1 cso: dividing by 5 to obtain sum r(r+1)(r+2)(r+3) = n(n+1)(n+2)(n+3)(n+4)/5
(c) Answer: n(n+1)(n+2)(n+3)(n+4)/5
(d) M1 substituting n = 10 into n(n+1)(n+2)(n+3)(n+4)/5
(d) A1 48048 cao
(d) Answer: 48048

Series and Summation - Worksheets, Questions and Revision