Mathematical Induction with MS

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1 Mathematical Induction with MS 1a. [4 marks] Using the definition of a derivative as, show that the derivative of. 1b. [9 marks] Prove by induction that the derivative of is. 2a. [3 marks] Consider a function f, defined by. Find an expression for. 2b. [8 marks] Let, where. Use mathematical induction to show that for any. 2c. [6 marks] Show that is an expression for the inverse of. 2d. [6 marks] (i) State. (ii) Show that, given 0 < x < 1,. (iii) For, let be the area of the region enclosed by the graph of, the x-axis and the line x = 1. Find the area of the region enclosed by and in terms of. 3a. [3 marks] Find the sum of the infinite geometric sequence 27, 9, 3, 1,.... 3b. [7 marks] Use mathematical induction to prove that for, 1

2 4a. [8 marks] Prove by mathematical induction that, for, 4b. [17 marks] (a) Using integration by parts, show that. (b) Solve the differential equation, given that y = 0 when x = 0, writing your answer in the form. (c) (i) Sketch the graph of, found in part (b), for. Determine the coordinates of the point P, the first positive intercept on the x-axis, and mark it on your sketch. (ii) The region bounded by the graph of and the x-axis, between the origin and P, is rotated 360 about the x-axis to form a solid of revolution. Calculate the volume of this solid. 5. [8 marks] Prove by mathematical induction,. 6a. [14 marks] (a) The sum of the first six terms of an arithmetic series is 81. The sum of its first eleven terms is 231. Find the first term and the common difference. (b) The sum of the first two terms of a geometric series is 1 and the sum of its first four terms is 5. If all of its terms are positive, find the first term and the common ratio. (c) The term of a new series is defined as the product of the term of the arithmetic series and the term of the geometric series above. Show that the term of this new series is. 6b. [7 marks] Using mathematical induction, prove that 2

3 7a. [3 marks] The function f is defined by. Show that. 7b. [4 marks] Obtain a similar expression for. 7c. [8 marks] Suggest an expression for,, and prove your conjecture using mathematical induction. 8. [20 marks] (a) Show that. (b) Hence prove, by induction, that for all. (c) Solve the equation. 9. [10 marks] (a) Consider the following sequence of equations. (i) Formulate a conjecture for the equation in the sequence. (ii) Verify your conjecture for n = 4. (b) A sequence of numbers has the term given by. Bill conjectures that all members of the sequence are prime numbers. Show that Bill s conjecture is false. (c) Use mathematical induction to prove that is divisible by 6 for all. 10. [7 marks] Use the method of mathematical induction to prove that. 3 is divisible by 576 for

4 11. [7 marks] Given that, use mathematical induction to prove that. 12. [8 marks] Prove, by mathematical induction, that, is divisible by 5. 13a. [7 marks] Consider. Use mathematical induction to prove that. 13b. [4 marks] Given and, (i) express and in modulus-argument form; (ii) hence find. 13c. [1 mark] The complex numbers and diagram. are represented by point A and point B respectively on an Argand Plot point A and point B on the Argand diagram. 13d. [3 marks] Point A is rotated through in the anticlockwise direction about the origin O to become point. Point B is rotated through in the clockwise direction about O to become point. Find the area of triangle O. 13e. [5 marks] Given that and are roots of the equation, where, find the values of and. 14. [7 marks] Prove by mathematical induction that is divisible by 3 for all. 4

5 Mathematical Induction MS 1a. [4 marks] let and using the result AG [4 marks] Even though the definition of the derivative was given in the question, solutions to (a) were often disappointing with algebraic errors fairly common, usually due to brackets being omitted or manipulated incorrectly. Solutions to the proof by induction in (b) were often poor. Many candidates fail to understand that they have to assume that the result is true for and then show that this leads to it being true for. Many candidates just write Let which is of course meaningless. The conclusion is often of the form True for therefore true by induction. Credit is only given for a conclusion which includes a statement such as True for true for. 1b. [9 marks] let we want to prove that let which is the same result as part (a) hence the result is true for R1 5

6 assume the result is true for () () hence if the result is true for since the result is true for, it is true for, the result is proved by mathematical induction R1 Note: Only award final R1 if all the M marks have been gained. [9 marks] Even though the definition of the derivative was given in the question, solutions to (a) were often disappointing with algebraic errors fairly common, usually due to brackets being omitted or manipulated incorrectly. Solutions to the proof by induction in (b) were often poor. Many candidates fail to understand that they have to assume that the result is true for and then show that this leads to it being true for. Many candidates just write Let which is of course meaningless. The conclusion is often of the form True for therefore true by induction. Credit is only given for a conclusion which includes a statement such as True for true for. 2a. [3 marks] [3 marks] Part a) proved to be an easy 3 marks for most candidates. 2b. [8 marks] 6

7 assume that P(k) is true, i.e., consider EITHER () OR () THEN P(k) true implies P(k + 1) true, P(1) true so P(n) true for all R1 [8 marks] Part b) was often answered well, and candidates were well prepared in this session for this type of question. Candidates still need to take care when showing explicitly that P(1) is true, and some are 7

8 still writing Let n = k which gains no marks. The inductive step was often well argued, and given in clear detail, though the final inductive reasoning step was incorrect, or appeared rushed, even from the better candidates. True for n =1, n = k and n = k + 1 is still disappointingly seen, as were some even more unconvincing variations. 2c. [6 marks] METHOD 1 METHOD 2 AG attempt Note: Award marks for numerators and denominators. METHOD 3 AG attempt Note: Award marks for numerators and denominators. AG 8

9 [6 marks] Part c) was again very well answered by the majority. A few weaker candidates attempted to find an inverse for the individual case n = 1, but gained no credit for this. 2d. [6 marks] (i) (ii) METHOD 1 () so R1 AG METHOD 2 () true in the interval R1 (iii) () [6 marks] Part d) was not at all well understood, with virtually no candidates able to tie together the hints given by connecting the different parts of the question. Rash, and often thoughtless attempts were made at each part, though by this stage some seemed to be struggling through lack of time. The inequality part of the question tended to be fudged, with arguments seen by examiners being largely unconvincing and lacking clarity. A tiny number of candidates provided the correct answer to the final part, though a surprising number persisted with what should have been recognised as fruitless working usually in the form of long-winded integration attempts. 3a. [3 marks] () 9

10 N1 [3 marks] Part (a) was correctly answered by the majority of candidates, although a few found r = 3. 3b. [7 marks] Attempting to show that the result is true for n = 1 LHS = a and Hence the result is true for n = 1 Assume it is true for n = k Consider n = k + 1: Note: Award for an equivalent correct intermediate step. Note: Illogical attempted proofs that use the result to be proved would gain A0A0 for the last three above marks. The result is true for it is true for and as it is true for, the result is proved by mathematical induction. R1 N0 Note: To obtain the final R1 mark a reasonable attempt must have been made to prove the k + 1 step. [7 marks] 10

11 Part (b) was often started off well, but a number of candidates failed to initiate the n = k + 1 step in a satisfactory way. A number of candidates omitted the P(1) is true part of the concluding statement. 4a. [8 marks] prove that for n = 1 so true for n = 1 R1 assume true for n = k so now for n = k +1 LHS: (or equivalent) (accept ) Therefore if it is true for n = k it is true for n = k + 1. It has been shown to be true for n = 1 so it is true for all. R1 Note: To obtain the final R mark, a reasonable attempt at induction must have been made. [8 marks] Part A: Given that this question is at the easier end of the proof by induction spectrum, it was disappointing that so many candidates failed to score full marks. The n = 1 case was generally well done. The whole point of the method is that it involves logic, so let n = k or put n = k, instead of assume... to be true for n = k, gains no marks. The algebraic steps need to be more convincing than some candidates were able to show. It is astonishing that the R1 mark for the final statement was so often not awarded. 4b. [17 marks] 11

12 (a) METHOD 1 AG METHOD 2 AG [6 marks] (b) when [5 marks] (c) 12

13 (i) P is (1.16, 0) Note: Award for 1.16 seen anywhere, for complete sketch. Note: Allow FT on their answer from (b) (ii) A2 Note: Allow FT on their answers from (b) and (c)(i). [6 marks] Part B: Part (a) was often well done, although some faltered after the first integration. Part (b) was also generally well done, although there were some errors with the constant of integration. In (c) the graph was often attempted, but errors in (b) usually led to manifestly incorrect plots. Many attempted the volume of integration and some obtained the correct value. 5. [8 marks] 13

14 let LHS RHS hence true for R1 assume true for hence if true for, true for R1 since the result is true for and the result is proved by mathematical induction R1 [8 marks] This question was done poorly on a number of levels. Many students knew the structure of induction but did not show that they understood what they were doing. The general notation was poor for both the induction itself and the sigma notation. In noting the case for too many stated the equation rather than using the LHS and RHS separately and concluding with a statement. There were also too many who did not state the conclusion for this case. Many did not state the assumption for as an assumption. Most stated the equation for and worked with the equation. Also common was the lack of sigma and inappropriate use of n and k in the statement. There were some very nice solutions however. The final conclusion was often not complete or not considered which would lead to the conclusion that the student did not really understand what induction is about. 6a. [14 marks] 14

15 (a) solving simultaneously, a = 6, d = 3 [6 marks] (b) obtaining (since all terms are positive) [5 marks] (c) AG [3 marks] Total [14 marks] Parts (a), (b) and (c) were answered successfully by a large number of candidates. Some, however, had difficulty with the arithmetic. 6b. [7 marks] prove: show true for n = 1, i.e. 15

16 assume true for n = k, i.e. consider n = k +1 hence true for n = k + 1 is true whenever is true, and is true, therefore is true R1 for [7 marks] In part (d) many candidates showed little understanding of sigma notation and proof by induction. There were cases of circular reasoning and using n, k and r randomly. A concluding sentence almost always appeared, even if the proof was done incorrectly, or not done at all. 7a. [3 marks] AG [3 marks] [N/A] 7b. [4 marks] 16

17 [4 marks] [N/A] 7c. [8 marks] the conjecture is that for n = 1, this formula gives which is correct let the result be true for n = k, consider therefore true for true for and since true for the result is proved by induction. R1 Note: Award the final R1 only if the two M marks have been awarded. [8 marks] [N/A] 8. [20 marks] 17

18 (a) AG [2 marks] (b) if n = 1 so LHS = RHS and the statement is true for n = 1 R1 assume true for n = k Note: Only award if the word true appears. Do not award for let n = k only. Subsequent marks are independent of this. so if n = k + 1 then so if true for n = k, then also true for n = k + 1 as true for n = 1 then true for all R1 Note: Final R1 is independent of previous work. [12 marks] (c) 18

19 but this is impossible for not including any answers outside the domain R1 Note: Award the first for correctly obtaining or equivalent and subsequent marks as appropriate including the answers. [6 marks] Total [20 marks] This question showed the weaknesses of many candidates in dealing with formal proofs and showing their reasoning in a logical manner. In part (a) just a few candidates clearly showed the result and part (b) showed that most candidates struggle with the formality of a proof by induction. The logic of many solutions was poor, though sometimes contained correct trigonometric work. Very few candidates were successful in answering part (c) using the unit circle. Most candidates attempted to manipulate the equation to obtain a cubic equation but made little progress. A few candidates guessed as a solution but were not able to determine the other solutions. 9. [10 marks] (a) (i) R1 (ii) LHS = 40; RHS = 40 [2 marks] (b) the sequence of values are: 5, 7, 11, 19, 35 or an example 35 is not prime, so Bill s conjecture is false R1AG [2 marks] (c) is divisible by 6 is divisible by true assume is true 19

20 Note: Do not award for statement starting let n = k. Subsequent marks are independent of this. consider () is true P(1) true and true true, so by MI is true for all R1 Note: Only award R1 if there is consideration of P(1), and in the final statement. Only award R1 if at least one of the two preceding A marks has been awarded. [6 marks] Total [10 marks] Although there were a good number of wholly correct solutions to this question, it was clear that a number of students had not been prepared for questions on conjectures. The proof by induction was relatively well done, but candidates often showed a lack of rigour in the proof. It was fairly common to see students who did not appreciate the idea that is assumed not given and this was penalised. Also it appeared that a number of students had been taught to write down the final reasoning for a proof by induction, even if no attempt of a proof had taken place. In these cases, the final reasoning mark was not awarded. 10. [7 marks] is divisible by 576 for for Zero is divisible by 576, (as every non-zero number divides zero), and so P(1) is true. R1 Note: Award R0 for P(1) = 0 shown and zero is divisible by 576 not specified. Note: Ignore P(2) = 576 if P(1) = 0 is shown and zero is divisible by 576 is specified. Assume is true for some. Note: Do not award for statements such as let n = k. consider 20

21 EITHER which is a multiple of 576 OR (or equivalent) which is a multiple of 576 THEN P(1) is true and true true, so is true for all R1 Note: Award R1 only if at least four prior marks have been awarded. [7 marks] This proof by mathematical induction challenged most candidates. While most candidates were able to show that P(1) = 0, a significant number did not state that zero is divisible by 576. A few candidates started their proof by looking at P(2). It was pleasing to see that the inductive step was reasonably well done by most candidates. However many candidates committed simple algebraic errors. The most common error was to state that. The concluding statement often omitted the required implication statement and also often omitted that P(1) was found to be true. 11. [7 marks] proposition is true for n = 1 since Note: Must see the 1! for the. assume true for n = k,, i.e. consider () 21

22 hence, is true whenever is true, and is true, and therefore the proposition is true for all positive integers R1 Note: The final R1 is only available if at least 4 of the previous marks have been awarded. [7 marks] Most candidates were awarded good marks for this question. A disappointing minority thought that the th derivative was the th derivative multiplied by the first derivative. Providing an acceptable final statement remains a perennial issue. 12. [8 marks] if which is divisible by 5, hence true for Note: Award A0 for using but do not penalize further in question. assume true for Note: Only award the if truth is assumed. so if hence if true for, then also true for. Since true for, then true for all R1 Note: Only award the R1 if the first two s have been awarded. [8 marks] [N/A] 22

23 13a. [7 marks] let be the proposition let is true R1 assume true for Note: Only award the if truth is assumed. now show true implies also true is true true implies true and is true, therefore by mathematical induction statement is true for R1 Note: Only award the final R1 if the first 4 marks have been awarded. [7 marks] [N/A] 13b. [4 marks] (i) Notes: Accept 3 sf answers only. Accept equivalent forms. Accept and. (ii) 23

24 () Notes: Award () for an attempt to find and. Accept equivalent forms. [4 marks] [N/A] 13c. [1 mark] Note: Award if A or and B or are in their correct quadrants, are aligned vertically and it is clear that. [1 mark] [N/A] 13d. [3 marks] Area 24

25 Notes: Award A0A0 for using. [3 marks] [N/A] 13e. [5 marks] Note: Award for recognition that a complex conjugate is also a root. Note: Award for an attempt to expand two quadratics. [5 marks] [N/A] 14. [7 marks] or a multiple of 3 assume the proposition is true for Note: Do not award for statements with Let. consider Note: Accept or statement that is a multiple of 3. true for, and 25

26 hence true for all R1 Note: Only award the final R1 if at least 4 of the previous marks have been achieved. [7 marks] It was pleasing to see a great many clear and comprehensive answers for this relatively straightforward induction question. The inductive step only seemed to pose problems for the very weakest candidates. As in previous sessions, marks were mainly lost by candidates writing variations on Let, rather than Assume true for. The final reasoning step still needs attention, with variations on evident, suggesting that mathematical induction as a technique is not clearly understood. 26

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