P4.2 Sequences

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Generally, candidates found this question accessible, with many scoring full marks or nearly full marks. In part (a) most candidates used the given recurrence relation along with u1 = 6 to find u2 = 6k – 5 and then u3 = 6k2 – 5k – 5, and then set u3 = – 1 and rearranged to obtain the given answer 6k2 – 5k – 4 = 0. Algebraic errors were few, but some candidates forgot to include the ‘= 0’ in their final line and lost the second mark in this part. There were occasional attempts at using arithmetic or geometric series which gained no credit.

Common mistakes

• u₂ = 6k − 5, u₃ = k(6k − 5) − 5 = −1
• must include "= 0" in the given answer
• AP/GP formulae score 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Generally, candidates found this question accessible, with many scoring full marks or nearly full marks. In part (b)(i) most candidates were able to solve the quadratic equation, but not all identified that k = 4 3 was the only correct answer, or subsequently used this value, so lost the mark in this part of the question. Some candidates spent more time than they needed using the quadratic formula, where a factorisation of the quadratic would have been more direct. Use of the calculator equation solver was also permitted here. Candidates must learn to check their answers against any limitations given in the question. When errors did occur, they were most likely to appear in part (b)(ii), although most still answered it well. Most candidates identified the requirement of the question to add the first three terms of the sequence together, and correctly used k = 4 3 to calculate u2 =6 ( 4 3 ) -5 before finding 6 + 3 + (-1) = 8. Others first simplified the sum to obtain 6k or 6k2 + k - 4 before substituting the value of k. A common incorrect approach was to use u3 = 6k2 – 5k – 4 in the sum. Some also used k = 4 3 and k = 1 2 - to calculate the sum, without identifying which was the required answer. Alternatively, a good number of candidates attempted, correctly and incorrectly, to work out the values of u1 and u3 even though they were given in the question. It was also noticeable that a number of candidates used the quadratic expression from part (a) in order to work out values in (b)(ii), usually assuming u3 = k2 – 5k – 4. They, therefore, did not clearly understand what this equation represented. A few candidates incorrectly attempted to use the recurrence relation to find things like ( 4 3 )(1) – 5, ( 4 3 )(2) – 5 and ( 4 3 )(3) – 5, using these expressions to find their sum, and there were again occasional attempts using arithmetic or geometric series which also gained no credit. There were also a few attempts at integration. Overall, this was a question that allowed candidates to gain a good number of marks with confidence.

Common mistakes

• k = 4/3 only (k positive, reject −½)
• sum = 6 + 3 + (−1) = 8
• do not use the quadratic 6k² − 5k − 4 as u₃
• AP/GP formulae score 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

For the most part candidates answered this question well. This was dependent on whether they correctly understood the iteration formula. In part (a) (i), although many scored this first mark, there was a common misunderstanding of how a recurrence relationship works with confusion as to the meaning of n. Rather than term position, it was often confused with 𝑢𝑛 so that the incorrect calculation ( ) 35 2 35 35 7cos 5 1 2 u = + - - was often seen, giving the correct answer but obviously not scoring the B mark. Also not uncommonly seen was n replaced by 1 n u + , with candidates stating ( ) 40 2 40 40 7cos 5 1 2 u = + - - . The errors stated above also affected the performance of some candidates in part (ii). Furthermore, a minority of candidates had difficulty with accuracy in the sign of the final term, struggling with the odd/even power of -1 . For example, when calculating 3 u , ( ) 2 3 2 40 7cos 5 1 40 7 5 38 2 u = + - - = - + = , was not uncommon. Occasionally, candidates were seen to be working with degrees rather than radians, also leading to loss of the accuracy mark. In part (ii) many candidates recognised how to construct the arithmetic calculation to find the required sum. The most common of these attempts were: ( ) 1 2 3 4 1 6 u u u u u + + + + and ( ) 2 3 4 1 6 7 u u u u + + + Credit was given to candidates who used such ideas with incorrect values from earlier in the question for 3 u and 4 u , although those who had a wrong answer in (a)(i) who used their incorrect value for 2 u (rather than the given 2 40 u = ), lost both marks. It is important that centres make candidates aware that they must use the ‘show that’ answer in subsequent calculations no matter what they achieve in their working. Occasionally, candidates used ( ) 25 1 2 3 4 1 6.25 r r u u u u u = = + + + but these usually did not add 1 to give a fully correct method. Some also incorrectly tried summing the series as though the terms formed an arithmetic progression. Some candidates incorrectly multiplied (u1 + u2 + u3 + u4) by 4 instead of by 6. A minority forgot to add 35 to their total. A very small minority of candidates generated all the terms of the whole sequence and added them together.

Common mistakes

• n is the term position: u₂ = 35 + 7cos(π/2) − 5(−1)¹ = 40 (using n = 35 scores 0 even though it gives 40)
• u₃ = 28, u₄ = 33
• work in radians
• sign of (−1)ⁿ

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

For the most part candidates answered this question well. This was dependent on whether they correctly understood the iteration formula. In part (b) (i) most candidates understood the order 4, periodic nature of the recurrence relationship and that 5 1 4 3, k u u u k Z + - = = . Occasionally, candidates unnecessarily calculated 5 u , using 4 u in the recurrence relationship, which sometimes led to the mark being lost either if 4 u had an incorrect value or if an error was made in the calculation.

Common mistakes

• order 4 so u₅ = u₁ = 35
• Σ to 25 = 6 × (35 + 40 + 28 + 33) + 35 = 851
• AP/GP formulas score 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question proved to be successful for many candidates and many scored full marks here. In part (a)(i), the vast majority were able to successfully write down the sequence of terms and hence demonstrate that the sequence was periodic. An algebraic proof was not required here, and one was only rarely seen. Perhaps surprisingly having identified the pattern in the sequence in (a)(i), not all were able to identify the order of the sequence as ‘2’ in (a)(ii). This perhaps highlights a lack here were: ‘first order’; ‘order 3, 5’; ‘±2’or just a re-writing of their sequence from part of awareness of what is meant by “order” of a sequence. Common incorrect answers (a)(i). In part b), candidates were most successful when they considered the nature of the sequence rather than trying to resort to a formula. A fairly common first step was to divide 85 by 2 to give 42 with remainder 1. Many students then correctly broke down 42 x 3 + 43 x 5 or equivalently (42 x 8) + 3 . Errors occurred when candidates the sum into either 42.5 x 5 + 42.5 x 3 resulting in an answer of 340, or when candidates mistakenly struggled with the odd number of terms and believed they should calculate errors included summing only 84 terms (42 x 8 = 336) or attempts to ‘chunk’ thought that the 85th term was ‘5’ rather than ‘3’ resulting in an answer of 341. Other incorrectly into blocks of e.g. size 5: in such attempts the sum of the first five terms was (3 + 5 + 3 + 5 + 3 = 19) followed by deduction that 5 = 17 leading to an incorrect 85 obtained answer of 19 x 17 = 323. A sizeable number of incorrect attempts involving use of the sum of arithmetic or geometric series were also commonly seen. Perhaps most surprising was the significant minority who were able to set up a correct approach and write down a correct method but who then made arithmetical errors to write down an incorrect total which cost them the final A mark.

Common mistakes

• a₂ = 5, a₃ = 3 is enough
• order is 2 (not ±2, first order or 3, 5)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question proved to be successful for many candidates and many scored full marks here.

Common mistakes

• 42 × (3 + 5) + 3 = 339
• 85 terms is odd and the 85th term is 3 (not 5, giving 341)
• 42.5 × 8 = 340 is wrong
• AP/GP formulae score 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This proved to be a little more demanding than the first two questions but over 40% of candidates still scored all 6 marks. Errors seen by candidates are summarised below. In part (a) several candidates did not simplify 𝑢𝑢2 sufficiently, giving 𝑢𝑢2 =𝑘𝑘-24/2 which then lead to a more complicated expression for 𝑢𝑢3 which they had difficulty simplifying. Some candidates did not use the given expression, but rather used 𝑢𝑢1+𝑢𝑢2+𝑢𝑢3 = 0.

Common mistakes

• u₂ = k − 12 (simplify 24/2), u₃ = k − 24/(k − 12)
• use the given u₁ + 2u₂ + u₃ = 0
• show how the fractions are removed before the given answer

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This proved to be a little more demanding than the first two questions but over 40% of candidates still scored all 6 marks. Errors seen by candidates are summarised below. In part (b) candidates usually found both roots of the given quadratic either by factorisation or by use of the quadratic formula, with a small number by completing the square. A significant minority failed to gain the A mark here as they did not state that 6 was an integer or that 40/3 was not an integer, with some excluding 40/3 on the grounds that it was irrational.

Common mistakes

• roots 6 and 40/3
• k = 6 because k must be an integer (40/3 is not an integer
• it is not irrational)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This proved to be a little more demanding than the first two questions but over 40% of candidates still scored all 6 marks. Errors seen by candidates are summarised below. Part (c) was invariably answered correctly by those who had solved the quadratic in part (b). although there were some candidates who gave a choice of answers.

Common mistakes

• u₃ = 10 from k = 6
• give one answer only

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question on sequences proved very discriminating, with weaker candidates scoring only the first mark. As with many questions on this paper however, there were some excellent and well written responses. In part (a) applying the iterative formula to find a2 was achieved by almost every candidate. A significant number then attempted to find a3 and also a4 in many cases but failed to simplify expressions for 2 4 2 k a = and 3 (2 2) 2 k k a k + = . Setting 4 1 a a = was often seen but fully correct proofs were rare.

Common mistakes

• a₂ = 2k, a₃ = k + 1, a₄ = k(k + 3)/(k + 1)
• periodic of order 3 so set a₄ = a₁ = 2
• simplify the terms

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question on sequences proved very discriminating, with weaker candidates scoring only the first mark. As with many questions on this paper however, there were some excellent and well written responses. In part (b) only a very small minority of candidates gave a correct and sufficient reason why k cannot be equal to 1. Although a larger minority did state the sequence would be 2,2,2, … only a small number of these pointed out that this is not a periodic sequence with order 3. Part (c) could be done independently to (a) and (b), and many candidates took advantage of this. It was not uncommon to see the terms 2, 4, 1 - - in (c) where only the first mark in (a) had been scored. Methods of finding the sum to 80 terms generally revolved around the calculation ( ) ( ) ( ) 26 2 4 1) 2 4 x + -+ - + + - . Incorrect attempts involved uses of both arithmetic or geometric sum formula or ones involving an average of the terms.

Common mistakes

• k = 1 gives 2, 2, 2, ... (all terms equal), which has order 1, not order 3
• must refer to the order

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question on sequences proved very discriminating, with weaker candidates scoring only the first mark. As with many questions on this paper however, there were some excellent and well written responses. Part (c) could be done independently to (a) and (b), and many candidates took advantage of this. It was not uncommon to see the terms 2, 4, 1 - - in (c) where only the first mark in (a) had been scored. Methods of finding the sum to 80 terms generally revolved around the calculation ( ) ( ) ( ) 26 2 4 1) 2 4 x + -+ - + + - . Incorrect attempts involved uses of both arithmetic or geometric sum formula or ones involving an average of the terms.

Common mistakes

• k = −2 (k ≠ 1) gives repeating terms 2, −4, −1
• 26 full cycles plus 2 + (−4) = −80
• arithmetic or geometric sum formulas do not apply

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In general, Q11 discriminated well between candidates of all abilities. Q11(a) was demanding with most candidates scoring no more than one mark. Q11(b), Q11(c) and Q11(d) proved accessible to candidates of all abilities. In Q11(a), most candidates took logs, as advised by the question, and many obtained either ln ln y x x or log log . y x x At this point a significant number of candidates made no further creditable progress. Some candidates used a complete method of applying the product rule on x and implicit differentiation to obtain a correct ln x Most of these candidates then applied d y 1 d y x 1 y x e . x 0 to obtain Some 1 ln . x d d x log ’ method to obtain candidates, who received partial credit, used a ‘ after 0.1, 10 1 d y x finding A few candidates used logarithms without reference to a 1 log . x 10 d y x base. Those who proceeded to obtain received partial credit and those who 0.1 1 e x obtained were allowed full marks. There were some alternative methods employed by a few candidates in Q11(a). A few d y y x ln e x x x ln x x y x candidates rewrote as and applied to give a (1 ln )e 0 x d 1 e . x Others differentiated both sides of the equation ln ln y x x correct with 1 e . x respect to y and usually proceeded to find A very few candidates applied the ln y , x quotient rule to the equation but were rarely successful with this method. ln x xx x x xx or 2 1.6. In Q11(b), most candidates evaluated either at both and 1.5 Many candidates compared their values with either 2 or 0 (depending on whether they xx xx or 2 ). Although many of these concluded that 1.5 1.6, evaluated only a few made any reference to the curve being continuous in this interval. A few candidates, x who made no creditable progress, used values of x outside the given range, e.g. 1.4 x x 1.6. and Very few candidates used values within the range e.g. and 1.51 x 1.59, which meant that they were able to earn both marks. x 1.673, 4x In Q11(c), most candidates found a correct although a few did not state 4 , x x and 3x as part of their correct to 3 decimal places. Some wrote down the iterates 1 2 4x with no intermediate work. solution, while others stated In Q11(d), some candidates gave correct descriptions such as ‘sequence oscillates between 1 and 2’ or ‘the sequence is periodic with period 2’. Other descriptions such as ‘alternates between 1 and 2’, ‘keep getting 1 and 2’ or ‘fluctuates between 1 and 2’ were condoned. Some credit was given for partial explanations such as ‘sequence is non-convergent’, ‘sequence oscillates’ or ‘sequence is divergent’.

Common mistakes

• x₂ = 1.633, x₃ = 1.466, x₄ = 1.673 (3 d.p.)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In general, Q11 discriminated well between candidates of all abilities. Q11(a) was demanding with most candidates scoring no more than one mark. Q11(b), Q11(c) and Q11(d) proved accessible to candidates of all abilities. In Q11(a), most candidates took logs, as advised by the question, and many obtained either ln ln y x x or log log . y x x At this point a significant number of candidates made no further creditable progress. Some candidates used a complete method of applying the product rule on x and implicit differentiation to obtain a correct ln x Most of these candidates then applied d y 1 d y x 1 y x e . x 0 to obtain Some 1 ln . x d d x log ’ method to obtain candidates, who received partial credit, used a ‘ after 0.1, 10 1 d y x finding A few candidates used logarithms without reference to a 1 log . x 10 d y x base. Those who proceeded to obtain received partial credit and those who 0.1 1 e x obtained were allowed full marks. There were some alternative methods employed by a few candidates in Q11(a). A few d y y x ln e x x x ln x x y x candidates rewrote as and applied to give a (1 ln )e 0 x d 1 e . x Others differentiated both sides of the equation ln ln y x x correct with 1 e . x respect to y and usually proceeded to find A very few candidates applied the ln y , x quotient rule to the equation but were rarely successful with this method. ln x xx x x xx or 2 1.6. In Q11(b), most candidates evaluated either at both and 1.5 Many candidates compared their values with either 2 or 0 (depending on whether they xx xx or 2 ). Although many of these concluded that 1.5 1.6, evaluated only a few made any reference to the curve being continuous in this interval. A few candidates, x who made no creditable progress, used values of x outside the given range, e.g. 1.4 x x 1.6. and Very few candidates used values within the range e.g. and 1.51 x 1.59, which meant that they were able to earn both marks. x 1.673, 4x In Q11(c), most candidates found a correct although a few did not state 4 , x x and 3x as part of their correct to 3 decimal places. Some wrote down the iterates 1 2 4x with no intermediate work. solution, while others stated In Q11(d), some candidates gave correct descriptions such as ‘sequence oscillates between 1 and 2’ or ‘the sequence is periodic with period 2’. Other descriptions such as ‘alternates between 1 and 2’, ‘keep getting 1 and 2’ or ‘fluctuates between 1 and 2’ were condoned. Some credit was given for partial explanations such as ‘sequence is non-convergent’, ‘sequence oscillates’ or ‘sequence is divergent’.

Common mistakes

• sequence is periodic with period 2 / oscillates between 1 and 2 (both marks)
• "non-convergent", "oscillates" or "divergent" alone gets one mark

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was an accessible question with part (ii) more successfully answered than part (i). 16 r (3 5 2 ) r In part (i), some students attempted to find by applying the strategy 1 r 16 16 16 16 r r (3 16) 5 2 , r (3 5 ) 2 . r while other students applied the strategy Most 1 1 r r 1 1 r r 5, 5 a d students applied the sum formula for an arithmetic series with to 16 16 8, 5 a d (3 16) 5 r (3 16) (2(5) 15(5)) to give while some applied to 2 1 r 16 n 1 ( to give 16 (2(8) 15(5)). (3 5 ) r 1) r n n A few students used to help them to 2 2 1 r 1 r 16 5 . r evaluate Students were generally less successful with using the formula for the sum 1 r 16 r r 2 . of a geometric series to find Common errors included assuming the first term, a, was 1 1 or writing down the value 131070 with no method (perhaps by working back from the given answer). A significant number of students listed all 16 terms and added them to achieve the given result. Those students who did not list all the terms, but just wrote e.g. 10 17 26 39 60 ... 65619 131798, did not receive any credit. Some students, who attempted to find the first, second and third differences between the terms of the series 10, 17, 26, 39, 60,..., usually made no creditable progress. Part (ii) was very well attempted, even by those who failed to gain any credit in part (i). Almost all students wrote down the next two terms u correctly. Many deduced that they , u 2 3 100 were working with an alternating sequence and found the correct exact value for . u r 1 r Some students attempted to round their answer, usually to 108.3. A few students attempted 100 to find by using the formula for the sum of an arithmetic series or the formula for the u r 1 r sum of a geometric series.

Common mistakes

• sequence is periodic (2/3, 3/2, 2/3, ...) of order 2
• sum = 50(2/3 + 3/2) = 325/3 exact