P4.3 Sigma notation

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 (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.

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 question was fairly well answered by a number of candidates, who understood the need to use the sum to infinity formula, but a sizeable majority made little or no progress, not realising for example, that the sum started from n = 2 and not n =1 – hence they often did not get the first mark for a correct first term and/or common ratio. 9/16 was seen fairly often, but not always attributed to the first term, with some candidates thinking this was the value of the sum. One candidate misread the infinity symbol as an 8 and applied the sum to 8 terms correctly, but was obviously not able to score any marks. A number of students found the sum to infinity, using n = 1 to infinity and then subtracting the first term. This was done well by candidates who went down this route. Unfortunately, many students achieved no marks on this part due to a wide variety of errors at the beginning. Calculating cos (360) on the calculator in radians instead of degrees was seen. Some tried to split the summation incorrectly into ∑ ቀ 3 4ቁ 𝑛𝑛 ∞ 2 x ∑ cos 180𝑛𝑛 ∞ 1 and find the sum to infinity of the trigonometric terms separately. A few wrote ( 3 4)2cos (360) but did not evaluate the cos(360). There were some very good solutions but in many responses there was often a lack of clear structure to their proof.

Common mistakes

• cos(180n)° = (−1)ⁿ so r = −3/4
• first term is (3/4)² = 9/16 because the sum starts at n = 2
• calculator in radians gives wrong values
• do not split into two separate sums

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

Overall very good attempts to this question were spoiled by a lack in accuracy . 5th kilometre 6 1.05 1 6th kilometre 6 1.05 2 7th kilometre 6 1.05 3 Although most candidates knew that for part (c) they needed to sum a sequence of terms, errors were frequent and many. A sizeable minority lacked any appreciation of the model and resorted to the calculation 20 6 1.05 1 1.05 1 . The majority of errors however were mainly due to a failure to link up the correct values for a and n in the formula for a geometric series. The most common of these was to use the strategy 24 + 16 6 1.05 1 1.05 1 , forgetting that the geometric sequence in this case should start at 6.3 instead of 6. It is often a good idea in such a question to write out the first few terms to gain an appreciation of the model and how the sequence is formed. (See below) Km 1st 2nd 3rd 4th 5th 6th 7th 20th Time 6 6 6 6 6 1.05 2 6 1.05 3 6 1.05 16 6 1.05 From here it should be a fairly straightforward task to write down 24 + 16 6.3 1.05 1 1.05 1

Common mistakes

• 24 + 6.3(1.05¹⁶ − 1)/0.05
• first term 6.3 not 6 and n = 16
• wrong 6(1.05²⁰ − 1)/0.05 ignores the model
• 173 min 3 s

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

Examiner's comment

Q08 was found challenging by many candidates, with Q08(i) more successfully answered than Q08(ii). The most successful candidates were the ones who listed the first few terms of the series. In other words, those candidates that wrote 4 5 6 3 4 5 1 1 1 log log log ... and gained a 20 20 20 ... 5 5 5 2 3 4 2 2 2 better understanding of the series that they were required to sum. There were, however, many attempts that made no creditable progress and a significant number of blank responses to either one or both parts of Q08. In Q08(i), many candidates identified that a sum to infinity was required in their 1 a r . solution and attempted to apply the formula Most used in this formula, 2 1 r 1 1 . r r although some used incorrect values such as or Disappointingly, a 4 2 r 4 significant number of candidates used from the expression given in the question a , and applied it to the formula even though the formula only works when 1 r 1 1. r There were a number of correct strategies that candidates could use. The 4 a 1 1 , most successful strategy was to apply to and many 20 , a r 1 r 2 2 candidates who used this strategy achieved the correct answer of 2.5. Other complete 3 r r 1 1 strategies included applying or 20 20 2 2 1 1 r r 3 r r 1 1 which led to (20 17.5) or (40 37.5) respectively. 20 20 2 2 0 0 r r Some candidates applied an incorrect overall strategy by subtracting an incorrect number of terms from their sum to infinity. These candidates gave incorrect answers such as (20 18.75), (40 38.75), (40 35) or (40 17.5). A few candidates obtained the correct answer of 2.5 by using their calculator. Only a few candidates produced a correct solution to Q08(ii). There were two main correct methods that were used by candidates in equal measure. One was to list the first few terms and the last few terms of the series and use the addition law of logarithms 3 4 50 log ... . to achieve Many candidates used cancelling to complete their 5 2 3 49 50 proof correctly by writing The other method was to use the log log (25) 2. 5 5 2 48 subtraction law of logarithms to give and list terms to (log ( 2) log ( 1)) n n 5 5 1 n (log 3 log 4 ...... log 50) (log 2 log 3 ...... log 49). give Again many 5 5 5 5 5 5 candidates used cancelling and completed their proof by writing A few candidates progressed by writing log (50) log (2) log (25) 2. 5 5 5 log (3 4 ....... 50) log (2 3 ....... 49) and used factorials to give 5 5 50! 50! log log 49! log log (25) 2. 5 5 5 5 2 2(49!) Some candidates incorrectly assumed an arithmetic series and attempted to find its sum 3 50 n a n 48, log , log n a l ( ) l [2 ( 1) ] a n d by using or with and 5 5 2 49 2 2 4 3 log log . d A few candidates incorrectly assumed a geometric series and 5 5 3 2 attempted to find its sum. Some candidates, who made no creditable progress, resorted to listing their terms as decimals, or attempted to add all 48 terms as decimals by writing 0.2519... 0.1787... 0.1386... ...... 0.0125... 2.

Common mistakes

• first term a = 20(½)⁴ = 1.25, r = ½, S∞ = 2.5
• listing the first terms helps
• subtracting the wrong number of terms from a sum from r = 0 or 1
• using r = 20 or ¼

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

Examiner's comment

Q08 was found challenging by many candidates, with Q08(i) more successfully answered than Q08(ii). The most successful candidates were the ones who listed the first few terms of the series. In other words, those candidates that wrote 4 5 6 3 4 5 1 1 1 log log log ... and gained a 20 20 20 ... 5 5 5 2 3 4 2 2 2 better understanding of the series that they were required to sum. There were, however, many attempts that made no creditable progress and a significant number of blank responses to either one or both parts of Q08. In Q08(i), many candidates identified that a sum to infinity was required in their 1 a r . solution and attempted to apply the formula Most used in this formula, 2 1 r 1 1 . r r although some used incorrect values such as or Disappointingly, a 4 2 r 4 significant number of candidates used from the expression given in the question a , and applied it to the formula even though the formula only works when 1 r 1 1. r There were a number of correct strategies that candidates could use. The 4 a 1 1 , most successful strategy was to apply to and many 20 , a r 1 r 2 2 candidates who used this strategy achieved the correct answer of 2.5. Other complete 3 r r 1 1 strategies included applying or 20 20 2 2 1 1 r r 3 r r 1 1 which led to (20 17.5) or (40 37.5) respectively. 20 20 2 2 0 0 r r Some candidates applied an incorrect overall strategy by subtracting an incorrect number of terms from their sum to infinity. These candidates gave incorrect answers such as (20 18.75), (40 38.75), (40 35) or (40 17.5). A few candidates obtained the correct answer of 2.5 by using their calculator. Only a few candidates produced a correct solution to Q08(ii). There were two main correct methods that were used by candidates in equal measure. One was to list the first few terms and the last few terms of the series and use the addition law of logarithms 3 4 50 log ... . to achieve Many candidates used cancelling to complete their 5 2 3 49 50 proof correctly by writing The other method was to use the log log (25) 2. 5 5 2 48 subtraction law of logarithms to give and list terms to (log ( 2) log ( 1)) n n 5 5 1 n (log 3 log 4 ...... log 50) (log 2 log 3 ...... log 49). give Again many 5 5 5 5 5 5 candidates used cancelling and completed their proof by writing A few candidates progressed by writing log (50) log (2) log (25) 2. 5 5 5 log (3 4 ....... 50) log (2 3 ....... 49) and used factorials to give 5 5 50! 50! log log 49! log log (25) 2. 5 5 5 5 2 2(49!) Some candidates incorrectly assumed an arithmetic series and attempted to find its sum 3 50 n a n 48, log , log n a l ( ) l [2 ( 1) ] a n d by using or with and 5 5 2 49 2 2 4 3 log log . d A few candidates incorrectly assumed a geometric series and 5 5 3 2 attempted to find its sum. Some candidates, who made no creditable progress, resorted to listing their terms as decimals, or attempted to add all 48 terms as decimals by writing 0.2519... 0.1787... 0.1386... ...... 0.0125... 2.

Common mistakes

• write out first and last terms: log₅(3/2 × 4/3 × … × 50/49) = log₅25 = 2 (telescoping/cancelling)
• it is not an arithmetic or geometric series
• adding decimals scores nothing

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

• split into Σ(3 + 5r) + Σ2^r
• AP sum 8(8 + 83) = 728
• GP sum with a = 2 (not 1), r = 2: 2(2^16 − 1) = 131070
• listing terms needs all 16 shown

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