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A-Level · Pure Mathematics

P4 Pure: Sequences and Series (incl. Binomial Expansion)

EDEXCEL 9MA0 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Priya's training plan for a marathon increases the distance she runs each week by a fixed amount. In week 1 she runs 8 km, and each following week she runs 1.5 km further than the previous week, for a total of 16 weeks.

(a)Find the distance Priya runs in week 16.(2)

(b)Find the total distance Priya runs over the 16 weeks.(3)

(c)After week 16, Priya changes to running a constant 30.5 km every week (the same distance as week 16) for a further m weeks. Given that her total distance run overall (the original 16 weeks plus the constant-distance weeks) must be at least 800 km, find the least value of m.(4)

(Total for Question 1 is 9 marks)

A sequence is arithmetic with first term a = 7 and common difference d = 4.

(a)Find the 15th term of the sequence.(2)

(b)Show that the sum of the first n terms, S_n, satisfies S_n = n(2n+5).(3)

(c)Given that S_n = 900, form a quadratic equation in n and hence find the value of n.(4)

(Total for Question 2 is 9 marks)

Ferndale Ltd's annual profit forms a geometric sequence. In year 1 the profit is 12000 pounds, and the profit grows by 8% each year, so the common ratio is r = 1.08.

(a)Find the profit in year 8, giving your answer to the nearest pound.(2)

(b)Find the total profit made over the first 10 years, giving your answer to the nearest pound.(3)

(c)State one limitation of using this geometric model to predict Ferndale Ltd's profit far into the future.(1)

(Total for Question 3 is 6 marks)

A geometric series has first term a = 40 and sum to infinity 100.

(a)Find the common ratio, r.(3)

(b)Find the sum of the first 6 terms of the series, giving your answer to 3 significant figures.(3)

(c)Explain why the series would not have a sum to infinity if the common ratio were r=1.5 instead.(1)

(Total for Question 4 is 7 marks)

Consider the sum sum_{r=1}^{n} (4r - 3).

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

(b)Hence find the value of sum_{r=1}^{25} (4r-3).(2)

(c)Show that sum_{r=n+1}^{2n} (4r-3) = 6n² - n.(4)

(Total for Question 5 is 9 marks)

The binomial expansion of (1+x)⁶ is to be used to answer this question.

(a)Write out the full expansion of (1+x)⁶ in ascending powers of x.(3)

(b)Hence find the coefficient of x³ in the expansion of (1+2x)^6.(2)

(c)Find the term independent of x in the expansion of (x² + 3/x)^6.(4)

(Total for Question 6 is 9 marks)

This question uses the binomial expansion of (2+x)^7.

(a)Find, in ascending powers of x, the first four terms of the binomial expansion of (2+x)⁷, giving each coefficient as an integer.(4)

(b)Hence find an estimate for 2.01⁷, giving your answer to 3 significant figures.(3)

(Total for Question 7 is 7 marks)

The Kingswood Theatre has rows of seats arranged in an arithmetic sequence. Row 1 (nearest the stage) has 18 seats, the common difference between consecutive rows is d, there are 30 rows, and the total seating capacity is 1410.

(a)Show that d = 2.(3)

(b)Find the number of seats in row 30 (the back row).(2)

(c)The theatre wants to add extra rows at the back, continuing the same arithmetic sequence (same first term and common difference), so that the total capacity is at least 2000 seats. Find the minimum number of extra rows needed.(5)

(Total for Question 8 is 10 marks)

In the binomial expansion, in ascending powers of x, of (1+cx)¹⁰, where c is a non-zero constant, the coefficient of x³ is 4 times the coefficient of x^2.

(a)Find the value of c.(4)

(b)Using this value of c, find the coefficient of x⁴ in the expansion, giving your answer as an exact fraction.(3)

(Total for Question 9 is 7 marks)

The first three terms of a geometric sequence, in order, are k, k+6 and 4k+9, where k is a positive constant.

(a)Show that k² - k - 12 = 0.(3)

(b)Hence show that k=4.(2)

(c)Using k=4, find the common ratio of the sequence and the sum of the first 12 terms, giving your answer to the nearest integer.(4)

(Total for Question 10 is 9 marks)

f(x) = (1-3x)^1/3.

(a)Find the binomial expansion of f(x), in ascending powers of x, up to and including the term in x³, simplifying each coefficient.(5)

(b)State the range of values of x for which this expansion is valid.(1)

(c)By choosing a suitable value of x, use your expansion to estimate 0.94^1/3 to 5 decimal places, and state, with a reason, whether your estimate is an over-estimate or an under-estimate.(4)

(Total for Question 11 is 10 marks)

A sequence u1, u2, u3, ... is defined by u1=3 and u_{n+1} = 1/(1-u_n) for n≥1.

(a)Find the values of u2, u3 and u4, and hence show that the sequence is periodic, stating the order (period) of the sequence.(4)

(b)Hence find sum_{r=1}^{100} u_r.(4)

(c)Find the value of u_2026.(2)

(Total for Question 12 is 10 marks)

An arithmetic sequence has first term a and common difference d, where a>0 and d>0. The 1st, 2nd and 4th terms of this sequence, taken in that order, form the first three terms of a geometric sequence.

(a)Show that d=a.(3)

(b)Given that a=5, write down the first four terms of the arithmetic sequence and verify that its 1st, 2nd and 4th terms form a geometric sequence, stating the common ratio.(3)

(c)A geometric sequence has first term 5 and common ratio 2 (as found in part (b)). Find the smallest value of n for which the sum of the first n terms of this geometric sequence exceeds 100000.(5)

(Total for Question 13 is 11 marks)

Mark scheme · P4 Pure: Sequences and Series (incl. Binomial Expansion)

Question 1

(a) M1 uses u_n = a + (n-1)d with a=8, d=1.5, n=16
(a) A1 30.5 km cao
(a) Answer: 30.5 km
(b) M1 correct use of S_n = n/2(2a+(n-1)d) with n=16, a=8, d=1.5
(b) A1 correct unsimplified value 8(16+22.5)
(b) A1 308 km cao
(b) Answer: 308 km
(c) M1 forms 308 + 30.5m ≥ 800
(c) M1 correct rearrangement m ≥ 492/30.5
(c) A1 m ≥ awrt 16.13
(c) A1 least integer m=17 cao, with reference to context (whole number of weeks)
(c) Answer: m = 17

Question 2

(a) M1 uses u_n = a + (n-1)d with n=15
(a) A1 63 cao
(a) Answer: 63
(b) M1 correct substitution into S_n = n/2(2a+(n-1)d)
(b) M1 correct expansion and simplification
(b) A1 S_n = n(2n+5) cso, answer given
(b) Answer: S_n = n(2n+5) (shown)
(c) M1 forms 2n²+5n-900=0
(c) M1 solves the quadratic using the formula or by factorising
(c) A1 n=20 cao
(c) A1 rejects n=-22.5 since n must be a positive integer
(c) Answer: n = 20

Question 3

(a) M1 uses u_n = a r^n-1 with n=8
(a) A1 awrt 20566
(a) Answer: 20566 pounds (nearest pound)
(b) M1 correct use of S_n = a(rⁿ - 1)/(r-1) with n=10
(b) M1 correct value of r¹⁰ used (awrt 2.1589)
(b) A1 awrt 173839
(b) Answer: 173839 pounds (nearest pound)
(c) B1 valid limitation, e.g. profit cannot keep growing at a constant percentage rate indefinitely / markets saturate / real profits are affected by unpredictable factors not captured by a constant ratio
(c) Answer: e.g. constant 8% growth is unrealistic long-term as markets saturate.

Question 4

(a) M1 uses S_infinity = a/(1-r)
(a) M1 forms 1-r = 40/100
(a) A1 r=0.6 cao
(a) Answer: r = 0.6
(b) M1 correct use of S_n = a(1-rⁿ)/(1-r) with n=6
(b) M1 correct value of r⁶ = 0.6⁶ = awrt 0.0467
(b) A1 awrt 95.3
(b) Answer: 95.3 (3sf)
(c) B1 states |r|>1 so the terms do not tend to zero and the series diverges, so no sum to infinity exists
(c) Answer: Because |r| = 1.5 > 1 the terms grow without bound so the series diverges.

Question 5

(a) M1 splits the sum as 4 x sum(r) - 3n
(a) M1 uses sum(r) = n(n+1)/2 and simplifies
(a) A1 n(2n-1) cso, answer given
(a) Answer: n(2n-1) (shown)
(b) M1 substitutes n=25 into n(2n-1)
(b) A1 1225 cao
(b) Answer: 1225
(c) M1 expresses the required sum as sum_{r=1}^{2n} - sum_{r=1}^{n}
(c) M1 substitutes 2n into n(2n-1) to get 2n(4n-1)
(c) M1 correct expansion and subtraction: (8n²-2n) - (2n²-n)
(c) A1 6n² - n cso, answer given
(c) Answer: 6n² - n (shown)

Question 6

(a) M1 correct binomial coefficients 1, 6, 15, 20, 15, 6, 1 (Pascal's triangle row 6, or nCr values)
(a) A1 correct powers of x attached to each coefficient
(a) A1 fully correct expansion cao
(a) Answer: 1 + 6x + 15x² + 20x³ + 15x⁴ + 6x⁵ + x⁶
(b) M1 uses C(6,3) x 2³
(b) A1 160 cao
(b) Answer: 160
(c) M1 correct general term C(6,k)(x²)^6-k(3/x)^k
(c) M1 correct simplified power of x: 12-3k
(c) M1 sets 12-3k=0 and solves k=4
(c) A1 1215 cao
(c) Answer: 1215

Question 7

(a) M1 correct structure using C(7,k) 2^7-k x^k
(a) A1 128 + 448x
(a) A1 672x² term correct
(a) A1 560x³ term correct
(a) Answer: 128 + 448x + 672x² + 560x³ (+...)
(b) M1 substitutes x=0.01 into the expansion from part (a)
(b) M1 correct evaluation of the sum (awrt 132.5)
(b) A1 awrt 133
(b) Answer: 133 (3sf)

Question 8

(a) M1 correct substitution into S_n = n/2(2a+(n-1)d) with a=18, n=30, S=1410
(a) M1 correct rearrangement to 870 = 435d
(a) A1 d=2 cso
(a) Answer: d = 2 (shown)
(b) M1 u30 = 18 + 29(2)
(b) A1 76 cao
(b) Answer: 76 seats
(c) M1 forms S_n = n/2(36+2(n-1)) = n(n+17) for the extended arrangement
(c) M1 sets up the inequality n(n+17) ≥ 2000
(c) dM1 solves the associated quadratic n²+17n-2000=0 (formula or trial), obtaining n=awrt 37.0
(c) A1 smallest integer total rows n=38 (since n=37 gives 1998<2000)
(c) A1 minimum extra rows = 8 (38-30) cao
(c) Answer: 8 extra rows

Question 9

(a) M1 writes the coefficient of x² as C(10,2)c² and the coefficient of x³ as C(10,3)c³
(a) M1 forms the equation 120c³ = 4(45c²), i.e. 120c³ = 180c²
(a) M1 solves 60c²(2c-3)=0
(a) A1 c=3/2 cao, rejecting c=0
(a) Answer: c = 3/2
(b) M1 uses C(10,4)c⁴
(b) M1 substitutes c=3/2 correctly, (3/2)⁴ = 81/16
(b) A1 8505/8 oe (awrt 1063.13)
(b) Answer: 8505/8 (=1063.125)

Question 10

(a) M1 forms (k+6)² = k(4k+9) using the geometric sequence property
(a) M1 correctly expands both sides
(a) A1 k²-k-12=0 cso, answer given
(a) Answer: k² - k - 12 = 0 (shown)
(b) M1 factorises (k-4)(k+3)=0 or uses the quadratic formula
(b) A1 k=4 cso, rejecting k=-3 since k is positive
(b) Answer: k = 4
(c) B1 r=2.5 (=10/4, confirmed also by 25/10)
(c) M1 correct use of S_n = a(rⁿ-1)/(r-1) with a=4, n=12
(c) M1 correct value of r¹² (awrt 59604.6)
(c) A1 awrt 158943
(c) Answer: r = 2.5; S12 = 158943 (nearest integer)

Question 11

(a) M1 correct structure (1+u)ⁿ = 1+nu+n(n-1)/2 u²+n(n-1)(n-2)/6 u³ with n=1/3, u=-3x
(a) A1 correct first two terms 1 - x
(a) M1 correct method for the x² coefficient
(a) A1 -x² term correct
(a) A1 -(5/3)x³ term correct
(a) Answer: 1 - x - x² - (5/3)x³ (+...)
(b) B1 |x| < 1/3 oe (e.g. -1/3 < x < 1/3)
(b) Answer: |x| < 1/3
(c) M1 identifies x=0.02 (from 1-3x=0.94)
(c) M1 correctly substitutes x=0.02 into the expansion from part (a)
(c) A1 awrt 0.97959
(c) B1 states over-estimate, with valid reason (e.g. the next term in the series is negative, so the truncated sum is larger than the true value)
(c) Answer: 0.97959 (over-estimate)

Question 12

(a) M1 correct u2=-1/2
(a) M1 correct u3=2/3
(a) A1 correct u4=3 (=u1), demonstrating repetition
(a) A1 states the sequence is periodic with period (order) 3
(a) Answer: u2 = -1/2, u3 = 2/3, u4 = 3; period 3
(b) M1 finds the sum of one period = 3 + (-1/2) + 2/3 = 19/6
(b) M1 recognises 100=3(33)+1, giving 33 complete periods plus one extra term
(b) dM1 computes 33(19/6) + u1
(b) A1 215/2 oe (=107.5) cao
(b) Answer: 215/2 (=107.5)
(c) M1 considers the remainder when 2026 is divided by 3 (2026=3(675)+1)
(c) A1 u_2026=3 cao
(c) Answer: u_2026 = 3

Question 13

(a) M1 forms (a+d)² = a(a+3d) using the given geometric sequence condition
(a) M1 correctly expands to a²+2ad+d²=a²+3ad and rearranges to d²=ad
(a) A1 d=a cso (dividing by d, valid since d>0)
(a) Answer: d = a (shown)
(b) B1 arithmetic terms 5, 10, 15, 20 (using d=a=5)
(b) M1 computes both ratios 10/5 and 20/10
(b) A1 both ratios equal 2, confirming a geometric sequence with common ratio r=2
(b) Answer: 5, 10, 15, 20; r = 2
(c) M1 correct use of S_n = a(rⁿ-1)/(r-1) = 5(2ⁿ-1)
(c) M1 forms the inequality 5(2ⁿ-1) > 100000, i.e. 2ⁿ > 20001
(c) M1 takes logs of both sides: n > log(20001)/log(2)
(c) A1 n > awrt 14.3
(c) A1 smallest integer n=15 cao
(c) Answer: n = 15

Pure: Sequences and Series (incl. Binomial Expansion) - Worksheets, Questions and Revision