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A-Level · Pure Mathematics

P12 Pure: Sequences, Series and Binomial Depth

EDEXCEL 9MA0 · Calculator allowed · about 140 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

An arithmetic sequence has 4th term equal to 17 and 11th term equal to 45.

(a)Show that the common difference of the sequence is 4.(2)

(b)Find the first term, a, of the sequence.(1)

(c)Find the sum of the first 30 terms of the sequence.(2)

(Total for Question 1 is 5 marks)

A geometric sequence has first term 6 and third term 54.

(a)Find the possible value(s) of the common ratio, r.(3)

(b)Given that all terms of the sequence are positive, find the 7th term.(2)

(Total for Question 2 is 5 marks)

A factory records the number of defective items produced each month. In month 1 there are 84 defective items, and this decreases by 3 each month compared with the previous month, forming an arithmetic sequence, while the process remains uncorrected.

(a)Find the number of defective items modelled in month 12.(2)

(b)Find the total number of defective items modelled over months 6 to 15 inclusive.(4)

(c)The process is considered corrected once the modelled number of defective items first reaches zero or fewer. Find the first month in which this occurs according to the model, and comment on whether the arithmetic model remains appropriate after this point.(3)

(Total for Question 3 is 9 marks)

The sum of the first n terms of an arithmetic series is given by S_n = 3n² + 5n.

(a)Find the first term and the common difference of the series.(4)

(b)Hence find the 25th term of the series.(2)

(Total for Question 4 is 6 marks)

Consider the binomial expansion of (2+3x)⁷ in ascending powers of x. Let T_k denote the term in x^k (so T₀ is the constant term).

(a)Find an expression for T_k in terms of k.(2)

(b)Show that T_(k+1)/T_k = 3(7-k) / (2(k+1)).(3)

(c)Hence find the value of k for which T_k is greatest, and find this greatest coefficient.(3)

(Total for Question 5 is 8 marks)

The first three terms of a geometric series are 4, 4r and 4r² respectively, where r is real and r ≠ 1. The sum of the first 6 terms of the series is 5 times the sum of the first 3 terms.

(a)Show that r³ = 4.(3)

(b)Find the value of r, giving your answer to 3 significant figures.(2)

(c)Find the sum of the first 6 terms of the series, giving your answer to the nearest whole number.(2)

(Total for Question 6 is 7 marks)

A geometric series has first term 20 and common ratio r = -0.4.

(a)Explain why this series has a sum to infinity.(1)

(b)Find the sum to infinity of the series, giving your answer as an exact fraction.(3)

(c)Find the smallest value of n such that the sum of the first n terms differs from the sum to infinity by less than 0.01.(4)

(Total for Question 7 is 8 marks)

A sequence is defined by u₁ = 10 and u_(n+1) = 0.5 u_n + 3 for n ≥ 1.

(a)Find the values of u2, u3 and u4.(3)

(b)Show that the sequence with terms u_n - 6 is a geometric sequence, and state its common ratio.(4)

(c)Hence show that u_n = 6 + 4(0.5)^n-1, and find the value that u_n approaches as n becomes very large.(4)

(Total for Question 8 is 11 marks)

g(x) = (1+2x)⁵ (1-x)^-2, for |x| < 1/2.

(a)Find the binomial expansion of (1+2x)⁵ in ascending powers of x, up to and including the term in x^2.(3)

(b)Find the binomial expansion of (1-x)^-2 in ascending powers of x, up to and including the term in x^2.(3)

(c)Hence show that g(x) = 1 + 12x + 63x² + ... for small values of x.(4)

(Total for Question 9 is 10 marks)

f(x) = 1/√4-3x = (4-3x)^-1/2.

(a)Show that f(x) can be written as (1/2)(1 - (3/4)x)^-1/2.(2)

(b)Hence find the binomial expansion of f(x) in ascending powers of x, up to and including the term in x^2. Give each coefficient as a simplified fraction.(5)

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

(d)By substituting x=0.2 into the expansion from part (b), find an estimate for 1/√3.4, giving your unrounded answer to 5 decimal places. Given that 1/√3.4 = 0.542326 to 6 decimal places, use your unrounded estimate to calculate the percentage error, giving your answer to 2 significant figures.(4)

(Total for Question 10 is 12 marks)

This question concerns arithmetic sequences and series.

(a)Prove, using the method of pairing terms, that the sum of the first n terms of an arithmetic sequence with first term a and common difference d is given by S_n = n/2 (2a+(n-1)d).(4)

(b)Two arithmetic sequences, A and B, have the same common difference d. The first term of sequence A is p and the first term of sequence B is q, where p ≠ q. Given that the sum of the first 10 terms of sequence A equals the sum of the first 6 terms of sequence B, show that 5p - 3q = -15d.(4)

(c)Given additionally that d=2 and q=3p, find the value of p.(3)

(Total for Question 11 is 11 marks)

Priya pays into a savings scheme. She pays 500 pounds in year 1. Each subsequent year, she pays 4% more than the year before, so her annual payments form a geometric sequence with first term 500 and common ratio 1.04.

(a)Find the amount Priya pays in year 8, giving your answer to the nearest penny.(3)

(b)Find the total amount Priya has paid into the scheme by the end of year 15 (the sum of the first 15 payments), giving your answer to the nearest pound.(3)

(c)Priya wants the total amount paid into the scheme to first exceed 20000 pounds. Using logarithms, find the smallest number of years, n, for which this occurs.(5)

(d)State one assumption made in this model that may not hold in reality.(1)

(Total for Question 12 is 12 marks)

Mark scheme · P12 Pure: Sequences, Series and Binomial Depth

Question 1

(a) M1 forms two equations a+3d=17 and a+10d=45 (or equivalent) and eliminates a
(a) A1 d=4 cso, with full working shown
(a) Answer: d = 4 (shown)
(b) B1 a=5 ft (using d=4 found in part (a))
(b) Answer: a = 5
(c) M1 correct substitution into S_n = n/2(2a+(n-1)d) with n=30, a=5, d=4
(c) A1 1890 cao
(c) Answer: 1890

Question 2

(a) M1 forms 6r²=54
(a) M1 solves r²=9
(a) A1 both r=3 and r=-3 stated (oe)
(a) Answer: r = 3 or r = -3
(b) M1 correct substitution u7=6(3)⁶
(b) A1 4374 cao
(b) Answer: 4374

Question 3

(a) M1 uses u_n=a+(n-1)d with a=84, d=-3, n=12
(a) A1 51 cao
(a) Answer: 51
(b) B1 identifies that months 6 to 15 inclusive is 10 terms
(b) M1 correct method, e.g. n/2(u6+u15) or S15-S5
(b) A1 correct intermediate values u6=69 and u15=42 (or S15=945, S5=390)
(b) A1 555 cao
(b) Answer: 555
(c) M1 forms 84-3(n-1)≤0 (oe) and solves to n≥29
(c) A1 cao month 29, with u28=3>0 and u29=0 checked
(c) B1 valid comment that the model breaks down beyond this point since a count of defective items cannot be negative in reality
(c) Answer: Month 29; the arithmetic model is no longer appropriate beyond this point since a count cannot go negative.

Question 4

(a) M1 finds the first term a=S₁=8
(a) M1 finds u2=S₂-S₁ (or equivalent method comparing coefficients) leading to d
(a) A1 a=8 cao
(a) A1 d=6 cao
(a) Answer: a = 8, d = 6
(b) M1 uses u_n=a+(n-1)d with n=25
(b) A1 152 cao
(b) Answer: 152

Question 5

(a) M1 correct structure using the general binomial term C(7,k)(2)^7-k(3x)^k
(a) A1 cao T_k = C(7,k) * 2^7-k * 3^k
(a) Answer: T_k = C(7,k) * 2^7-k * 3^k
(b) M1 uses C(7,k+1)/C(7,k) = (7-k)/(k+1)
(b) M1 correctly finds the ratio of powers of 2 and 3 (a factor of 1/2 and a factor of 3)
(b) A1 cso combines to 3(7-k)/(2(k+1))
(b) Answer: T_(k+1)/T_k = 3(7-k)/(2(k+1)) (shown)
(c) M1 sets T_(k+1)/T_k > 1 and solves the resulting inequality to k < 3.8 (oe)
(c) M1 deduces that the coefficients increase up to k=4 and decrease after, so the greatest term is T₄
(c) A1 cao T₄ = 22680
(c) Answer: k = 4; greatest coefficient = 22680

Question 6

(a) M1 uses the sum formulae for S6 and S3 with the given condition S6=5S3 (or recognises terms 4 to 6 equal r³ times terms 1 to 3)
(a) M1 reaches r⁶-5r³+4=0 and factorises to (r³-1)(r³-4)=0
(a) A1 cso rejects r³=1 (since r ≠ 1 is given) leaving r³=4
(a) Answer: r³ = 4 (shown)
(b) M1 takes the cube root of both sides
(b) A1 awrt 1.59
(b) Answer: r = 1.59 (awrt, 3sf)
(c) M1 correct method, e.g. S6 = 4(1-r⁶)/(1-r) using r⁶=(r³)²=16, or S6=5*S3
(c) A1 awrt 102
(c) Answer: 102

Question 7

(a) B1 states |r| = 0.4 < 1 (or -1 < r < 1), so the series converges
(a) Answer: Because |r| = 0.4 < 1, the series converges (has a sum to infinity).
(b) M1 uses S_infinity = a/(1-r)
(b) A1 ft correct substitution 20/(1-(-0.4))
(b) A1 cao 100/7
(b) Answer: 100/7 (= 14 2/7)
(c) M1 forms S_infinity - S_n = a rⁿ/(1-r) (oe)
(c) M1 sets up the inequality (100/7)(0.4)ⁿ < 0.01 (oe)
(c) M1 dM1 takes logs and solves for n, correctly reversing the inequality when dividing by a negative log value
(c) A1 cao n=8, with a supporting check
(c) Answer: n = 8

Question 8

(a) M1 correct method substituting into the recurrence at least once
(a) A1 u2=8 and u3=7 correct
(a) A1 cao u4=6.5
(a) Answer: u2 = 8, u3 = 7, u4 = 6.5
(b) M1 lets v_n = u_n - 6 and substitutes u_(n+1)=0.5u_n+3 into v_(n+1)=u_(n+1)-6
(b) M1 correctly simplifies to v_(n+1) = 0.5(u_n - 6)
(b) A1 cso identifies v_(n+1) = 0.5 v_n, confirming a geometric sequence
(b) A1 states common ratio 0.5
(b) Answer: u_n - 6 is geometric with common ratio 0.5 (shown)
(c) M1 writes the general term of the geometric sequence v_n = v₁ r^n-1 = 4(0.5)^n-1
(c) A1 cso u_n = v_n + 6 = 6 + 4(0.5)^n-1, matching the given result
(c) M1 recognises that since |0.5| < 1, (0.5)^n-1 tends to 0 as n tends to infinity
(c) A1 cao limit = 6
(c) Answer: u_n = 6 + 4(0.5)^n-1; u_n tends to 6 as n tends to infinity

Question 9

(a) M1 correct structure using binomial coefficients C(5,1) and C(5,2)
(a) A1 10x term correct
(a) A1 40x² term correct
(a) Answer: 1 + 10x + 40x² (+...)
(b) M1 correct general expansion structure (1+u)ⁿ = 1+nu+n(n-1)/2 u² with n=-2, u=-x
(b) A1 2x term correct
(b) A1 3x² term correct
(b) Answer: 1 + 2x + 3x² (+...)
(c) M1 attempts to multiply the two expansions, considering only terms up to x²
(c) A1 correct x coefficient 12 (from 1x2 + 10x1)
(c) M1 dM1 correct method for the x² coefficient, combining all three contributing products (1x3, 10x2, 40x1)
(c) A1 cso 63, matching the given result
(c) Answer: g(x) = 1 + 12x + 63x² + ... (shown)

Question 10

(a) M1 factorises 4-3x as 4(1-(3/4)x)
(a) A1 cso correctly simplifies 4^-1/2=1/2 to reach the given form
(a) Answer: f(x) = (1/2)(1-(3/4)x)^-1/2 (shown)
(b) M1 correct general structure (1+u)ⁿ = 1+nu+n(n-1)/2 u² with n=-1/2
(b) A1 correct term1 in u: -(1/2)u
(b) A1 correct term2 in u: (3/8)u²
(b) M1 substitutes u=-(3/4)x throughout and multiplies by the (1/2) factor from part (a)
(b) A1 cao 1/2 + (3/16)x + (27/256)x²
(b) Answer: f(x) is approximately 1/2 + (3/16)x + (27/256)x² (+...)
(c) B1 |x| < 4/3 oe (e.g. -4/3 < x < 4/3)
(c) Answer: |x| < 4/3
(d) M1 substitutes x=0.2 into the expansion from part (b)
(d) M1 correct unrounded estimate awrt 0.54172
(d) M1 correct method for percentage error, |estimate - true|/true x 100
(d) A1 cao awrt 0.11%
(d) Answer: Estimate is approximately 0.54172; percentage error is approximately 0.11%

Question 11

(a) B1 writes S_n = a+(a+d)+(a+2d)+...+(a+(n-1)d)
(a) M1 writes S_n again in reverse order, (a+(n-1)d)+(a+(n-2)d)+...+a
(a) M1 adds the two expressions term by term, showing each pair sums to 2a+(n-1)d, giving 2S_n = n(2a+(n-1)d)
(a) A1 cso correctly divides by 2 to reach S_n = n/2(2a+(n-1)d)
(a) Answer: S_n = n/2(2a+(n-1)d) (proved)
(b) M1 correct expression for the sum of the first 10 terms of A, 10p+45d (or equivalent unsimplified form)
(b) M1 correct expression for the sum of the first 6 terms of B, 6q+15d
(b) M1 dM1 equates the two sums and rearranges to isolate terms in d
(b) A1 cso simplifies (dividing by 2) to reach 5p-3q=-15d exactly
(b) Answer: 5p - 3q = -15d (shown)
(c) M1 substitutes q=3p and d=2 into 5p-3q=-15d
(c) M1 solves the resulting linear equation in p
(c) A1 cao p=7.5
(c) Answer: p = 7.5

Question 12

(a) M1 uses u_n = a r^n-1 with a=500, r=1.04, n=8
(a) M1 correct value of 1.04⁷ (awrt 1.3159)
(a) A1 awrt 657.97 (nearest penny)
(a) Answer: 657.97 pounds (nearest penny)
(b) M1 correct use of S_n = a(rⁿ-1)/(r-1) with n=15
(b) M1 correct value of 1.04¹⁵ (awrt 1.8009)
(b) A1 awrt 10012 (nearest pound)
(b) Answer: 10012 pounds (nearest pound)
(c) M1 forms the inequality 500(1.04ⁿ-1)/0.04 > 20000, i.e. 12500(1.04ⁿ-1) > 20000
(c) M1 rearranges to 1.04ⁿ > 2.6
(c) M1 dM1 takes logs of both sides, n > ln(2.6)/ln(1.04) (oe base)
(c) A1 correct unrounded value awrt 24.36 (or better)
(c) A1 cao smallest integer n=25, with supporting check
(c) Answer: n = 25
(d) B1 valid assumption, e.g. that Priya's payments increase by exactly 4% every year without fail, or that she is able and willing to keep increasing her payments indefinitely, which may not be realistic
(d) Answer: e.g. it assumes the 4% annual increase in payments continues exactly every year, which may not be realistic since Priya's circumstances or the scheme's terms could change.

Pure: Sequences, Series and Binomial Depth - Worksheets, Questions and Revision