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school. “Paper 42” means Paper 4, variant 2.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Sequences & inequalities · Core · Paper 1 style · 5 marks
(a)Prove by mathematical induction that, for all positive integers n,Σ (from r = 1 to n) r(r + 2) = (1/6)n(n + 1)(2n + 7).[5]
Show answer
(a) Proof by induction (see working).
B1: n = 1: LHS = 1 × 3 = 3, RHS = (1/6)(1)(2)(9) = 3.M1: assume true for n = k and add the (k + 1)th term (k + 1)(k + 3).M1: factorise (1/6)(k + 1)[k(2k + 7) + 6(k + 3)] = (1/6)(k + 1)(2k² + 13k + 18).A1: = (1/6)(k + 1)(k + 2)(2k + 9) = (1/6)(k + 1)((k + 1) + 1)(2(k + 1) + 7).A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q2Sequences & inequalities · Core · Paper 1 style · 5 marks
(a)Prove by mathematical induction that, for all positive integers n,Σ (from r = 1 to n) 1/((2r − 1)(2r + 1)) = n/(2n + 1).[5]
Show answer
(a) Proof by induction (see working).
B1: n = 1: LHS = 1/(1 × 3) = 1/3, RHS = 1/3.M1: assume true for n = k; add 1/((2k + 1)(2k + 3)).M1: common denominator: [k(2k + 3) + 1]/((2k + 1)(2k + 3)) = (2k² + 3k + 1)/((2k + 1)(2k + 3)).A1: = (k + 1)(2k + 1)/((2k + 1)(2k + 3)) = (k + 1)/(2(k + 1) + 1).A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q3Sequences & inequalities · Core · Paper 1 style · 5 marks
(a)The sequence u₁, u₂, u₃, … is defined by u₁ = 3 and u_(n+1) = 2u_n − 1 for n ≥ 1. Find u₅.[1]
(b)Prove by mathematical induction that u_n = 2ⁿ + 1 for all positive integers n.[4]
Show answer
(a) u₅ = 33
B1: u₂ = 5, u₃ = 9, u₄ = 17, u₅ = 33.
(b) Proof by induction (see working).
B1: n = 1: 2¹ + 1 = 3 = u₁.M1: assume u_k = 2ᵏ + 1; then u_(k+1) = 2(2ᵏ + 1) − 1.A1: = 2^(k+1) + 1.A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q4Sequences & inequalities · Core · Paper 1 style · 5 marks
(a)Prove by mathematical induction that, for all positive integers n,Σ (from r = 1 to n) r × r! = (n + 1)! − 1.[5]
Show answer
(a) Proof by induction (see working).
B1: n = 1: LHS = 1 × 1! = 1, RHS = 2! − 1 = 1.M1: assume true for n = k; add (k + 1)(k + 1)!.M1: (k + 1)! − 1 + (k + 1)(k + 1)! = (k + 1)!(1 + k + 1) − 1.A1: = (k + 2)(k + 1)! − 1 = (k + 2)! − 1.A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q5Sequences & inequalities · Core · Paper 1 style · 5 marks
(a)Prove by mathematical induction that, for all positive integers n,Σ (from r = 1 to n) (2r − 1)² = (1/3)n(2n − 1)(2n + 1).[5]
Show answer
(a) Proof by induction (see working).
B1: n = 1: LHS = 1, RHS = (1/3)(1)(1)(3) = 1.M1: assume true for n = k; add (2k + 1)².M1: (1/3)(2k + 1)[k(2k − 1) + 3(2k + 1)] = (1/3)(2k + 1)(2k² + 5k + 3).A1: = (1/3)(2k + 1)(k + 1)(2k + 3) = (1/3)(k + 1)(2(k + 1) − 1)(2(k + 1) + 1).A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q6Sequences & inequalities · Standard · Paper 1 style · 6 marks
(a)The sequence u₁, u₂, u₃, … is defined by u₁ = 1 and u_(n+1) = 3u_n + 2. Prove by mathematical induction that u_n = 2 × 3^(n−1) − 1 for all positive integers n.[4]
(b)Find the least value of n for which u_n > 10 000.[2]
Show answer
(a) Proof by induction (see working).
B1: n = 1: 2 × 3⁰ − 1 = 1 = u₁.M1: assume u_k = 2 × 3^(k−1) − 1; u_(k+1) = 3(2 × 3^(k−1) − 1) + 2.A1: = 2 × 3ᵏ − 3 + 2 = 2 × 3ᵏ − 1.A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q7Sequences & inequalities · Standard · Paper 1 style · 7 marks
(a)Prove by mathematical induction that, for all positive integers n,Σ (from r = 1 to n) r × 2ʳ = (n − 1)2^(n+1) + 2.[5]
(b)Hence find the value of Σ (from r = 11 to 20) r × 2ʳ.[2]
Show answer
(a) Proof by induction (see working).
B1: n = 1: LHS = 2, RHS = 0 + 2 = 2.M1: assume true for n = k; add (k + 1)2^(k+1).M1: (k − 1)2^(k+1) + (k + 1)2^(k+1) + 2 = 2k × 2^(k+1) + 2.A1: = k × 2^(k+2) + 2 = ((k + 1) − 1)2^((k+1)+1) + 2.A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q8Sequences & inequalities · Standard · Paper 1 style · 6 marks
(a)Prove by mathematical induction that 2ⁿ > n² for all integers n ≥ 5.[5]
(b)State the values of n, with 1 ≤ n ≤ 4, for which 2ⁿ > n² is false.[1]
Show answer
(a) Proof by induction (see working).
B1: n = 5: 2⁵ = 32 > 25.M1: assume 2ᵏ > k² for some k ≥ 5; then 2^(k+1) = 2 × 2ᵏ > 2k².M1: 2k² − (k + 1)² = k² − 2k − 1 = (k − 1)² − 2 > 0 for k ≥ 5 (indeed for k ≥ 3).A1: so 2^(k+1) > (k + 1)².A1: true for n = 5 and true for k + 1 whenever true for k ≥ 5, so true for all n ≥ 5.
Q9Sequences & inequalities · Standard · Paper 1 style · 6 marks
(a)The sequence u₁, u₂, u₃, … is defined by u₁ = 2 and u_(n+1) = u_n/(1 + u_n). Prove by mathematical induction that u_n = 2/(2n − 1) for all positive integers n.[5]
(b)Find the value of u₁₀₀ as a fraction.[1]
Show answer
(a) Proof by induction (see working).
B1: n = 1: 2/(2 − 1) = 2 = u₁.M1: assume u_k = 2/(2k − 1); u_(k+1) = [2/(2k − 1)] ÷ [1 + 2/(2k − 1)].M1: = [2/(2k − 1)] ÷ [(2k + 1)/(2k − 1)].A1: = 2/(2k + 1) = 2/(2(k + 1) − 1).A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
(b) u₁₀₀ = 2/199
B1: 2/(200 − 1).
Q10Sequences & inequalities · Standard · Paper 1 style · 6 marks
(a)Prove by mathematical induction that 3ⁿ ≥ 2n² + 1 for all positive integers n.[6]
Show answer
(a) Proof by induction (see working).
B1: n = 1: 3 ≥ 3 true.M1: assume 3ᵏ ≥ 2k² + 1; then 3^(k+1) ≥ 3(2k² + 1) = 6k² + 3.M1: compare with 2(k + 1)² + 1 = 2k² + 4k + 3.M1: (6k² + 3) − (2k² + 4k + 3) = 4k² − 4k = 4k(k − 1) ≥ 0 for k ≥ 1.A1: so 3^(k+1) ≥ 2(k + 1)² + 1.A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q11Sequences & inequalities · Challenge · Paper 1 style · 8 marks
(a)The sequence u₁, u₂, u₃, … is defined by u₁ = 1 and u_(n+1) = √(u_n + 6). Prove by mathematical induction that 0 < u_n < 3 for all positive integers n.[4]
(b)Prove that u_(n+1) > u_n for all positive integers n.[3]
(c)Find u₃ correct to 4 decimal places.[1]
Show answer
(a) Proof by induction (see working).
B1: n = 1: 0 < 1 < 3.M1: assume 0 < u_k < 3; then 6 < u_k + 6 < 9.A1: so √6 < u_(k+1) < 3, hence 0 < u_(k+1) < 3.A1: conclusion for all n.
(b) Proof (see working).
M1: u_(n+1)² − u_n² = u_n + 6 − u_n² = −(u_n − 3)(u_n + 2).A1: since 0 < u_n < 3, (u_n − 3) < 0 and (u_n + 2) > 0, so u_(n+1)² > u_n².A1: both terms positive, so u_(n+1) > u_n.
(c) u₃ = 2.9404
B1: u₂ = √7 = 2.64575…, u₃ = √8.64575… = 2.9404.
Q12Sequences & inequalities · Challenge · Paper 1 style · 7 marks
(a)A sequence is defined by u₁ = 1, u₂ = 5 and u_(n+2) = 5u_(n+1) − 6u_n for n ≥ 1. Prove by mathematical induction that u_n = 3ⁿ − 2ⁿ for all positive integers n.[6]
(b)Find u₈.[1]
Show answer
(a) Proof by induction (see working).
B1: n = 1: 3 − 2 = 1; n = 2: 9 − 4 = 5 (both base cases needed).M1: assume u_k = 3ᵏ − 2ᵏ and u_(k+1) = 3^(k+1) − 2^(k+1).M1: u_(k+2) = 5(3^(k+1) − 2^(k+1)) − 6(3ᵏ − 2ᵏ).M1: = 3ᵏ(15 − 6) − 2ᵏ(10 − 6) = 9 × 3ᵏ − 4 × 2ᵏ.A1: = 3^(k+2) − 2^(k+2).A1: true for n = 1, 2 and if true for n = k and k + 1 then true for k + 2, so true for all positive integers n.
(b) u₈ = 6305
B1: 3⁸ − 2⁸ = 6561 − 256 = 6305.
Q13Sequences & inequalities · Challenge · Paper 1 style · 7 marks
(a)Prove by mathematical induction that, for all positive integers n,Σ (from r = 1 to n) 1/r² ≤ 2 − 1/n.[6]
(b)Deduce an upper bound for Σ (from r = 1 to ∞) 1/r².[1]
Show answer
(a) Proof by induction (see working).
B1: n = 1: 1 ≤ 2 − 1 = 1.M1: assume Σ (1 to k) 1/r² ≤ 2 − 1/k; then Σ (1 to k + 1) ≤ 2 − 1/k + 1/(k + 1)².M1: need −1/k + 1/(k + 1)² ≤ −1/(k + 1), i.e. 1/(k + 1)² ≤ 1/k − 1/(k + 1) = 1/(k(k + 1)).A1: true since k(k + 1) < (k + 1)².A1: so Σ (1 to k + 1) 1/r² ≤ 2 − 1/(k + 1).A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
(b) 2 (the sum is at most 2)
B1: 2 − 1/n < 2 for every n, so the partial sums are bounded above by 2.
Q14Sequences & inequalities · Challenge · Paper 1 style · 8 marks
(a)A calculator algorithm for √3 uses u₁ = 2 and u_(n+1) = (u_n² + 3)/(2u_n). Find u₃ as an exact fraction.[2]
(b)Show that u_(n+1) − √3 = (u_n − √3)²/(2u_n).[2]
(c)Hence prove by mathematical induction that u_n > √3 for all positive integers n.[4]
B1: n = 1: 2 > √3.M1: assume u_k > √3 (so u_k > 0); by (b) u_(k+1) − √3 = (u_k − √3)²/(2u_k).A1: numerator > 0 (as u_k ≠ √3) and denominator > 0, so u_(k+1) > √3.A1: conclusion for all n.
Q15nth derivatives · Core · Paper 1 style · 5 marks
(a)Given that y = x eˣ, prove by mathematical induction that dⁿy/dxⁿ = (x + n)eˣ for all positive integers n.[5]
Show answer
(a) Proof by induction (see working).
B1: n = 1: dy/dx = eˣ + x eˣ = (x + 1)eˣ.M1: assume dᵏy/dxᵏ = (x + k)eˣ; differentiate using the product rule.A1: d^(k+1)y/dx^(k+1) = eˣ + (x + k)eˣ = (x + k + 1)eˣ.A1: form (x + (k + 1))eˣ clearly stated.A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q16nth derivatives · Core · Paper 1 style · 6 marks
(a)Given that f(x) = 1/(2x + 1), prove by mathematical induction that f⁽ⁿ⁾(x) = (−2)ⁿ n!/(2x + 1)^(n+1) for all positive integers n.[5]
(b)Hence find the value of f⁽⁴⁾(0).[1]
Show answer
(a) Proof by induction (see working).
B1: n = 1: f′(x) = −2(2x + 1)⁻² = (−2)¹ 1!/(2x + 1)².M1: assume f⁽ᵏ⁾(x) = (−2)ᵏ k!(2x + 1)^(−(k+1)); differentiate.A1: f⁽ᵏ⁺¹⁾(x) = (−2)ᵏ k! × (−(k + 1)) × 2 × (2x + 1)^(−(k+2)).A1: = (−2)^(k+1)(k + 1)!/(2x + 1)^(k+2).A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
(b) 384
B1: (−2)⁴ × 4! / 1 = 16 × 24 = 384.
Q17nth derivatives · Standard · Paper 1 style · 7 marks
(a)Given that y = ln(1 + x), x > −1, prove by mathematical induction that dⁿy/dxⁿ = (−1)^(n−1)(n − 1)!/(1 + x)ⁿ for all positive integers n.[5]
(b)Hence write down the coefficient of x⁶ in the Maclaurin series for ln(1 + x).[2]
Show answer
(a) Proof by induction (see working).
B1: n = 1: dy/dx = 1/(1 + x) = (−1)⁰ 0!/(1 + x)¹.M1: assume dᵏy/dxᵏ = (−1)^(k−1)(k − 1)!(1 + x)^(−k); differentiate.A1: = (−1)^(k−1)(k − 1)! × (−k)(1 + x)^(−k−1).A1: = (−1)ᵏ k!/(1 + x)^(k+1).A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q18nth derivatives · Standard · Paper 1 style · 7 marks
(a)Given that y = sin 3x, prove by mathematical induction that dⁿy/dxⁿ = 3ⁿ sin(3x + nπ/2) for all positive integers n.[5]
(b)Find the exact value of d⁵y/dx⁵ when x = π/9.[2]
Show answer
(a) Proof by induction (see working).
B1: n = 1: dy/dx = 3 cos 3x = 3 sin(3x + π/2).M1: assume dᵏy/dxᵏ = 3ᵏ sin(3x + kπ/2); differentiate.A1: d^(k+1)y/dx^(k+1) = 3^(k+1) cos(3x + kπ/2).A1: = 3^(k+1) sin(3x + kπ/2 + π/2) = 3^(k+1) sin(3x + (k + 1)π/2).A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q19nth derivatives · Standard · Paper 1 style · 6 marks
(a)Given that y = x² e^(2x), prove by mathematical induction thatdⁿy/dxⁿ = 2^(n−2) e^(2x) (4x² + 4nx + n(n − 1)) for all positive integers n.[6]
Show answer
(a) Proof by induction (see working).
B1: n = 1: dy/dx = 2x e^(2x) + 2x² e^(2x) = 2⁻¹ e^(2x)(4x² + 4x) ✓.M1: assume true for n = k; differentiate the product.M1: d/dx[2^(k−2) e^(2x)(4x² + 4kx + k(k − 1))] = 2^(k−2) e^(2x)[2(4x² + 4kx + k² − k) + 8x + 4k].A1: = 2^(k−1) e^(2x)[4x² + 4(k + 1)x + k² + k].A1: k² + k = (k + 1)k, giving the formula with n = k + 1.A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q20nth derivatives · Challenge · Paper 1 style · 7 marks
(a)Given that y = eˣ cos x, prove by mathematical induction that dⁿy/dxⁿ = 2^(n/2) eˣ cos(x + nπ/4) for all positive integers n.[6]
(b)Hence find the value of d⁸y/dx⁸ when x = 0.[1]
Show answer
(a) Proof by induction (see working).
B1: n = 1: dy/dx = eˣ(cos x − sin x) = √2 eˣ cos(x + π/4).M1: assume dᵏy/dxᵏ = 2^(k/2) eˣ cos(x + kπ/4); differentiate.A1: = 2^(k/2) eˣ[cos(x + kπ/4) − sin(x + kπ/4)].M1: use cos A − sin A = √2 cos(A + π/4).A1: = 2^((k+1)/2) eˣ cos(x + (k + 1)π/4).A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
(b) 16
B1: 2⁴ × e⁰ × cos 2π = 16.
Q21nth derivatives · Challenge · Paper 1 style · 7 marks
(a)Given that y = x ln x, x > 0, prove by mathematical induction that, for all integers n ≥ 2,dⁿy/dxⁿ = (−1)ⁿ (n − 2)!/x^(n−1).[6]
(b)Find the value of d⁶y/dx⁶ when x = 2.[1]
Show answer
(a) Proof by induction (see working).
M1: dy/dx = ln x + 1.B1: n = 2: d²y/dx² = 1/x = (−1)² 0!/x¹.M1: assume dᵏy/dxᵏ = (−1)ᵏ (k − 2)! x^(−(k−1)); differentiate.A1: = (−1)ᵏ (k − 2)! × (−(k − 1)) x^(−k).A1: = (−1)^(k+1) (k − 1)!/xᵏ, the formula with n = k + 1.A1: true for n = 2 and true for k + 1 when true for k ≥ 2, so true for all n ≥ 2.
(b) 3/4
B1: (−1)⁶ 4!/2⁵ = 24/32 = 3/4.
Q22Divisibility · Core · Paper 1 style · 5 marks
(a)Prove by mathematical induction that 5ⁿ + 3 is divisible by 4 for all positive integers n.[5]
Show answer
(a) Proof by induction (see working).
B1: n = 1: 5 + 3 = 8 = 4 × 2.M1: assume 5ᵏ + 3 = 4m for some integer m.M1: 5^(k+1) + 3 = 5 × 5ᵏ + 3 = 5(4m − 3) + 3.A1: = 20m − 12 = 4(5m − 3), divisible by 4.A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q23Divisibility · Core · Paper 1 style · 5 marks
(a)Prove by mathematical induction that 7ⁿ − 3ⁿ is divisible by 4 for all positive integers n.[5]
Show answer
(a) Proof by induction (see working).
B1: n = 1: 7 − 3 = 4.M1: assume 7ᵏ − 3ᵏ = 4m.M1: 7^(k+1) − 3^(k+1) = 7 × 7ᵏ − 3 × 3ᵏ = 7(7ᵏ − 3ᵏ) + 4 × 3ᵏ.A1: = 28m + 4 × 3ᵏ = 4(7m + 3ᵏ), divisible by 4.A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q24Divisibility · Standard · Paper 1 style · 6 marks
(a)Let f(n) = n³ + 5n. Show that f(n + 1) − f(n) = 3n(n + 1) + 6.[2]
(b)Hence prove by mathematical induction that n³ + 5n is divisible by 6 for all positive integers n.[4]
B1: n = 1: f(1) = 6.M1: assume 6 | f(k); f(k + 1) = f(k) + 3k(k + 1) + 6.A1: k(k + 1) is a product of consecutive integers, so even; hence 3k(k + 1) is divisible by 6, and so f(k + 1) is divisible by 6.A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q25Divisibility · Standard · Paper 1 style · 7 marks
(a)Prove by mathematical induction that 3^(2n) + 7 is divisible by 8 for all positive integers n.[5]
(b)Find the remainder when 3^(2n) + 7 is divided by 16 in the case n = 2. Hence explain why the statement '3^(2n) + 7 is divisible by 16 for all positive integers n' is false, even though it is true for n = 1.[2]
Show answer
(a) Proof by induction (see working).
B1: n = 1: 9 + 7 = 16 = 8 × 2.M1: assume 3^(2k) + 7 = 8m.M1: 3^(2k+2) + 7 = 9 × 3^(2k) + 7 = 9(8m − 7) + 7.A1: = 72m − 56 = 8(9m − 7).A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
(b) Remainder 8; n = 2 is a counter-example (88 is not a multiple of 16).
M1: n = 2: 81 + 7 = 88 = 5 × 16 + 8.A1: remainder 8 ≠ 0, so one counter-example disproves the statement; a true base case alone proves nothing.
Q26Divisibility · Challenge · Paper 1 style · 6 marks
(a)Prove by mathematical induction that 2^(n+2) + 3^(2n+1) is divisible by 7 for all positive integers n.[6]
Show answer
(a) Proof by induction (see working).
B1: n = 1: 2³ + 3³ = 8 + 27 = 35 = 7 × 5.M1: assume 2^(k+2) + 3^(2k+1) = 7m.M1: 2^(k+3) + 3^(2k+3) = 2 × 2^(k+2) + 9 × 3^(2k+1).M1: = 2(2^(k+2) + 3^(2k+1)) + 7 × 3^(2k+1).A1: = 14m + 7 × 3^(2k+1) = 7(2m + 3^(2k+1)).A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q27Divisibility · Challenge · Paper 1 style · 6 marks
(a)Prove by mathematical induction that 4^(n+1) + 5^(2n−1) is divisible by 21 for all positive integers n.[6]
Show answer
(a) Proof by induction (see working).
B1: n = 1: 4² + 5¹ = 21.M1: assume 4^(k+1) + 5^(2k−1) = 21m.M1: 4^(k+2) + 5^(2k+1) = 4 × 4^(k+1) + 25 × 5^(2k−1).M1: = 4(4^(k+1) + 5^(2k−1)) + 21 × 5^(2k−1).A1: = 21(4m + 5^(2k−1)), divisible by 21.A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q28Matrix powers · Core · Paper 1 style · 6 marks
(a)The matrix A is given by A = (2 1; 0 1), where rows are separated by a semicolon. Prove by mathematical induction that, for all positive integers n,Aⁿ = (2ⁿ 2ⁿ − 1; 0 1).[5]
(b)Hence write down A¹⁰.[1]
Show answer
(a) Proof by induction (see working).
B1: n = 1: (2 1; 0 1) = A.M1: assume Aᵏ = (2ᵏ 2ᵏ − 1; 0 1); form A^(k+1) = Aᵏ A.A1: = (2^(k+1) 2ᵏ + 2ᵏ − 1; 0 1).A1: = (2^(k+1) 2^(k+1) − 1; 0 1).A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
(b) A¹⁰ = (1024 1023; 0 1)
B1: 2¹⁰ = 1024.
Q29Matrix powers · Standard · Paper 1 style · 7 marks
(a)The matrix M is given by M = (4 −3; 2 −1), rows separated by a semicolon. Prove by mathematical induction that, for all positive integers n,Mⁿ = (3 × 2ⁿ − 2 3 − 3 × 2ⁿ ; 2^(n+1) − 2 3 − 2^(n+1)).[6]
(b)Find M⁵.[1]
Show answer
(a) Proof by induction (see working).
B1: n = 1: (6 − 2, 3 − 6; 4 − 2, 3 − 4) = (4 −3; 2 −1) = M.M1: assume formula for Mᵏ; form M^(k+1) = Mᵏ M.M1: top-left: 4(3 × 2ᵏ − 2) + 2(3 − 3 × 2ᵏ) = 6 × 2ᵏ − 2 = 3 × 2^(k+1) − 2; top-right: −3(3 × 2ᵏ − 2) − (3 − 3 × 2ᵏ) = 3 − 6 × 2ᵏ = 3 − 3 × 2^(k+1).A1: bottom-left: 4(2^(k+1) − 2) + 2(3 − 2^(k+1)) = 2^(k+2) − 2.A1: bottom-right: −3(2^(k+1) − 2) − (3 − 2^(k+1)) = 3 − 2^(k+2).A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
Q30Matrix powers · Challenge · Paper 1 style · 8 marks
(a)The 3 × 3 matrix B has rows (1, 1, 0), (0, 1, 1) and (0, 0, 1). Prove by mathematical induction that, for all positive integers n, Bⁿ has rows(1, n, n(n − 1)/2), (0, 1, n) and (0, 0, 1).[6]
(b)Hence find the value of n for which the top-right entry of Bⁿ equals 190.[2]
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(a) Proof by induction (see working).
B1: n = 1: rows (1, 1, 0), (0, 1, 1), (0, 0, 1) = B.M1: assume formula for Bᵏ; form B^(k+1) = Bᵏ B.A1: first row: (1, 1 + k, k + k(k − 1)/2).M1: k + k(k − 1)/2 = k(k + 1)/2 = (k + 1)((k + 1) − 1)/2.A1: second row (0, 1, 1 + k), third row (0, 0, 1).A1: correct conclusion stating that the result is true for n = k + 1 when true for n = k, and true for n = 1, so true for all positive integers n.
