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Edexcel IGCSE Maths (4MA1) · Unit 7: Vectors & Proof

Edexcel IGCSE 4MA1 Proof: Exam Questions and Answers

Proof questions with full solutions for Pearson Edexcel International GCSE Mathematics A (4MA1), Higher tier. Read the 3 worked examples first, then try the 8 exam-style questions yourself and check your working against the mark scheme.

Practise proof in Unit 7 →Proof revision notes

Try it first

1 of 5

  1. Question 1 of 5
  2. Question 2 of 5
  3. Question 3 of 5
  4. Question 4 of 5
  5. Question 5 of 5

Start with one question on proof and counterexamples. Stuck? Take a hint. No sign-in needed.

Which value is the smallest counterexample to the statement “2p−1 is prime for every prime number p”?
Choose your answer · keys 1–4

What this topic covers

Proof worked examples

Worked example 1: Algebraic proof

Prove that the sum of any three consecutive integers is a multiple of 3.

Solution

4 steps
Tip: Use algebra, not examples, in a proof.

Worked example 2: Geometric proof

ABC is a triangle with AB = AC. M is the midpoint of BC. Prove that angle ABC = angle ACB.

Triangle ABC with AB = AC (single ticks) and M the midpoint of BC (double ticks); AM dashed

Solution

4 steps
Tip: Give a reason for every statement.

Worked example 3: Harder algebraic proof

Prove that (n+3)2−(n−3)2 is a multiple of 12 for all integers n.

Solution

3 steps
Tip: Bracket the second expansion to keep signs correct.

Proof exam-style questions

Try each question before you open the answer (one has its full worked solution open). The mark schemes use Edexcel-style marks: M for method, A for accuracy (after the method mark), B for an independent result.

Question 1 · easy · 2 marks

Prove that the product of an even number and an odd number is always even.

Show the answer and mark scheme
  • M1 uses 2k for the even number and 2n + 1 for the odd number and multiplies: 2k(2n + 1)
  • A1 2k(2n + 1) = 2(2kn + k), which is 2 × an integer, so even

Answer: 2k(2n+1)=2(2kn+k), 2 × an integer, so even (proved)

Question 2 · easy · 2 marks

Prove that the sum of any four consecutive integers is always 2 more than a multiple of 4.

Show the answer and mark scheme
  • M1 n + (n + 1) + (n + 2) + (n + 3) = 4n + 6
  • A1 4n + 6 = 4(n + 1) + 2, which is 2 more than a multiple of 4

Answer: 4n+6=4(n+1)+2, 2 more than a multiple of 4 (proved)

Question 3 · medium · 3 marks

Prove that the sum of any three consecutive odd numbers is always 3 more than a multiple of 6.

Full worked solution Free, no sign-in

  • M1 (2n + 1) + (2n + 3) + (2n + 5)
  • M1 6n + 9
  • A1 6n + 9 = 6(n + 1) + 3, which is 3 more than a multiple of 6

Answer: 6n+9=6(n+1)+3, 3 more than a multiple of 6 (proved)

Every question here has a full worked solution. See them all →

Question 4 · medium · 3 marks

Two consecutive odd numbers are written down. The product of the two numbers is subtracted from the square of the larger number. Prove that the result is always twice the larger number.

Show the answer and mark scheme
  • M1 uses 2n − 1 and 2n + 1 (or 2n + 1 and 2n + 3): (2n + 1)² − (2n − 1)(2n + 1)
  • M1 4n² + 4n + 1 − (4n² − 1) = 4n + 2
  • A1 4n + 2 = 2(2n + 1), which is twice the larger number

Answer: (2n+1)2−(2n−1)(2n+1)=4n+2=2(2n+1), twice the larger number (proved)

Question 5 · medium · 2 marks

n is an integer. Show that 5(n+3)−3(n+5) is always even.

Show the answer and mark scheme
  • M1 5n + 15 − 3n − 15
  • A1 2n, a multiple of 2, so even

Answer: 2n, so even (shown)

Question 6 · medium · 3 marks

Prove that the mean of the squares of any two consecutive odd numbers is always 1 more than a multiple of 4.

Show the answer and mark scheme
  • M1 (2n − 1)² + (2n + 1)² = 8n² + 2
  • M1 (8n² + 2) ÷ 2 = 4n² + 1
  • A1 4n² is a multiple of 4, so the mean is 1 more than a multiple of 4

Answer: mean =4n2+1, 1 more than a multiple of 4 (proved)

Question 7 · hard · 3 marks

Prove that when any odd number is squared and 7 is added, the result is always a multiple of 8.

Show the answer and mark scheme
  • M1 (2n + 1)² + 7 = 4n² + 4n + 8
  • M1 4n(n + 1) + 8
  • A1 n(n + 1) is a product of consecutive integers, so even; so 4n(n + 1) is a multiple of 8, and adding 8 gives a multiple of 8

Answer: (2n+1)2+7=4n(n+1)+8, with n(n+1) even, so a multiple of 8 (proved)

Question 8 · very hard · 5 marks

(a) Expand and simplify (n+2)2−n2. (b) Hence prove that the difference between the squares of two consecutive odd numbers is always a multiple of 8. (c) Is the difference between the squares of two consecutive even numbers always a multiple of 8? Give a reason for your answer.

Show the answer and mark scheme
  • M1 n² + 4n + 4 − n²
  • A1 4n + 4
  • M1 odd n = 2m + 1: 4(2m + 1) + 4 = 8m + 8
  • A1 8m + 8 = 8(m + 1), a multiple of 8
  • B1 No, with a counter-example, e.g. 4² − 2² = 12, not a multiple of 8 (for even n = 2m, 8m + 4 is never a multiple of 8)

Answer: (a) 4n+4 (b) with n=2m+1: 8m+8=8(m+1), a multiple of 8 (proved) (c) No, e.g. 42−22=12, which is not a multiple of 8

Every question and worked example here is original, written for IGCSE Math Revision rather than copied from Pearson papers, and each answer was re-solved independently before publishing.

Revise proof for the exam

Questions students ask

Which Edexcel IGCSE Maths papers test proof?

Either Higher tier paper, 1H or 2H, can test proof: both cover the whole Pearson Edexcel International GCSE Mathematics A (4MA1) specification and both allow a calculator. Show your method, because most marks are for working.

Where can I practise more proof questions?

In the IGCSE student hub: Unit 7 (Vectors & Proof) has more exam-style questions on this topic, each with a mark scheme. Practice starts with a 7-day free trial; the worked examples and questions on this page are free.

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