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Algebraic proof Quiz

Grade 8+ • 8C • 10 Questions (27 Marks)

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Algebraic proof — Notes & Worked Examples Show / Hide

Explanation:

Never use numerical examples to prove an algebraic statement—you must use general algebra for any integer n: an even number is 2n, an odd number is 2n + 1, consecutive integers are n, n + 1, n + 2, and consecutive odd numbers are 2n + 1, 2n + 3.

Expand and simplify the expression, then factorise out the required multiple (e.g. write 8n + 4 as 4(2n + 1) to prove it is a multiple of 4) and finish with a concluding sentence explaining why the result holds for all integers n.

Example 1: Prove algebraically that the difference between the squares of any two consecutive odd numbers is always a multiple of 8.

Solution:

Let the consecutive odd numbers be 2n + 1 and 2n + 3:
(2n + 3)2 − (2n + 1)2 = (4n2 + 12n + 9) − (4n2 + 4n + 1)
= 8n + 8 = 8(n + 1)
Since n + 1 is an integer, 8(n + 1) is always a multiple of 8.

Example 2: Prove that n2 − 6n + 13 is positive for all real values of n.

Solution:

Complete the square on n2 − 6n + 13:
= (n − 3)2 − 9 + 13 = (n − 3)2 + 4
Since any squared real number is ≥ 0, (n − 3)2 ≥ 0
Therefore (n − 3)2 + 4 ≥ 4 > 0, so it is always positive.
Question 1
[1 mark]
Expand and simplify (2n + 4)2 - (2n - 4)2, and hence state which integer it is always a multiple of for all positive integers n.
Worked Solution & Answer
(4n2 + 16n + 16) - (4n2 - 16n + 16) = 32n B1
Answer: 32n (multiple of 32)
Question 2
[1 mark]
Expand and simplify (2n + 2)2 - (2n - 2)2, and hence state which integer it is always a multiple of for all positive integers n.
Worked Solution & Answer
(4n2 + 8n + 4) - (4n2 - 8n + 4) = 16n B1
Answer: 16n (multiple of 16)
Question 3
[2 marks]
Work out
n is an integer.
Prove algebraically that the sum of 12n(n + 1) and 12(n + 1)(n + 2) is always a square number.
Worked Solution & Answer
Factorising 1/2(n + 1)[n + (n + 2)] = 1/2(n + 1)(2n + 2) = (n + 1)(n + 1) = (n + 1)², which is the square of an integer. M1 C1
Answer: 12(n2 + n) + 12(n2 + 3n + 2) = 12(2n2 + 4n + 2) = n2 + 2n + 1 = (n + 1)2, which is a square number
Question Paper (PDF) • Mark Scheme (PDF)
Question 4
[3 marks]
Prove algebraically that the sum of the squares of any two consecutive even numbers is always a multiple of 4
Worked Solution & Answer
Let the two consecutive even numbers be 2n and 2n + 2 M2
(2n)2 + (2n + 2)2 = 4n2 + 4n2 + 8n + 4 = 8n2 + 8n + 4 = 4(2n2 + 2n + 1), which is a multiple of 4. C1
Answer: (2n)2 + (2n + 2)2 = 8n2 + 8n + 4 = 4(2n2 + 2n + 1)
Question Paper (PDF) • Mark Scheme (PDF)
Question 5 IGCSE Jan 2023 2HR Q19
[3 marks]
Prove algebraically that, for any three consecutive even numbers,
the sum of the squares of the smallest even number and the largest even number is 8 more than twice the square of the middle even number.
Worked Solution & Answer
Let the three consecutive even numbers be 2n - 2, 2n, 2n + 2 M2
Sum of squares of smallest and largest: (2n - 2)2 + (2n + 2)2 = (4n2 - 8n + 4) + (4n2 + 8n + 4) = 8n2 + 8 = 2(2n)2 + 8, which is 8 more than twice the square of the middle even number. C1
Answer: Let the numbers be 2n, 2n + 2, 2n + 4. Then (2n)2 + (2n + 4)2 = 8n2 + 16n + 16 = 2(4n2 + 8n + 4) + 8 = 2(2n + 2)2 + 8.
Question Paper (PDF) • Mark Scheme (PDF)
Question 6 AQA Nov 2018 2H Q20
[3 marks]
n is a positive integer.
Prove algebraically that
2n2(3n + n) + 6n(n2 - 1)
is a cube number.
Worked Solution & Answer
1. Expand each part of the expression: M1
2n2(3n + n) = 2n2 × 3n + 2n2 × n = 6n + 2n3
6n(n2 - 1) = 6n3 - 6n
2. Add the two expressions together: M1
(6n + 2n3) + (6n3 - 6n) = 2n3 + 6n3 + 6n - 6n = 8n3
3. Rewrite as a perfect cube:
8n3 = (2n)3
Since n is a positive integer, 2n is an integer, so (2n)3 is a cube number. A1
Answer: Proof that expression simplifies to (2n)³
Question Paper (PDF) • Mark Scheme (PDF)
Question 7
[3 marks]
Prove algebraically that the difference between the squares of any two consecutive odd numbers is always a multiple of 8
Worked Solution & Answer
Let two consecutive odd numbers be 2n – 1 and 2n + 1. (2n + 1)² – (2n – 1)² = (4n² + 4n + 1) – (4n² – 4n + 1) = 8n, which is a multiple of 8 for all integers n. M2 C1
Answer: (2n + 1)² – (2n – 1)² = (4n² + 4n + 1) – (4n² – 4n + 1) = 8n, which is a multiple of 8
Question Paper (PDF) • Mark Scheme (PDF)
Question 8 AQA Jun 2023 3H Q19
[3 marks]
Two integers have a difference of 6
The integers are multiplied together.
9 is then added.
Prove algebraically that the result is always a square number.
Worked Solution & Answer
Let the integers be n and n + 6. M2
n(n + 6) + 9 = n² + 6n + 9 = (n + 3)², which is a square number for any integer n. C1
Answer: Let the integers be n and n + 6: n(n + 6) + 9 = n² + 6n + 9 = (n + 3)²
Question Paper (PDF) • Mark Scheme (PDF)
Question 9 IGCSE Jun 2019 2HR Q17
[4 marks]
The table gives information about the first six terms of a sequence of numbers.
Term numberTerm of sequence11 × 2222 × 3233 × 4244 × 5255 × 6266 × 72
Prove algebraically that the sum of any two consecutive terms of this sequence is always a square number.
Worked Solution & Answer
nth term + (n+1)th term = ½n(n+1) + ½(n+1)(n+2) = ½(n² + n + n² + 3n + 2) = ½(2n² + 4n + 2) = n² + 2n + 1 = (n + 1)² M3 C1
Answer: n(n+1)2 + (n+1)(n+2)2 = (n+1)(2n+2)2 = (n+1)2, which is a square number
Question Paper (PDF) • Mark Scheme (PDF)
Question 10
[4 marks]
Work out
n is an integer greater than 1
Prove algebraically that n2 - 2 - (n - 2)2 is always an even number.
Worked Solution & Answer
Expand (n – 2)² = n² – 4n + 4. Then n² – 2 – (n² – 4n + 4) = n² – 2 – n² + 4n – 4 = 4n – 6 = 2(2n – 3) M3
Since 2n – 3 is an integer, 2(2n – 3) is always a multiple of 2 (even). C1
Answer: n2 - 2 - (n2 - 4n + 4) = 4n - 6 = 2(2n - 3), which is a multiple of 2 and therefore always even.
Question Paper (PDF) • Mark Scheme (PDF)
Score: 0 / 10 (0 / 27 marks)
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Grade 8+ Quizzes

14 Topics
8A - Quad. sim. equations8B - Transforming graphs y=f(x)
8C - Algebraic proof8D - Completing the square
8E - Quadratic inequalities8F - Quadratic sequences
8G - Velocity time graphs8H - Circles and tangents
8J - Circle theorem proof8K - Vector proof
8L - Probability equation questions8M - Invariant points
8N - Spheres, cones and cylinders8P - Volume of a frustum