Proof
Algebra
- Higher
- Non-calculator
- 8 questions
- 29 marks
These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.
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Prove that the sum of two even numbers is even.
2 marks
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Prove that the sum of two odd numbers is even.
2 marks
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Prove that the product of two odd numbers is odd.
3 marks
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Prove that the difference of the squares of two consecutive integers is odd.
3 marks
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Prove that $(3n + 1)^{2} - (3n - 1)^{2}$ is a multiple of $12$.
4 marks
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Prove that $n^{2} - n$ is even for every integer $n$.
4 marks
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Prove that the sum of three consecutive integers is a multiple of $3$.
5 marks
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Prove that $(2n + 1)^{2} - (2n - 1)^{2} = 8n$. Then prove that the sum of two consecutive triangular numbers is a square number. The $n$th triangular number is $\dfrac{n(n + 1)}{2}$.
6 marks
Mark scheme
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Answer. The sum is $2(m + n)$.
- Write the numbers as $2m$ and $2n$.
- The sum is $2(m + n)$, which is even.
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Answer. The sum is $2(m + n + 1)$.
- Write the numbers as $2m + 1$ and $2n + 1$.
- The sum is $2m + 2n + 2 = 2(m + n + 1)$.
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Answer. The product is one more than a multiple of $2$.
- $(2m + 1)(2n + 1) = 4mn + 2m + 2n + 1$
- $= 2(2mn + m + n) + 1$, which is odd.
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Answer. The difference is $2n + 1$.
- $(n + 1)^{2} - n^{2} = 2n + 1$
- $2n + 1$ is odd.
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Answer. The difference is $12n$.
- Expand: $(9n^{2} + 6n + 1) - (9n^{2} - 6n + 1)$.
- The difference is $12n$, a multiple of $12$.
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Answer. $n(n - 1)$ is the product of two consecutive integers.
- $n^{2} - n = n(n - 1)$
- Of two consecutive integers, one is even, so the product is even.
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Answer. The sum is $3(n + 1)$.
- The integers are $n$, $n + 1$ and $n + 2$.
- The sum is $3n + 3 = 3(n + 1)$.
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Answer. The difference is $8n$. The triangular sum is $n^{2}$.
- $(4n^{2} + 4n + 1) - (4n^{2} - 4n + 1) = 8n$.
- $T_{n-1} + T_{n} = \dfrac{(n - 1)n}{2} + \dfrac{n(n + 1)}{2}$.
- $= \dfrac{n}{2}\left((n - 1) + (n + 1)\right) = n^{2}$.