Kai is proving that $n^3 - n$ is a multiple of 3 for all positive integer values of $n$. Kai begins a proof by exhaustion. Step 1 $n^3 - n = n(n^2 - 1)$ Step 2 When $n = 3m$, where $m$ is a non-negative integer, $n^3 - n = 3m(9m^2 - 1)$, which is a multiple of 3. Step 3 When $n = 3m + 1$, $n^3 - n = (3m + 1)^3 - (3m + 1)$ Step 4 $= (3m + 1)(9m^2 + 6m + 1) = 3(m + 1)(3m^2 + 1)$, which is a multiple of 3. Step 5 Therefore $n^3 - n$ is a multiple of 3 for all positive integer values of $n$.

A-Level AQA Maths Pure Proof: Kai is proving that $n^3

Kai is proving that $n^3 - n$ is a multiple of 3 for all positive integer values of $n$ - AQA - A-Level Maths Pure - Question 8 - 2021 - Paper 2

Question 8

Kai is proving that $n^3 - n$ is a multiple of 3 for all positive integer values of $n$. Kai begins a proof by exhaustion. Step 1 $n^3 - n = n(n^2 - 1)$ Step 2 ... show full transcript

Worked Solution & Example Answer:Kai is proving that $n^3 - n$ is a multiple of 3 for all positive integer values of $n$ - AQA - A-Level Maths Pure - Question 8 - 2021 - Paper 2

Step 1

8 (a) Explain the two mistakes that Kai has made after Step 3.

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Answer

Algebraic Mistake: In Step 3, Kai uses the expression $(3m + 1)^3 - (3m + 1)$ but fails to expand it correctly, which leads to an incorrect formulation in Step 4. He should calculate this expression more carefully.
Incomplete Cases: Kai has only considered the cases where $n = 3m$ and $n = 3m + 1$ . He has not examined the situation where $n = 3m + 2$ , which is necessary to prove the statement for all positive integers.

Step 2

8 (b) Correct Kai's argument from Step 4 onwards.

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Answer

In Step 4, the correct expansion should show: $n^3 - n = (3m + 1)(3m^2 + 3m + 1 - 1) = (3m + 1)(9m^2 + 6m)$ .

For when $n = 3m + 2$ , we expand:

Step 4: For $n = 3m + 2$ , we have: $n^3 - n = (3m + 2)^3 - (3m + 2)$ $= (3m + 2)(9m^2 + 12m + 4 - 1)$ $= (3m + 2)(9m^2 + 12m + 3)$ . This also clearly results in terms that include a factor of 3.

Consequently, for each case:

When $n = 3m$ , $n = 3m + 1$ , and $n = 3m + 2$ , all lead to $n^3 - n$ being a multiple of 3.

Thus, we conclude that $n^3 - n$ is consistent as a multiple of 3 for all positive integer values of $n$ .

Kai is proving that $n^3 - n$ is a multiple of 3 for all positive integer values of $n$ - AQA - A-Level Maths Pure - Question 8 - 2021 - Paper 2

Worked Solution & Example Answer:Kai is proving that $n^3 - n$ is a multiple of 3 for all positive integer values of $n$ - AQA - A-Level Maths Pure - Question 8 - 2021 - Paper 2

8 (a) Explain the two mistakes that Kai has made after Step 3.

8 (b) Correct Kai's argument from Step 4 onwards.

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