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 $= (3m + 1)(9m^2 + 6m + 1 - 1)$ $= (3m + 1)(9m^2 + 6m)$ 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 Mechanics Quantities, Units & Modelling: 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 Mechanics - 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 Mechanics - Question 8 - 2021 - Paper 2

Step 1

Explain the two mistakes that Kai has made after Step 3.

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Answer

Kai made two mistakes after Step 3:

Expansion Mistake: He did not correctly expand the expression $(3m + 1)^3$ . The correct expression should be: $n^3 = (3m + 1)^3 = 27m^3 + 27m^2 + 9m + 1$ which results in ${n^3 - n} = (27m^3 + 27m^2 + 9m + 1) - (3m + 1)$ .
Exhaustion Mistake: He has only considered cases where $n = 3m$ and $n = 3m + 1$ . He failed to account for integers of the form $n = 3m + 2$ , which is necessary to complete the proof by exhaustion.

Step 2

Correct Kai's argument from Step 4 onwards.

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Answer

To correct Kai's argument:

Step 4: Re-evaluating the expression:

When $n = 3m + 1$ , we should compute: $n^3 - n = (3m + 1)^3 - (3m + 1) = 27m^3 + 27m^2 + 9m + 1 - (3m + 1)$ $= 27m^3 + 27m^2 + 6m$

This can be factored: $= 3(9m^3 + 9m^2 + 2m)$

which is a multiple of 3.

Step 5: Evaluating the case for $n = 3m + 2$ :

When $n = 3m + 2$ : $n^3 - n = (3m + 2)^3 - (3m + 2)$ $= 27m^3 + 54m^2 + 36m + 8 - (3m + 2)$ $= 27m^3 + 54m^2 + 33m + 6$ $= 3(9m^3 + 18m^2 + 11m + 2)$

This shows that $n^3 - n$ is also a multiple of 3 in this case.

Conclusion: Since the expression holds for all three forms of $n$ ( $3m$ , $3m + 1$ , and $3m + 2$ ), we conclude that $n^3 - n$ is 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 Mechanics - 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 Mechanics - Question 8 - 2021 - Paper 2

Explain the two mistakes that Kai has made after Step 3.

Correct Kai's argument from Step 4 onwards.

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