A2 June 2024 Paper 1 Q8

Notes

B1: Base case shown to be true – condone when \(n = 0\), 11 – 1 – 1 = 9 or just 9, must say that 9 is divisible by 3 or show explicitly e.g. \(9 = 3 \times 3\).
Allow base case with \(n = 1\), \(11 \times 7 - 13 - 1 = 63\) or just 63, must say that 63 is divisible by 3 or show explicitly e.g. \(63 = 3 \times 21\).

M1*: Forms correct inductive hypothesis as an equation in terms of two different letters (oe in words e.g. \(11 \times 7^k - 13^k - 1\) is divisible by 3) – condone using \(n\) e.g. \(11 \times 7^n - 13^n - 1 = 3m\) for this mark.

M1dep*: Considers \(11 \times 7^{k+1} - 13^{k+1} - 1\) and use inductive hypothesis that \(11 \times 7^k - 13^k - 1 = 3m\) - must not be using \(n\) for \(k\).

A1: Shows using the inductive hypothesis that \(11 \times 7^{k+1} - 13^{k+1} - 1\) is a multiple of 3. Some common answers are \(3(-22 \times 7^k + 4 + 13m)\) or \(3(7m + 2 - 2 \times 13^k)\) or \(3(m + 22 \times 7^k - 4 \times 13^k)\).

A1: Conclusion, dependent on B1M1M1A1 and no errors seen in their proof. Do not award this mark if \(n = 0\) not used as base case. However, this mark can be awarded if \(n = 1\) used as base case and \(n = 0\) considered separately. Must mention, ‘\(n = k\) or P(\(k\))’, ‘\(n = k + 1\) or P(\(k + 1\))’, ‘\(n = 0\) or base case or P(0)’ (provided P(\(n\)) is defined) and ‘\(n \geqslant 0\) or all \(n\) oe’ condone ‘all integers’ but not ‘positive integers’ or similar incorrect statement.

Alternative method

Scheme	Marks
Base case: when \(n = 0, 11 \times 7^0 - 13^0 - 1 = 9\) which is divisible by 3.	B1
Assume that, when \(n = k, 11 \times 7^n - 13^n - 1\) is divisible by 3. That is, \(\mathrm{f}(k) = 11 \times 7^k - 13^k - 1\) is divisible by 3.	M1*
\(\mathrm{f}(k + 1) - \mathrm{f}(k) = (11 \times 7^{k+1} - 13^{k+1} - 1) - (11 \times 7^k - 13^k - 1)\) \(= 6 \times 11 \times 7^k - 12 \times 13^k\)	M1dep*
\(\mathrm{f}(k + 1) = \mathrm{f}(k) + 3(22 \times 7^k - 4 \times 13^k)\) (which is divisible by 3).	A1
So, if true for \(n = k\) this implies true for \(n = k + 1\). True when \(n = 0\) so therefore true for all integers \(n \geqslant 0\).	A1
	[5]

B1: Base case shown to be true – condone when \(n = 0\), 11 – 1 – 1 = 9 or just 9, must say that 9 is divisible by 3 or show explicitly e.g. \(9 = 3 \times 3\).
Allow base case with \(n = 1\), \(11 \times 7 - 13 - 1 = 63\) or just 63, must say that 63 is divisible by 3 or show explicitly e.g. \(63 = 3 \times 21\).

M1*: Forms correct inductive hypothesis - condone using \(n\) e.g. \(\mathrm{f}(n) = 11 \times 7^n - 13^n - 1\) for this mark. Do not need to use \(\mathrm{f}(k)\) – can just state that \(11 \times 7^k - 13^k - 1\) is divisible by 3.

M1dep*: Considers correct expression for \(\mathrm{f}(k + 1) - \mathrm{f}(k)\) and simplifies to an expression of the form \(\pm a \times 7^k \pm b \times 13^k\) - must not be using \(n\) for \(k\).

A1: Shows using the inductive hypothesis that \(\mathrm{f}(k + 1)\) is divisible by 3 so must be considering an expression for \(\mathrm{f}(k + 1)\) and not just an expression for \(\mathrm{f}(k + 1) - \mathrm{f}(k)\).

A1: Conclusion, dependent on B1M1M1A1 and no errors seen in their proof. Do not award this mark if \(n = 0\) not used as base case. However, this mark can be awarded if \(n = 1\) used as base case and \(n = 0\) considered separately. Must mention, ‘\(n = k\) or P(\(k\))’, ‘\(n = k + 1\) or P(\(k + 1\))’, ‘\(n = 0\) or base case or P(0)’ (provided P(\(n\)) is defined) and ‘\(n \geqslant 0\) or all \(n\) oe’ condone ‘all integers’ but not ‘positive integers’ or similar incorrect statement.