AS June 2023 Paper 1 Q6
6 Prove by induction that \(4 \times 8^n + 66\) is divisible by 14 for all integers \(n \geqslant 0\). [6]
| Scheme | Marks | AO |
|---|---|---|
| \(n = 0\), \(4 \times 1 + 66 = 70 = 5 \times 14\) is divisible by 14 | B1 | 2.1 |
| Assume true for \(n = k\) ie that \(4 \times 8^k + 66\) is divisible by 14 oe | M1 | 2.1 |
| Considering \(4 \times 8^{k+1} + 66\) and rewriting it as \(4 \times 8 \times 8^k + 66\) or \(32 \times 8^k + 66\) oe | M1 | 1.1 |
| \(= 8(14p - 66) + 66\) from inductive hypothesis | M1 | 1.1 |
| \(= 112p - 462\) or \(14 \times 8p - 14 \times 33\) oe \(= 14(8p - 33)\) which is divisible by 14 | A1 | 2.2a |
| So true for \(n = k \Rightarrow\) true for \(n = k + 1\). But true for \(n = 0\). So true for all integers \(n \geqslant 0\) | A1 | 2.4 |
| [6] |
Notes
B1: Basis case. Must explicitly see 70 and state, or show, divisibility.
If \(n = 1\) (leading to \(98 = 7 \times 14\)) used and not corrected then allow this mark but withhold final mark.
M1: Statement of inductive hypothesis. Allow “\(= 14p\)” without further qualification
M1: Uses law of indices correctly to obtain expression in terms of \(8^k\) and no other exponential term
Could consider \(\mathrm{f}(k + 1) - \mathrm{f}(k)\) or similar
M1: Uses inductive hypothesis properly. Do not allow if eg \(32^k\) (ie law of indices must have been correctly used)
Allow for substitution of their \(n = k\) case into their \(n = k + 1\) case
A1: Simplification with sufficient working to establish and state divisibility for \(k + 1\).
Must either show \(14\times\) explicitly in each term or 14 is a factor
A1: Clear conclusion for induction process. See note for basis case above. Do not allow “true for all positive integers”.
Need to see if true for \(n = k\), then true for \(n = k + 1\) i.e. \(k\) and \(k + 1\)th case linked to \(n\).
Could see Proposition notation \((P_k \Rightarrow P_{k+1})\)
A formal proof by induction, with no gaps in logic, is required for full marks.
Full marks can be gained by using induction to prove that \(2 \times 8^n + 33\) is divisible by 7 for all \(n \geqslant 0\) and observing that \(4 \times 8^n + 66 = 2(2 \times 8^n + 33)\) is divisible by 2 and \(2 \times 8^n + 33\) and hence by both 2 and 7 and hence by 14