A2 June 2023 Paper 2 Q12
12 The function \(\mathrm{f}\) is defined by
\[\mathrm{f}(n) = 3^{3n+1} + 2^{3n+4} \qquad \left(n \in \mathbb{Z}^+\right)\]Prove by induction that \(\mathrm{f}(n)\) is divisible by 19 for \(n \geqslant 1\) [6 marks]
| Scheme | Marks | AO |
|---|---|---|
| Shows that \(\mathrm{f}(n)\) is divisible by 19 for \(n = 1\) | B1 | 1.1b |
| States the assumption that \(\mathrm{f}(n)\) is divisible by 19 for \(n = k\) | M1 | 2.4 |
| Expresses \(\mathrm{f}(k + 1)\) in terms of \(k\) | M1 | 3.1a |
| Expresses \(\mathrm{f}(k + 1)\) or \(\mathrm{f}(k + 1) - \mathrm{f}(k)\) in the form \(a\left(3^{3k+1}\right) + b\left(2^{3k+4}\right)\) | M1 | 3.1a |
| Completes reasoned working to correctly deduce that \(\mathrm{f}(k + 1)\) is divisible by 19 | R1 | 2.2a |
| Concludes a reasoned argument by stating that \(\mathrm{f}(n)\) is divisible by 19 for \(n = 1\); if true for \(n = k\), then it’s also true for \(n = k + 1\) and hence by induction \(\mathrm{f}(n)\) is divisible by 19 for \(n \geqslant 1\) | R1 | 2.1 |
| (6 marks) |
Typical solution
Let \(n = 1\); then the formula gives
\[\mathrm{f}(1) = 3^4 + 2^7 = 209 = 11 \times 19\]so the result is true for \(n = 1\)
Assume the result is true for \(n = k\), so
\[\mathrm{f}(k) = 3^{3k+1} + 2^{3k+4} = 19m \quad (m \in \mathbb{Z})\]Then
\[\begin{aligned} \mathrm{f}(k + 1) &= 3^{3k+4} + 2^{3k+7} \\ &= 27\left(3^{3k+1}\right) + 8\left(2^{3k+4}\right) \\ &= 19\left(3^{3k+1}\right) + 8\left(3^{3k+1}\right) + 8\left(2^{3k+4}\right) \\ &= 19\left(3^{3k+1}\right) + 8\mathrm{f}(k) \\ &= 19\left(3^{3k+1} + 8m\right) \end{aligned}\]and the result also holds for \(n = k + 1\)
\(\mathrm{f}(n)\) is divisible by 19 for \(n = 1\); if true for \(n = k\), then it’s also true for \(n = k + 1\) and hence by induction \(\mathrm{f}(n)\) is divisible by 19 for \(n \geqslant 1\)