AS June 2022 Paper 1 Q4
4 Prove that \(3^n \gt 10n\) for all integers \(n \geqslant 4\). [5]
| Scheme | Marks | AO |
|---|---|---|
| If \(n = 4\), LHS \(= 3^4 = 81\) RHS \(= 10 \times 4 = 40 \lt 81 =\) LHS So true for \(n = 4\) | B1 | 2.5 |
| (Assume that) \(3^k \gt 10k\) (for some integer \(k \geqslant 4\)). | M1 | 2.1 |
| \(3^{k+1} = 3 \times 3^k \gt 3 \times 10k\ldots\) | M1 | 1.1 |
| \(\ldots = 30k = 10(k + 1) + 10(2k - 1) \gt 10(k + 1)\) since \(2k - 1 \gt 0\) since \(k \geqslant 4\) i.e. if \(3^k \gt 10k\) then \(3^{k+1} \gt 10(k + 1)\) | A1 | 2.2a |
| So true for \(n = k \Rightarrow\) true for \(n = k + 1\). But true for \(n = 4\). So true for all integers \(n \geqslant 4\) | A1 | 2.4 |
| [5] |
Notes
B1: Basis case. Comparison must be explicit and correct
An assertion without calculation such as e.g. \(3^4 \gt 10 \times 4\) is insufficient for B1.
BOD statements such as “therefore true when \(n = 1\)”
M1: (1st) Inductive hypothesis set up
M1: (2nd) Considering for \(k + 1\) and using inductive hypothesis correctly
Asserting \(3^{k+1} \gt 10(k + 1)\) without justification gets M0.
Could compare \(3k\) and \(k + 1\)
A1: (1st) Showing enough working to establish statement for \(k + 1\). Must be justified but justification could be e.g. \(k \gt 1\).
A1: (2nd) Clear and complete conclusion, following a correct and complete proof with no incorrect statements. Must be \(n \geqslant 4\) not eg 0 or 1 for A1.
This mark must only be awarded if the language and notation in the whole proof and conclusion is correct.