AS October 2020 Paper 1 Q6
6 Prove that \(n! \gt 2^{2n}\) for all integers \(n \geqslant 9\). [5]
| Scheme | Marks | AO |
|---|---|---|
| If \(n = 9\), LHS \(= 9! = 362880\) RHS \(= 2^{18} = 262144 \lt\) LHS [So true for \(n = 9\)] | B1 | 2.1 |
| Assume that \(k! \gt 2^{2k}\) for some \(k \geqslant 9\). | M1 | 2.1 |
| \((k + 1)! = (k + 1)k! \gt (k + 1) \times 2^{2k}\ldots\) | M1 | 1.1 |
| \(\ldots \gt 9 \times 2^{2k} \gt 4 \times 2^{2k} = 2^2 \times 2^{2k} = 2^{2 + 2k} = 2^{2(k+1)}\) ie \((k + 1)! \gt 2^{2(k+1)}\) | A1 | 2.2a |
| So true for \(n = k \Rightarrow\) true for \(n = k + 1\). But true for \(n = 9\). So true for all integers \(n \geqslant 9\) | A1 | 2.4 |
| [5] |
Notes
B1: Basis case. Comparison must be explicit and correct
Bod sight of “true for \(n=1\)”
M1: (1st) Inductive hypothesis set up
Might see \(2^{2k} = 4^k\) throughout
M1: (2nd) Considering for \(k + 1\) and using inductive hypothesis correctly
Allow M1 for use of inductive \(n = k\) step when showing \(P_{k+1} \to P_k\)
A1: (1st) Showing enough working to establish statement for \(k + 1\)
Might see \(\geqslant 10\) or \(\geqslant 9\) for \(\gt 9\).
A0 here if \(k \gt 9\) stated earlier
Do not allow if implication shown in the direction \(P_{k+1} \to P_k\)
A1: (2nd) Clear and complete conclusion