AS June 2025 Paper 1 Q7
7 Prove by induction that \(n! \gt 20^n\) for all integers \(n \geqslant 52\). [5]
| Scheme | Marks | AO |
|---|---|---|
| Basis case: RHS \(= 20^{52} = 4.50 \times 10^{67}\) LHS \(= 52! = 8.07 \times 10^{67} \gt 4.50 \times 10^{67} =\) RHS So \(n! \gt 20^n\) when \(n = 52\) | B1 | 2.5 |
| Assume true for \(n = k\) so \(k! \gt 20^k\) (where \(k \geqslant 52\)) | M1 | 2.1 |
| Then \((k + 1)! = (k + 1) \times k! \gt (k + 1) \times 20^k\) | M1 | 3.1a |
| \(\gt 20 \times 20^k\) since \(k + 1 \gt 20\) since \(k \geqslant 52\) \(= 20^{k+1}\) | A1 | 2.2a |
| So true for \(n = k \Rightarrow\) true for \(n = k + 1\). But true for \(n = 52\) so true for all integers \(n \geqslant 52\). | A1 | 2.4 |
| [5] |
Notes
B1: BC.
The conclusion for LHS>RHS might appear in final statement.
Ignore calculations and statements relating to \(n = 51\).
Allow \(8 \times 10^{67} \gt 4.5 \times 10^{67}\)
M1: Setting up inductive hypothesis properly.
M1: Correctly using the inductive hypothesis
Could see:
\(k! \gt 20^k\)
\((k + 1)k! \gt (k + 1)20^k\)
Allow \(k\) to “slip” in \(20^k\) as long as intention is clear
A1: Must be sufficient justification
Withhold if \(k\) slipped
A1: Clear conclusion for inductive process. Must be 52.
A formal proof is required for full marks but other complete and correct proofs can get full marks.