Standard +0.3 This is a straightforward proof by induction with a clear base case (n=4) and a routine inductive step that requires showing (k+1) > 2 when k! > 2^k. The inequality manipulation is simple since multiplying by (k+1) ≥ 5 easily dominates multiplying by 2. While induction is a Further Maths topic making it inherently more demanding, this particular question follows a standard template with no conceptual subtleties.
Let \(n=4\): \(4!=24\) and \(2^4=16\) so \(4!>2^4\)
B1
Basis case for proof by induction
Assume true for \(n=r\): \(r!>2^r\) for \(r\geq4\)
M1
Assumption
Then for \(n=r+1\): \((r+1)!=(r+1)\times r!>(r+1)\times2^r\) by assumption
M1
Add next statement
Since \(r+1>2\), \((r+1)\times2^r>2\times2^r=2^{r+1}\), so \((r+1)!>2^{r+1}\)
E1
Sufficient working to establish true for \(r+1\); must state \(r+1>2\); a *formal* proof required for full marks
If true for \(r\) then true for \(r+1\). Hence, given basis case, statement is true for all positive integers.
E1
Clear conclusion for induction process; accept other *complete* methods
## Question 8:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $n=4$: $4!=24$ and $2^4=16$ so $4!>2^4$ | B1 | Basis case for proof by induction |
| Assume true for $n=r$: $r!>2^r$ for $r\geq4$ | M1 | Assumption |
| Then for $n=r+1$: $(r+1)!=(r+1)\times r!>(r+1)\times2^r$ by assumption | M1 | Add next statement |
| Since $r+1>2$, $(r+1)\times2^r>2\times2^r=2^{r+1}$, so $(r+1)!>2^{r+1}$ | E1 | Sufficient working to establish true for $r+1$; must state $r+1>2$; a *formal* proof required for full marks |
| If true for $r$ then true for $r+1$. Hence, given basis case, statement is true for all positive integers. | E1 | Clear conclusion for induction process; accept other *complete* methods |
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