Standard +0.3 This is a straightforward proof by induction with a simple inequality involving factorials. The base case n=4 is trivial (24≥24), and the inductive step requires only basic algebraic manipulation showing that if k!≥6k then (k+1)!≥6(k+1), which follows immediately since (k+1)!=(k+1)·k!≥(k+1)·6k≥6(k+1) for k≥4. While it's a Further Maths question, it's a standard textbook induction exercise requiring no novel insight.
Must have statement in terms of some other variable than \(n\)
Answer: \((k + 1)! = (k + 1)k! \geq (k + 1)×6k\)
M1
Uses inductive hypothesis properly
Answer: \(\geq 6(k + 1)\) since \(k \geq 1\)
A1
Sufficient working to establish truth for \(k + 1\). Must justify with \(k \geq 1\) or \(k \geq 4\) oe
Answer: So true for \(n = k \Rightarrow\) true for \(n = k + 1\). But true for \(n = 4\). So true for all \(n \geq 4\)
E1
Clear conclusion for induction process. A formal proof by induction is required for full marks.
Answer: $n = 4: 4! = 24 \geq 6×4 = 24$ | B1 | Clear demonstration of equality is sufficient
Answer: Assume true for $n = k$ ie $k! \geq 6k$ | M1 | Must have statement in terms of some other variable than $n$
Answer: $(k + 1)! = (k + 1)k! \geq (k + 1)×6k$ | M1 | Uses inductive hypothesis properly
Answer: $\geq 6(k + 1)$ since $k \geq 1$ | A1 | Sufficient working to establish truth for $k + 1$. Must justify with $k \geq 1$ or $k \geq 4$ oe
Answer: So true for $n = k \Rightarrow$ true for $n = k + 1$. But true for $n = 4$. So true for all $n \geq 4$ | E1 | Clear conclusion for induction process. A formal proof by induction is required for full marks.
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