Standard +0.3 This is a straightforward proof by induction with a simple inequality. The base case n=4 is trivial (81>40), and the inductive step requires only basic algebraic manipulation showing 3^(k+1) > 10(k+1) from 3^k > 10k, which works cleanly since multiplying by 3 gives enough 'room' to absorb the extra 10. This is slightly easier than average as it's a standard textbook-style induction with no tricky inequalities or creative insights needed.
If \(n=4\), LHS \(= 3^4 = 81\); RHS \(= 10\times4 = 40 < 81 =\) LHS. So true for \(n=4\)
B1
Basis case. Comparison must be explicit and correct. An assertion without calculation such as e.g. \(3^4 > 10\times4\) is insufficient for B1. BOD statements such as "therefore true when \(n=1\)"
Considering for \(k+1\) and using inductive hypothesis correctly. Asserting \(3^{k+1}>10(k+1)\) without justification gets M0.
\(\ldots = 30k = 10(k+1) + 10(2k-1) > 10(k+1)\) since \(2k-1>0\) since \(k\geq4\), i.e. if \(3^k > 10k\) then \(3^{k+1} > 10(k+1)\)
A1
Showing enough working to establish statement for \(k+1\). Must be justified but justification could be e.g. \(k>1\). Could compare \(3k\) and \(k+1\)
So true for \(n=k \Rightarrow\) true for \(n=k+1\). But true for \(n=4\). So true for all integers \(n \geq 4\)
A1
Clear and complete conclusion, following a correct and complete proof with no incorrect statements. Must be \(n\geq4\) 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.
[5]
# Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| If $n=4$, LHS $= 3^4 = 81$; RHS $= 10\times4 = 40 < 81 =$ LHS. So true for $n=4$ | B1 | Basis case. Comparison must be explicit and correct. An assertion without calculation such as e.g. $3^4 > 10\times4$ is insufficient for B1. BOD statements such as "therefore true when $n=1$" |
| (Assume that) $3^k > 10k$ (for some integer $k \geq 4$) | M1 | Inductive hypothesis set up |
| $3^{k+1} = 3\times3^k > 3\times10k\ldots$ | M1 | Considering for $k+1$ and using inductive hypothesis correctly. Asserting $3^{k+1}>10(k+1)$ without justification gets M0. |
| $\ldots = 30k = 10(k+1) + 10(2k-1) > 10(k+1)$ since $2k-1>0$ since $k\geq4$, i.e. if $3^k > 10k$ then $3^{k+1} > 10(k+1)$ | A1 | Showing enough working to establish statement for $k+1$. Must be justified but justification could be e.g. $k>1$. Could compare $3k$ and $k+1$ |
| So true for $n=k \Rightarrow$ true for $n=k+1$. But true for $n=4$. So true for all integers $n \geq 4$ | A1 | Clear and complete conclusion, following a **correct and complete** proof with no incorrect statements. **Must** be $n\geq4$ 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. |
| **[5]** | | |
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