2 Prove, by mathematical induction, that, for integers \(n \geqslant 2\),
$$4 ^ { n } > 2 ^ { n } + 3 ^ { n }$$
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Question 2:
Answer Marks
Guidance
Answer/Working Marks
Guidance
\((P_n: 4^n > 2^n + 3^n)\) States proposition
Let \(n=2\), \(16 > 4+9 \Rightarrow P_2\) is true B1
Proves base case
Assume \(P_k\) is true \(\Rightarrow 4^k > 2^k + 3^k\) B1
States inductive hypothesis
\(4^{k+1} = 4\cdot4^k > 4(2^k+3^k) = 4\cdot2^k + 4\cdot3^k\) M1
Proves inductive step
\(> 2\cdot2^k + 3\cdot3^k = 2^{k+1} + 3^{k+1}\) A1
\(\therefore P_k \Rightarrow P_{k+1}\) Hence result true, by PMI, for all integers \(n \geq 2\) A1 (CWO)
States conclusion
Total: [5]
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## Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(P_n: 4^n > 2^n + 3^n)$ | | States proposition |
| Let $n=2$, $16 > 4+9 \Rightarrow P_2$ is true | B1 | Proves base case |
| Assume $P_k$ is true $\Rightarrow 4^k > 2^k + 3^k$ | B1 | States inductive hypothesis |
| $4^{k+1} = 4\cdot4^k > 4(2^k+3^k) = 4\cdot2^k + 4\cdot3^k$ | M1 | Proves inductive step |
| $> 2\cdot2^k + 3\cdot3^k = 2^{k+1} + 3^{k+1}$ | A1 | |
| $\therefore P_k \Rightarrow P_{k+1}$ | | |
| Hence result true, by PMI, for all integers $n \geq 2$ | A1 (CWO) | States conclusion |
**Total: [5]**
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2 Prove, by mathematical induction, that, for integers $n \geqslant 2$,
$$4 ^ { n } > 2 ^ { n } + 3 ^ { n }$$
\hfill \mbox{\textit{CAIE FP1 2012 Q2 [5]}}