Easy -1.8 This is a straightforward proof by exhaustion requiring only four simple numerical checks (n=0,1,2,3) with basic arithmetic: 4>1, 8>3, 16>9, 32>27. No algebraic manipulation, proof technique, or mathematical insight is needed—just direct calculation.
Selects suitable proof method; for exhaustion must check at least two correct values and make at least two correct comparisons
Hence \(2^{n+2} > 3^n\) for integer values of \(n\) such that \(0 \leq n < 4\)
R1 (AO 2.1)
Must include fully correct concluding statement referring to 'integer' or listing four integers; if direct/contradiction proof used, must use laws of logs correctly
## Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Table showing $n$, $2^{n+2}$, $3^n$ with values: $n=0$: $4>1$; $n=1$: $8>3$; $n=2$: $16>9$; $n=3$: $32>27$ | M1 (AO 3.1a) | Selects suitable proof method; for exhaustion must check at least two correct values and make at least two correct comparisons |
| Hence $2^{n+2} > 3^n$ for integer values of $n$ such that $0 \leq n < 4$ | R1 (AO 2.1) | Must include fully correct concluding statement referring to 'integer' or listing four integers; if direct/contradiction proof used, must use laws of logs correctly |
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