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Let $f(n)=2^{n+1}, g(n)=1+(n+1)^{2 n}$ for all $n \in \mathbb{N}$. Then
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Verified Answer
The correct answer is:
$f(n) < g(n)$
$g(n)-f(n)=1+(n+1) 2^n-2 \cdot 2^n=1+n \cdot 2^n-2^n=1+(n-1) 2^n > 0$
$\because \mathrm{n} \geq 1 \Rightarrow(\mathrm{n}-1) \geq 0$
$\because \mathrm{n} \geq 1 \Rightarrow(\mathrm{n}-1) \geq 0$
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