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For all possible integers $n$ satisfying $2.25 \leq 2 + 2^{n+2} \leq 202,$ the number of integer values of $3 + 3^{n+1}$ is

Given that, $2.25 \leq 2 + 2^{n+2} \leq 202$

$\Rightarrow 0.25 \leq 2^{n+2} \leq 200$

$\Rightarrow 2^{-2} \leq 2^{n+2} < 2^{8} \quad[{\color {Blue}{\because 2^{7} = 128, 2^{8} = 256 \Rightarrow 2^{7} < 200 < 2^{8}}}]$

$\Rightarrow\; – 2 \leq n+2 < 8$

$\Rightarrow \;– 4 \leq n < 6$

$\Rightarrow n \in \{ – 4 , – 3, – 2, – 1, 0, 1, 2, 3, 4, 5 \}$

The integer value of $3 + 3^{n+1} :$

$\Rightarrow n + 1 \geq 0$

$\Rightarrow \boxed{ n \geq\; – 1}$

$\Rightarrow n \in \{– 1, 0, 1, 2, 3, 4, 5 \}$

$\therefore$ The number of integer values of $3 + 3^{n+1}$ is $7.$

Correct Answer $: 7$
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