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

Updated On: Jan 15, 2026
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Solution and Explanation

Given the inequality:

\[ 2.25 \leq 2 + 2^{n+2} \leq 202 \]

Step 1: Isolate the exponential term

Subtract 2 from all parts of the inequality:

\[ 2.25 - 2 \leq 2^{n+2} \leq 202 - 2 \\ \Rightarrow 0.25 \leq 2^{n+2} \leq 200 \]

Step 2: Determine the range of n

Find integer values of \( n \) satisfying \( 0.25 \leq 2^{n+2} \leq 200 \).

  • Lower bound: \( 2^{n+2} \geq 0.25 \). Since \( 0.25 = 2^{-2} \), we have \( n+2 \geq -2 \), which simplifies to \( n \geq -4 \).
  • Upper bound: \( 2^{n+2} \leq 200 \). Testing powers of 2: \( 2^7 = 128 \) (valid) and \( 2^8 = 256 \) (exceeds 200). Thus, \( n+2 \leq 7 \), which simplifies to \( n \leq 5 \).

Combining the bounds, the possible integer values for \( n \) are: \[ n \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4, 5\} \quad \text{(10 values)} \]

Step 3: Incorporate the condition for \( 3 + 3^{n+1} \)

The expression \( 3 + 3^{n+1} \) is integral for all integers \( n \). If \( 3^{n+1} \) must also be an integer (which it is for integer \( n \)), and positive (requiring \( n+1 \geq 0 \)), then \( n \geq -1 \).

The intersection of the range \( -4 \leq n \leq 5 \) and \( n \geq -1 \) is: \[ n \in \{-1, 0, 1, 2, 3, 4, 5\} \quad \text{(7 values)} \]

Final Answer:

There are \( \boxed{7} \) possible values of \( n \).

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