MCQ
For every integer $\text{n} ≥ 1, ({3^2}^{\text{n}}-1)$ is always divisible by.
- A$2^{\text{n}^2}$
- B$2^{\text{n + 4}}$
- ✓$2^{\text{n + 2}}$
- D$2^{\text{n + 3}}$
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