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
    $2^{\text{n + 3}}$

Answer

Correct option: C.
$2^{\text{n + 2}}$
For $\text{n = 1},\ 3^{\text{2}^1}-1=8,$ which is divisible by $2^{\text{n + 2}}$
Let us assume that $3^{2^{\text{m}}}-1$ is divisible by $2^\text{m + 2}$ for some integral value of $m.$
Let us consider the expression for $m+1$
$3^{2^{\text{m+1}}}-1$
$=(3^{2^{\text{m}}}-1)\ \times ​​(3^{2^{\text{m}}}+1)$
The first term is divisible $2^{\text{m+2}}$ and the second term is also an even number.
Hence, the term is divisible by $2^{\text{m+2}}$
Hence, by induction we can prove that it is true for all $m.$

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