For every integer n ≥ 1, ( 3^2 ^ n -1) is always divisible by. - Vidyadip

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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