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**Hint:**We will have to start the question by hit trial. That is we have to put values for different n and check for divisibility and further can be proved by principle of mathematical induction.

**Complete step-by-step answer:**

We have to check whether \[{3^{2n}} - 2n + 1\] is divisible by any of the options for all values of n which belong to natural numbers.

We will start by putting 1 in place of n and check the divisibility –

=\[{3^{2n}} - 2n + 1\]

=\[{3^{2(1)}} - 2(1) + 1\]

=\[9 - 2 + 1\]

=\[8\]

Therefore, it is divisible by 2, 4 and 8 from the options.

Now, we will put 2 in place of n, we get –

=\[{3^{2n}} - 2n + 1\]

=\[{3^{2(2)}} - 2(2) + 1\]

=\[81 - 4 + 1\]

=\[78\]

This is divisible by only 2.

Therefore, 2 is the correct answer.

**So, the correct answer is “Option A”.**

**Note:**This question can be solved by an alternative method by principle of mathematical induction.

Consider $f(n)$ = \[{3^{2n}} - 2n + 1\]

Let us check if it is divisible by 2 for that it must be true for n=1,

=\[{3^{2n}} - 2n + 1\]

=\[{3^{2(1)}} - 2(1) + 1\]

=\[9 - 2 + 1\]

=\[8\]

Therefore, it is divisible by 2.

Let us assume that \[{3^{2n}} - 2n + 1\] is divisible by 2 for any value of n=m.

=\[{3^{2n}} - 2n + 1\]

For n=m we have,

$F(m)$ =\[{3^{2(m)}} - 2(m) + 1\] = $2p……….. (1)$ (Since, it is assumed to be divisible by 2)

Now, we check for n=m+1,

$F(m+1)$ =\[{3^{2(m + 1)}} - 2(m + 1) + 1\]

$F(m+1)$ =\[{9.3^{2}} - 2(m) - 1\]

Rearranging and adding 1 and subtracting 1 we get,

=\[{3^{2n}} - 2(n) - 1 + ( - 1 - 1) + {8.3^{2n}}\]

=\[{3^{2n}} - 2(n) - 1 + {8.3^{2n}} - 2\]

From (1) we have,

=\[2p + 2({4.3^{2n}} - 1)\]

=\[2[p + ({4.3^{2n}} - 1)]\]

From above equation it is divisible by 2

Since, F(m + 1) is true, whenever F(m) is true.

Thus, F(1) is true and F(k + 1) is true, whenever F(k) is true.

Hence, by the principle of mathematical induction, F(n) is true for all n ∈ N.

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