Check whether (39)n is divisible by 3 for any n which is a natural number (N).

Let p (n) = (39)n

We will prove the result by the principle of mathematical induction.

For n = 1, p (1) = 39 which is divisible by 3.

So, the result is true for n = 1.

Suppose that result is true for n = k.

p (k) = (39)k is divisible by 3  ...  (1)

Now we will prove that result holds for n = k + 1.

p (k + 1) = (39)k+1 = (39)k × 39

Since 39 is divisible by 3 and from (1) (39)k is also divisible by 3

⇒ Product of (39)k and 39 is also divisible by 3.

p (k + 1) = (39)k+1 is divisible by 3.

Thus result is true for n = k + 1 if it is true for n = k.

Hence By principle of mathematical induction, result is true for all n∈N

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