Question

prove that (21)^n is an odd number for any natural number number​

Answer 1

(21)^n

Let us check for n=1

(21)^1= 21 which is odd..

Let us now check for 2

(21)^2=441

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Similarly check for more 2 to 3 natural numbers and state that it is true for all value of n where n belongs to natural no.

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Answer 2

Answer:

21 is an odd number and by the fundamental theorem of arithmetic it multiplied by itself would give only higher exponents of 3 and 7. As 2 is not its prime factor therefore any exponent of it cannot be an even number .