Q. 255.0( 3 Votes )

Prove the following by the principle of mathematical induction:

72n + 23n – 3 . 3n – 1 is divisible by 25 for all n ϵ N

Answer :

Let P(n): 72n + 23n - 3.3n - 1 is divisible by 25


For n=1


= 72 + 20.30


= 49 + 1


= 50


Therefor it is divisible by 25


So, P(n) is true for n = 1


Now, P(n) is true For n = k,


So, we have to show that 72n + 23n - 3.3n - 1 is divisible by 25


= 72k + 23k - 3.3k - 1 = 25λ - - - - - - - (1)


Now, P(n) is true For n = k + 1,


So, we have to show that 72k + 1 + 23k.3k is divisible by 25


= 72k + 2 + 23k.3k = 25μ


Now,


= 72(k + 1) + 23k.3k


= 72k.71 + 23k.3k


= (25λ – 23k - 3.3k - 1)49 + 23k.3k from eq 1


=


= 24×25×49λ - 23k.3k..49 + 24.23k.3k


= 24×25×49λ - 25.23k.3k


= 25(24.49λ - 23k.3k)


= 25μ


Therefore, P(n) is true for n = k + 1


Hence, P(n) is true for all n N

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