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Prove the following by the principle of mathematical induction: 

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

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

For n = 1 

= 72 + 2°.3°

= 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

= 25μ 

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

Hence, P(n) is true for all n ∈ N

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