Answer :

Let P(n) : (ab)n = an bn


Let check for n = 1 is true


= (ab)1 = a1b1


= ab = ab


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


Let P(n) is true for n=k,


= (ab)k=ak.bk - - - - - - (1)


We have to show that,


= (ab)k + 1=ak + 1.bk + 1


Now,


= (ab)k + 1


=(ab)k (ab)


= (akbk)(ab) using equation (1)


= (ak + 1)(bK + 1)


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


Hence, P(n) is true for all nN by PMI


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