# Prove the followi

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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