Answer :

Let P(n): 1.2 + 2.22 + 3.23 + … + n.2n=(n–1) 2n + 1 + 2


For n = 1


= 1.2 = 0.20 + 2


= 2 = 2


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


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


P(k): 1.2 + 2.22 + 3.23 + … + k.2k=(k–1) 2k + 1 + 2 - - - - - - (1)


We have to show that,


{1.2 + 2.22 + 3.23 + … + k.2k + (k + 1) 2k + 1 = k.2k + 2 + 2


Now,


{1.2 + 2.22 + 3.23 + … + k.2k} + (k + 1)2k + 1


= [(k - 1)2k + 1 + 2] + (k + 1)2k + 1 using equation (1)


= (k - 1)2k + 1 + 2 + (k + 1)2k + 1


= 2k + 1(k - 1 + k + 1) + 2


= 2k + 1.2k + 2


= k.2k + 2 + 2


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


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


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