Let P(n): 1.2 + 2.2² + 3.2³ + … + n.2ⁿ=(n–1) 2^{n + 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.2^k=(k–1) 2^{k + 1} + 2 - - - - - - (1)

We have to show that,

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

Now,

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

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

= (k - 1)2^{k + 1} + 2 + (k + 1)2^{k + 1}

= 2^{k + 1}(k - 1 + k + 1) + 2

= 2^{k + 1}.2k + 2

= k.2^{k + 2} + 2

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

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