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

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

Let us check for n = 1,

P (1):1.2 = 0.2⁰ + 2

: 2 = 2

P (n) is true for n = 1.

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

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

So,

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

Now, substituting the value of P (k) we get,

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

= (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

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

Hence, P (n) is true for all n ∈ N.