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

Asked by Sakshi | 1 year ago |  91

Solution :-

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

So,

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

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

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

= (k – 1)2k + 1 + 2 + (k + 1)2k + 1

= 2k + 1(k – 1 + k + 1) + 2

= 2k + 1 × 2k + 2

= k × 2k + 2 + 2

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

Hence, P (n) is true for all n ∈ N.

Answered by Aaryan | 1 year ago

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