Question -
Answer -
Let P (n) = 1.2 + 2.22 + 3.23 +… + n.2n = (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 toprove that P (k + 1) is true.
P (k): 1.2 + 2.22 + 3.23 +… + k.2k = (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 usingequation (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.