Question

Answer the following:

Prove, by method of induction, for all n ∈ N

2 + 3.2 + 4.2² + ... + (n + 1)2^n–1 = n.2ⁿ 

Answer 1

Let P(n) ≡ 2 + 3.2 + 4.2² + ... + (n + 1)2^n–1 = n.2^n-1 = n.2ⁿ, for all n ∈ N

Step I:

Put n = 1

L.H.S. = 2

R.H.S. = 1(2¹) = 2 = L.H.S.

∴ P(n) is true for n = 1

Step II:

Let us consider that P(n) is true for n = k

∴ 2 + 3.2 + 4.2² + … + (k + 1)2^k–1 = k.2^k  …(i)

Step III:

We have to prove that P(n) is true for n = k + 1

i.e., to prove that

2 + 3.2 + 4.2² + …. + (k + 2)2^k = (k + 1)2^k+1

L.H.S. = 2 + 3.2 + 4.2² + …. + (k + 2)2^k

= 2 + 3.2 + 4.2² + …. + (k + 1)2^k–1 + (k + 2)2^k

= k.2^k + (k + 2).2^k …[From (i)]

= (k + k + 2).2^k  

= (2k + 2). 2^k

= (k + 1).2.2^k

= (k + 1). 2^k+1

= R.H.S.

∴ P(n) is true for n = k + 1

Step IV:

From all steps above by the principle of mathematical induction, P(n) is true for all n ∈ N.