Maharashtra State BoardHSC Science (General) 11th
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Answer the following: Prove, by method of induction, for all n ∈ N 2 + 3.2 + 4.22 + ... + (n + 1)2n–1 = n.2n - Mathematics and Statistics

Sum

Answer the following:

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

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

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Solution

Let P(n) ≡ 2 + 3.2 + 4.22 + ... + (n + 1)2n–1 = n.2n-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)2k–1 = k.2k  …(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)2k = (k + 1)2k+1

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

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

= k.2k + (k + 2).2k …[From (i)]

= (k + k + 2).2k  

= (2k + 2). 2k

= (k + 1).2.2k

= (k + 1). 2k+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.

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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Miscellaneous Exercise 4 | Q II. (1) (iii) | Page 85
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