Answer the following: Prove, by method of induction, for all n ∈ N 2 + 3.2 + 4.22 + ... + (n + 1)2n–1 = n.2n

प्रश्न

Answer the following:

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

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

योग

उत्तर

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.

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Principle of Mathematical Induction

क्या इस प्रश्न या उत्तर में कोई त्रुटि है?

Q II. (1) (iii)Q II. (1) (ii)Q II. (1) (iv)

अध्याय 4: Methods of Induction and Binomial Theorem - Miscellaneous Exercise 4.2 [पृष्ठ ८५]

APPEARS IN

बालभारती Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board

अध्याय 4 Methods of Induction and Binomial Theorem
Miscellaneous Exercise 4.2 | Q II. (1) (iii) | पृष्ठ ८५

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