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

बेरीज

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.

shaalaa.com

Principle of Mathematical Induction

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

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

पाठ 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

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

संबंधित प्रश्‍न

Prove the following by using the principle of mathematical induction for all n ∈ N:

`a + ar + ar^2 + ... + ar^(n -1) = (a(r^n - 1))/(r -1)`

Prove the following by using the principle of mathematical induction for all n ∈ N:

`1/1.4 + 1/4.7 + 1/7.10 + ... + 1/((3n - 2)(3n + 1)) = n/((3n + 1))`

Prove the following by using the principle of mathematical induction for all n ∈ N: 10²ⁿ^{– 1}+ 1 is divisible by 11

1.2 + 2.2² + 3.2³ + ... + n.2ⁿ= (n − 1) 2ⁿ⁺¹+2

2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]

\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]

3²ⁿ+7 is divisible by 8 for all n ∈ N.

\[\frac{n^{11}}{11} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{62}{165}n\] is a positive integer for all n ∈ N.

x²ⁿ⁻¹ + y²ⁿ⁻¹ is divisible by x + y for all n ∈ N.

\[\text{ Given } a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right) \text{ for } n \geq 2, \text{ where } a > 0, A > 0 . \]
\[\text{ Prove that } \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1} .\]

Prove that the number of subsets of a set containing n distinct elements is 2ⁿ, for all n \[\in\] N .

\[\text{ A sequence } a_1 , a_2 , a_3 , . . . \text{ is defined by letting } a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]