हिंदी

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

Advertisements
Advertisements

प्रश्न

Answer the following:

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

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

योग
Advertisements

उत्तर

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.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Methods of Induction and Binomial Theorem - Miscellaneous Exercise 4.2 [पृष्ठ ८५]

APPEARS IN

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

`1+ 1/((1+2)) + 1/((1+2+3)) +...+ 1/((1+2+3+...n)) = (2n)/(n +1)`

Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2


Prove the following by using the principle of mathematical induction for all n ∈ N: `1/2 + 1/4 + 1/8 + ... + 1/2^n = 1 - 1/2^n`

 

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

`1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/((2n + 1)(2n +3)) = n/(3(2n +3))`

Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n– 9 is divisible by 8.


Given an example of a statement P (n) such that it is true for all n ∈ N.

 

Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer.


12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .

 

\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]


1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.

 

\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]

 


\[\frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}\]


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

 

12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]

 

\[\sin x + \sin 3x + . . . + \sin (2n - 1)x = \frac{\sin^2 nx}{\sin x}\]

 


\[\text{ Prove that } \cos\alpha + \cos\left( \alpha + \beta \right) + \cos\left( \alpha + 2\beta \right) + . . . + \cos\left[ \alpha + \left( n - 1 \right)\beta \right] = \frac{\cos\left\{ \alpha + \left( \frac{n - 1}{2} \right)\beta \right\}\sin\left( \frac{n\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} \text{ for all n } \in N .\]

 


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

2 + 4 + 6 + ..... + 2n = n (n+1)


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

(24n−1) is divisible by 15


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

5 + 52 + 53 + .... + 5n = `5/4(5^"n" - 1)`


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

Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 1


Answer the following:

Given that tn+1 = 5tn − 8, t1 = 3, prove by method of induction that tn = 5n−1 + 2


Answer the following:

Prove by method of induction

`[(3, -4),(1, -1)]^"n" = [(2"n" + 1, -4"n"),("n", -2"n" + 1)], ∀  "n" ∈ "N"`


Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:

2n + 1 < 2n, for all natual numbers n ≥ 3.


Let P(n): “2n < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is ______.


Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer


Give an example of a statement P(n) which is true for all n. Justify your answer. 


Prove the statement by using the Principle of Mathematical Induction:

`sqrt(n) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(n)`, for all natural numbers n ≥ 2.


Prove the statement by using the Principle of Mathematical Induction:

2 + 4 + 6 + ... + 2n = n2 + n for all natural numbers n.


A sequence a1, a2, a3 ... is defined by letting a1 = 3 and ak = 7ak – 1 for all natural numbers k ≥ 2. Show that an = 3.7n–1 for all natural numbers.


Prove that for all n ∈ N.
cos α + cos(α + β) + cos(α + 2β) + ... + cos(α + (n – 1)β) = `(cos(alpha + ((n - 1)/2)beta)sin((nbeta)/2))/(sin  beta/2)`.


Prove that, cosθ cos2θ cos22θ ... cos2n–1θ = `(sin 2^n theta)/(2^n sin theta)`, for all n ∈ N.


Show that `n^5/5 + n^3/3 + (7n)/15` is a natural number for all n ∈ N.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×