मराठी

(Ab)N = Anbn for All N ∈ N. - Mathematics

Advertisements
Advertisements

प्रश्न

(ab)n = anbn for all n ∈ N. 

 
Advertisements

उत्तर

Let P(n) be the given statement.
Now, 

\[P(n): (ab )^n = a^n b^n \text{ for all } n \in N . \]
\[\text{ Step } 1: \]
\[P(1): (ab )^1 = a^1 b^1 = ab\]
\[\text{ Thus, P(1) is true }  . \]
\[\text{ Step 2 } : \]
\[\text{ Let P(m) be true } . \]
\[\text{ Then } , \]
\[(ab )^m = a^m b^m \]
\[ \text{ We need to show that P(m + 1) is true whenever P(m) is true }  . \]
\[\text{ Now, }  \]
\[ P(m + 1): (ab )^{m + 1} = (ab )^m . ab\]
\[ = a^m b^m . ab\]
\[ = a^m a . b^m b\]
\[ = a^{m + 1} b^{m + 1} \]
\[\text{ Hence, P(m + 1) is true .}  \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n }  \in N .\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 12: Mathematical Induction - Exercise 12.2 [पृष्ठ २८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 12 Mathematical Induction
Exercise 12.2 | Q 23 | पृष्ठ २८

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2)  = `(n(n+1)(n+2)(n+3))/(4(n+3))`


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


Prove the following by using the principle of mathematical induction for all n ∈ N: 41n – 14n is a multiple of 27.


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


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

 

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

 


\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]


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

 

1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]

 

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


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

 

52n+2 −24n −25 is divisible by 576 for all n ∈ N.

 

n(n + 1) (n + 5) is a multiple of 3 for all n ∈ N.

 

Prove that n3 - 7+ 3 is divisible by 3 for all n \[\in\] 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

 


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

 


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


\[\text{ Let } P\left( n \right) \text{ be the statement } : 2^n \geq 3n . \text{ If } P\left( r \right) \text{ is true, then show that } P\left( r + 1 \right) \text{ is true . Do you conclude that } P\left( n \right)\text{  is true for all n }  \in N?\]


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


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:

3 + 7 + 11 + ..... + to n terms = n(2n+1)


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

1.2 + 2.3 + 3.4 + ..... + n(n + 1) = `"n"/3 ("n" + 1)("n" + 2)`


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

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


Answer the following:

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

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


Answer the following:

Prove by method of induction 152n–1 + 1 is divisible by 16, for all n ∈ N.


Answer the following:

Prove by method of induction 52n − 22n is divisible by 3, for all n ∈ N


Define the sequence a1, a2, a3 ... as follows:
a1 = 2, an = 5 an–1, for all natural numbers n ≥ 2.

Use the Principle of Mathematical Induction to show that the terms of the sequence satisfy the formula an = 2.5n–1 for all natural numbers.


Prove by the Principle of Mathematical Induction that 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! – 1 for all natural numbers n.


A student was asked to prove a statement P(n) by induction. He proved that P(k + 1) is true whenever P(k) is true for all k > 5 ∈ N and also that P(5) is true. On the basis of this he could conclude that P(n) is true ______.


Prove the statement by using the Principle of Mathematical Induction:

4n – 1 is divisible by 3, for each natural number n.


Prove the statement by using the Principle of Mathematical Induction:

32n – 1 is divisible by 8, for all natural numbers n.


Prove the statement by using the Principle of Mathematical Induction:

n2 < 2n for all natural numbers n ≥ 5.


Prove the statement by using the Principle of Mathematical Induction:

2n < (n + 2)! for all natural number 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, sinθ + sin2θ + sin3θ + ... + sinnθ = `((sin ntheta)/2 sin  ((n + 1))/2 theta)/(sin  theta/2)`, for all n ∈ N.


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


If xn – 1 is divisible by x – k, then the least positive integral value of k is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×