English

Prove the Following by Using the Principle of Mathematical Induction for All N ∈ N (2n +7) < (N + 3)2 - Mathematics

Advertisements
Advertisements

Question

Prove the following by using the principle of mathematical induction for all n ∈ N (2+7) < (n + 3)2

Advertisements

Solution

Let the given statement be P(n), i.e.,

P(n): (2+7) < (n + 3)2

It can be observed that P(n) is true for n = 1 since 2.1 + 7 = 9 < (1 + 3)2 = 16, which is true.

Let P(k) be true for some positive integer k, i.e.,

(2k + 7) < (k + 3)2 … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Principle of Mathematical Induction - Exercise 4.1 [Page 95]

APPEARS IN

NCERT Mathematics [English] Class 11
Chapter 4 Principle of Mathematical Induction
Exercise 4.1 | Q 24 | Page 95

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ...+ `1/(n(n+1)(n+2)) = (n(n+3))/(4(n+1) (n+2))`

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

`1^2 + 3^2 + 5^2 + ... + (2n -1)^2 = (n(2n - 1) (2n + 1))/3`

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))`

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

 

If P (n) is the statement "n2 − n + 41 is prime", prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.


1 + 2 + 3 + ... + n =  \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .


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

 

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

 

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


\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\] 


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

 

52n −1 is divisible by 24 for all n ∈ N.


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

 

2.7n + 3.5n − 5 is divisible by 24 for all n ∈ N.


11n+2 + 122n+1 is divisible by 133 for all n ∈ N.

 

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

  

Let P(n) be the statement : 2n ≥ 3n. If P(r) is true, show that P(r + 1) is true. Do you conclude that P(n) is true for all n ∈ N


\[\text { A sequence  } x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_1 = 2 \text{ and }  x_k = \frac{x_{k - 1}}{k} \text{ for all natural numbers } k, k \geq 2 . \text{ Show that }  x_n = \frac{2}{n!} \text{ for all } n \in N .\]


\[\text{ Using principle of mathematical induction, prove that } \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . + \frac{1}{\sqrt{n}} \text{ for all natural numbers } n \geq 2 .\]

 


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:

12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`


Answer the following:

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

8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`


Answer the following:

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


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

`(1 - 1/2^2).(1 - 1/3^2)...(1 - 1/n^2) = (n + 1)/(2n)`, for all natural numbers, n ≥ 2. 


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

22n – 1 is divisible by 3.


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

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


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


Prove the statement by using the Principle of Mathematical Induction:

n3 – n is divisible by 6, for each natural number n ≥ 2.


Prove the statement by using the Principle of Mathematical Induction:

n(n2 + 5) is divisible by 6, for each natural number n.


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 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)`.


If P(n): 2n < n!, n ∈ N, then P(n) is true for all n ≥ ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×