English

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

Advertisements
Advertisements

Question

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

Options

  • 1

  • 2

  • 3

  • 4

MCQ
Fill in the Blanks
Advertisements

Solution

Let P(n): “2n < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is 4.

Explanation:

P(1): 2 < 1 is false.

P(2): 22 < 1 × 2 is false.

P(3): 23 < 1 × 2 × 3 is false.

But P(4): 24 < 1 × 2 × 3 × 4 is true.

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Principle of Mathematical Induction - Solved Examples [Page 69]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 4 Principle of Mathematical Induction
Solved Examples | Q 11 | Page 69

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

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


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: 102n – 1 + 1 is divisible by 11


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.


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


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


a + ar + ar2 + ... + arn−1 =  \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]

 

a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = \[\frac{n}{2}\left[ 2a + (n - 1)d \right]\]

 


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

 

72n + 23n−3. 3n−1 is divisible by 25 for all n ∈ N.

 

Prove that 1 + 2 + 22 + ... + 2n = 2n+1 - 1 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? 


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


\[\text{ A sequence } x_0 , x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_0 = 5 and x_k = 4 + x_{k - 1}\text{  for all natural number k . } \]
\[\text{ Show that } x_n = 5 + 4n \text{ for all n }  \in N \text{ using mathematical induction .} \]


\[\text{ The distributive law from algebra states that for all real numbers}  c, a_1 \text{ and }  a_2 , \text{ we have }  c\left( a_1 + a_2 \right) = c a_1 + c a_2 . \]
\[\text{ Use this law and mathematical induction to prove that, for all natural numbers, } n \geq 2, if c, a_1 , a_2 , . . . , a_n \text{ are any real numbers, then } \]
\[c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n\]


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:

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


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

12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`


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

(23n − 1) is divisible by 7


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:

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

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


Answer the following:

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


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.


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


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:

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


Prove the statement by using the Principle of Mathematical Induction:

1 + 2 + 22 + ... + 2n = 2n+1 – 1 for all natural numbers n.


A sequence d1, d2, d3 ... is defined by letting d1 = 2 and dk = `(d_(k - 1))/"k"` for all natural numbers, k ≥ 2. Show that dn = `2/(n!)` for all n ∈ N.


If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.


State whether the following statement is true or false. Justify.

Let P(n) be a statement and let P(k) ⇒ P(k + 1), for some natural number k, then P(n) is true for all n ∈ N.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×