English

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

Advertisements
Advertisements

Question

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

Options

  • 1

  • 2

  • 3

  • 4

MCQ
Fill in the Blanks
Advertisements

Solution

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

Explanation:

Let P(n) = xn – 1 is divisible by x – k.

P(1) = x – 1 is divisible by x – k.

Since k = 1 is the possible least integral value of k.

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Principle of Mathematical Induction - Exercise [Page 72]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 4 Principle of Mathematical Induction
Exercise | Q 28 | Page 72

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: `1/2 + 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

`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: 102n – 1 + 1 is divisible by 11


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


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


If P (n) is the statement "2n ≥ 3n" and if P (r) is true, prove that P (r + 1) is 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 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]

 

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

 

1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]

 

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

 

1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]

 

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


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

 

32n+2 −8n − 9 is divisible by 8 for all n ∈ N.


(ab)n = anbn for all n ∈ 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


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

 

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/(1.3) + 1/(3.5) + 1/(5.7) + ... + 1/((2"n" - 1)(2"n" + 1)) = "n"/(2"n" + 1)`


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

(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)


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, for all n ∈ N

`1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "n"/(("n" + 2)("n" + 3)("n" + 4)) = ("n"("n" + 1))/(6("n" + 3)("n" + 4))`


Prove by induction that for all natural number n sinα + sin(α + β) + sin(α + 2β)+ ... + sin(α + (n – 1)β) = `(sin (alpha + (n - 1)/2 beta)sin((nbeta)/2))/(sin(beta/2))`


State whether the following proof (by mathematical induction) is true or false for the statement.

P(n): 12 + 22 + ... + n2 = `(n(n + 1) (2n + 1))/6`

Proof By the Principle of Mathematical induction, P(n) is true for n = 1,

12 = 1 = `(1(1 + 1)(2*1 + 1))/6`. Again for some k ≥ 1, k2 = `(k(k + 1)(2k + 1))/6`. Now we prove that

(k + 1)2 = `((k + 1)((k + 1) + 1)(2(k + 1) + 1))/6`


Prove the statement by using the Principle of Mathematical Induction:

For any natural number n, xn – yn is divisible by x – y, where x and y are any integers with x ≠ y.


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:

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


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.


A sequence b0, b1, b2 ... is defined by letting b0 = 5 and bk = 4 + bk – 1 for all natural numbers k. Show that bn = 5 + 4n for all natural number n using mathematical induction.


Show that `n^5/5 + n^3/3 + (7n)/15` is a natural number 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 ______.


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


Consider the statement: “P(n) : n2 – n + 41 is prime." Then which one of the following is true?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×