English

Prove the statement by using the Principle of Mathematical Induction: 32n – 1 is divisible by 8, for all natural numbers n.

Advertisements
Advertisements

Question

Prove the statement by using the Principle of Mathematical Induction:

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

Sum
Advertisements

Solution

P(n) = 32n – 1 is divisible by 8.

So, substituting different values for n, we get,

P(0) = 30 – 1 = 0 which is divisible by 8.

P(1) = 32 – 1 = 8 which is divisible by 8.

P(2) = 34 – 1 = 80 which is divisible by 8.

P(3) = 36 – 1 = 728 which is divisible by 8.

Let P(k) = 32k – 1 be divisible by 8.

So, we get,

⇒ 32k – 1 = 8x

Now, we also get that,

⇒ P(k + 1) = `3^(2("k" + 1))` – 1

= 32(8x + 1) – 1

= 72x + 8 is divisible by 8.

⇒ P(k + 1) is true when P(k) is true.

Therefore, by Mathematical Induction, P(n) = 32n – 1 is divisible by 8, for all natural numbers n.

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Principle of Mathematical Induction - Exercise [Page 70]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 11
Chapter 4 Principle of Mathematical Induction
Exercise | Q 6 | Page 70

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Prove the following by using the principle of mathematical induction for all n ∈ Nn (n + 1) (n + 5) is a multiple of 3.


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 "n(n + 1) is even", then what is P(3)?


If P (n) is the statement "n3 + n is divisible by 3", prove that P (3) is true but P (4) is not true.


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.

 

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


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

 

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

 

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


(ab)n = anbn for all n ∈ N. 

 

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

 

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

 

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

 

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

  

\[\frac{n^7}{7} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{n^2}{2} - \frac{37}{210}n\] is a positive integer for all n ∈ 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

 


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:

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:

(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))`


Answer the following:

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


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

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


The distributive law from algebra says that for all real numbers c, a1 and a2, we have c(a1 + a2) = ca1 + ca2.

Use this law and mathematical induction to prove that, for all natural numbers, n ≥ 2, if c, a1, a2, ..., an are any real numbers, then c(a1 + a2 + ... + an) = ca1 + ca2 + ... + can.


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:

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


Prove the statement by using the Principle of Mathematical Induction:

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


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.


Prove that `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24`, for all natural numbers n > 1.


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


For all n ∈ N, 3.52n+1 + 23n+1 is divisible by ______.


By using principle of mathematical induction for every natural number, (ab)n = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×