English

Answer the following: Prove by method of induction 52n − 22n is divisible by 3, for all n ∈ N - Mathematics and Statistics

Advertisements
Advertisements

Question

Answer the following:

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

Sum

Solution

Let P(n) ≡ 52n – 22n is divisible by 3, for all n ∈ N.

Step 1:

For n = 1, 52n – 22n = 52 – 22 = 25 – 4 = 21, which is divisible by 3.

∴ P(1) is true.

Step 2:

Let us assume that for some k ∈ N, P(k) is true, i.e. 52k – 22k is divisible by 3.

52k-22k3 = m (Say), whre m ∈ N

∴ 52k –  22k = 3m

∴ 52k = 22k + 3m   ...(1)

Step 3:

To prove that P(k + 1) is true, i.e., to prove that 52(k+1)-22(k+1) is divisible by 3.

Now, 52(k+1)-22(k+1) = 52k+2 – 22k+2 

= 52k .52 – 22k . 22

= (22k + 3m)25 – 22k . 4    ...[By (1)]

= 25(22k) + 75m – 4(22k)

= 21(22k) + 75m

= 3[7.22k + 25m]

52(k+1)-22(k+1)3 = 7.22k + 25m, where (7.22k + 25m) ∈ N

52(k+1)-22(k+1) is divisible by 3

∴ P(k + 1) is true.

Step 4:

From all the above steps and by the principle of mathematical induction P(n) is true for all n ∈ N,

i.e., 52n – 22n is divisible by 3, for all n ∈ N.

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Methods of Induction and Binomial Theorem - Miscellaneous Exercise 4.2 [Page 86]

APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) [English] 11 Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Miscellaneous Exercise 4.2 | Q II. (11) (iii) | Page 86

RELATED QUESTIONS

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

1+3+32+...+3n1=(3n-1)2


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

1.3 + 3.5 + 5.7 + ...+(2n -1)(2n + 1) = n(4n2+6n-1)3

Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2


Prove the following by using the principle of mathematical induction for all n ∈ N: 12+14+18+...+12n=1-12n

 

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

(1+11)(1+12)(1+13)...(1+1n)=(n+1)


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

13.5+15.7+17.9+...+1(2n+1)(2n+3)=n3(2n+3)

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: 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.


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


If P (n) is the statement "2n ≥ 3n" and if P (r) is true, prove that P (r + 1) is true.

 

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

 

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.


13.5+15.7+17.9+...+1(2n+1)(2n+3)=n3(2n+3)


13.7+17.11+111.5+...+1(4n1)(4n+3)=n3(4n+3) 


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


Prove that n3 - 7+ 3 is divisible by 3 for all n N .

  

 Prove that 1n+1+1n+2+...+12n>1324, for all natural numbers n>1.

 


 A sequence a1,a2,a3,... is defined by letting a1=3 and ak=7ak1 for all natural numbers k2. Show that an=37n1 for all nN.


 A sequence x1,x2,x3,... is defined by letting x1=2 and xk=xk1k for all natural numbers k,k2. Show that xn=2n! for all nN.


 The distributive law from algebra states that for all real numbersc,a1 and a2, we have c(a1+a2)=ca1+ca2.
 Use this law and mathematical induction to prove that, for all natural numbers, n2,ifc,a1,a2,...,an are any real numbers, then 
c(a1+a2+...+an)=ca1+ca2+...+can


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

1.3 + 3.5 + 5.7 + ..... to n terms = n3(4n2+6n-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, for all n ∈ N

13.4.5+24.5.6+35.6.7+...+n(n+2)(n+3)(n+4)=n(n+1)6(n+3)(n+4)


Answer the following:

Prove by method of induction loga xn = n logax, x > 0, n ∈ N


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

t=1n-1t(t+1)=n(n-1)(n+1)3, 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.


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.


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:

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:

1 + 5 + 9 + ... + (4n – 3) = n(2n – 1) for all natural numbers n.


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


Prove that, cosθ cos2θ cos22θ ... cos2n–1θ = sin2nθ2nsinθ, 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 ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×
Our website is made possible by ad-free subscriptions or displaying online advertisements to our visitors.
If you don't like ads you can support us by buying an ad-free subscription or please consider supporting us by disabling your ad blocker. Thank you.