Principle of Mathematical Induction – Class XI – Exercise 4.1[Part 1]

  1. Prove the following by using the principle of mathematical induction for all n € N:

1 + 3 + 32+ ………..+3n-1 = (3n – 1)/2

Solution:

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

P(n): 1 + 3 + 32+ ………..+3n-1 = (3n – 1)/2

For n = 1, we have

P(1): (31 – 1)/2 = 3-1/2 = 2/2 = 1 , which is true.

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

1 + 3 + 32+ ………..+3k-1 = (3k – 1)/2

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

Consider

1 + 3 + 32+ ………..+3n-1 + 3(k+1)-1 = 1 + 3 + 32+ ………..+3k-1 + 3k

=(3k – 1)/2 +3k

=[(3k – 1)+2.3k]/2

=[(1+2)3k – 1]/2

= [3.3k – 1]/2

= [3k+1 – 1]/2

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.


  1. Prove the following by using the principal of mathematical induction for all n €N

13 + 23 + 33 +  … + n3 = [n(n+1)/2]2

Solution:

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

P(n): 13 + 23 + 33 +  … + n3 = [n(n+1)/2]2

For n = 1, we have

P(1): 13 = 1 = (1(1+1)/2)2 = (1.2/2)2 = 12 = 1, which is true.

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

13 + 23 + 33 +  … + k3 = [k(k+1)/2]2 —————–(1)

Let us prove that P(k+1) is true.

Consider

13 + 23 + 33 +  … + k3 +(k+1)3 = (13 + 23 + 33 +  … + k3)+(k+1)3

Principle of Mathematical Induction - Class XI - Exercise 4.1

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

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


  1. Prove the following by using the principal of mathematical induction for all n€N

1 + 1/(1+2) + 1/(1+2+3) + … + 1/(1+2+3+ … +n) = 2n/(n+1)

Solution:

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

P(n): 1 + 1/(1+2) + 1/(1+2+3) + … + 1/(1+2+3+ … +n) = 2n/(n+1)

For n = 1, we have

P(1): 1 = 2(1)/(1+1) = 2/2 = 1. which is true.

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

1 + 1/(1+2) + 1/(1+2+3) + … + 1/(1+2+3+ … +k) = 2k/(k+1) —————-(1)

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

Consider

1 + 1/(1+2) + 1/(1+2+3) + … + 1/(1+2+3+ … +k) + 1/(1+2+3+…+k+(k+1)) = (1 + 1/(1+2) + 1/(1+2+3) + … + 1/(1+2+3+ … +k))+ 1/(1+2+3+…+k+(k+1))

= 2k/k+1 + 1/(1+2+3+…+k+(k+1))                [using(i)]

Principle of Mathematical Induction - Class XI - Exercise 4.1

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

Hence by the principal of mathematical induction, statement P(n) is true for all natural numbers ie., N


  1. Prove the following by using the principal of mathematical induction for all n€N

1.2.3 + 2.3.4 + … + n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4

Solution:

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

P(n): 1.2.3 + 2.3.4 + … +n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4

For n = 1, we have

P(1): 1.2.3 = 6 = 1(1+1)(1+2)(1+3)/4 = 1.2.3.4/4 = 6, which is true.

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

1.2.3 + 2.3.4 + … +k(k+1)(k+2) = k(k+1)(k+2)(k+3)/4 ————(i)

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

Consider

1.2.3 + 2.3.4 + … + k(k + 1) (k + 2) + (k + 1) (k + 2) (k + 3) = {1.2.3 + 2.3.4 + … + k(k + 1) (k + 2)} + (k + 1) (k + 2) (k + 3)

Principle of Mathematical Induction - Class XI - Exercise 4.1

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

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


  1. Prove that the following by using the principal of mathematical induction for all n€N

1.3 + 2.32 + 3.32 + … + n.3n = {(2n-1)(3n+1+3)} / 4

Solution:

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

1.3 + 2.32 + 3.32 + … + n.3n = {(2n-1)(3n+1+3)} / 4

For n = 1, we have

P(1): 1.3 = 3

Principle of Mathematical Induction - Class XI - Exercise 4.1

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.


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

1.2 + 2.3 + 3.4 + …+n(n+1) = n(n+1)(n+2)/3

Solution:

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

P(n) : 1.2 + 2.3 + 3.4 + …+n(n+1) = n(n+1)(n+2)/3

For n = 1, we have

P(1): 1.2 = 2 = 1(1+1)(1+2)/3 = 1.2.3/3 = 2 , which is true.

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

1.2 + 2.3 + 3.4 + …+k(k+1) = k(k+1)(k+2)/3  ————-(i)

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

Consider

1.2 + 2.3 + 3.4 + … + k.(k + 1) + (k + 1).(k + 2) = [1.2 + 2.3 + 3.4 + … + k.(k + 1)] + (k + 1).(k + 2)

Principle of Mathematical Induction - Class XI - Exercise 4.1

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.


7: 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

Solution:

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

P(n): 1.3 + 3.5 + 5.7 + … + (2n-1)(2n+1) = {n(4n2 + 6n – 1)} / 3

For n = 1, we have

P(1): 1.3 = 3 = {1(4.1 + 6.1 -1)}/3 = (4+6-1)/3 = 9/3 = 3, which is true.

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

Principle of Mathematical Induction - Class XI - Exercise 4.1

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.


  1. 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

Solution:

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

P(n): 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2

For n = 1, we have

P(1): 1.2 = 2 = (1 – 1) 21+1 + 2 = 0 + 2 = 2, which is true.

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

1.2 + 2.22 + 3.22 + … + n.2k = (k – 1) 2k+1 + 2 ————-(i)

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

Consider

{1.2 + 2.22 + 3.22 + … + k.2k }+ (k+ 1) 2k+1 + 2k+1 = (k-1)2k+1 + 2+(k+1)2k+1

=2k+1{(k-1)+(k+1)}+2

= 2k+1.2k + 2

=k.2(k+1)+1+2

={(k+1)-1}2(k+1)+1+2

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.


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

1/2 + 1/4 + 1/8 + … +1/2n = 1 – 1/2n

Solution:

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

P(n): 1/2 + 1/4 + 1/8 + … +1/2n = 1 – 1/2n

P(1): 1/2 = 1 – 1/21 = 1/2, which is true.

Principle of Mathematical Induction - Class XI - Exercise 4.1

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.


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

1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3n-1)(3n+2) = n/(6n+4)

Solution:

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

P(n):
1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3n-1)(3n+2) = n/(6n+4)

For n = 1, we have

Principle of Mathematical Induction - Class XI - Exercise 4.1

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.


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