NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction Ex 4.1

# NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction Ex 4.1

## NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction Ex 4.1

NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction Ex 4.1 are the part of NCERT Solutions for Class 11 Maths. Here you can find the NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction Ex 4.1.

Solution:

### Ex 4.1 Class 11 Maths Question 2:

Solution:

Let the given statement be P(n), then,

P(n) = 13 + 23 + 33 + …………. + n3 = [n(n  + 1)/2]2.
For n = 1, we have
P(1): 13 = [1(1 + 1)/2)2 = (2/2)2
= 12 = 1, which is true.
Let P(k) be true for some positive integer k, then,
13 + 23 + 33 + …………. + k3 = [k(k + 1)/2]2               …(i)
We shall now prove that P(k + 1) is true.
Consider, 13 + 23 + 33 + …………. + k3 = [k(k + 1)/2]2

Solution:

Solution:

Solution:

### Ex 4.1 Class 11 Maths Question 6:

Solution:

k(k + 1) (k + 2)/3 + (k + 1) (k + 2)

= (k + 1) (k + 2) + (k/3 + 1)

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

(k + 1) (k + 1 + 1) (k + 1 + 2)/3
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.

### Ex 4.1 Class 11 Maths Question 7:

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

Solution:

### Ex 4.1 Class 11 Maths Question 9:

Solution:
Let the given statement be P(n), i.e.,
P(n) : ½ + ¼ + 1/8 +
⋅⋅⋅⋅⋅⋅⋅⋅ + 1/2n = 1 1/2n

Solution:

Solution:

### Ex 4.1 Class 11 Maths Question 12:

Solution:

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

P(n) : a + ar + ar2 + …………… + arn – 1 = a(rn 1)/(r 1)

For n = 1, we have

P(1) : a = a(r11)/(r 1) = a, which is true.

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

a + ar + ar2 + …………… + ark – 1 = a(rk1)/(r 1) …………..(i)

We shall now prove that P(k + 1) is true.
Consider
{a + ar + ar2 + …………… + ark – 1} + ar(k + 1) – 1 = a(rk1)/(r 1) + ark  [Using equation (i)]

= [a(rk1) + ark(r 1)]/(r 1)

= [arka + ark+1 ark]/(r 1)

= [ark+1 a]/(r 1)

a(rk+11)/(r 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.

Solution:

### Ex 4.1 Class 11 Maths Question 14:

Solution:

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.

### Ex 4.1 Class 11 Maths Question 15:

Prove the following by using the principle of mathematical induction for all n N:
12 + 32 + 52 + … + (2n – 1)2 = n(2n 1) (2n + 1)/3

Solution:

Solution:

Solution:

Solution:

### Ex 4.1 Class 11 Maths Question 19:

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

n (n + 1) (n + 5) is a multiple of 3.

Solution:
Let the given statement be P(n), then,
P(n) : n (n + 1) (n + 5), which is a multiple of 3.
It can be noted that P(n) is true for n = 1, since 1 (1 + 1) (1 + 5) = 12, which is a multiple of 3.
Let P(k) be true for some positive integer k, then, k (k + 1) (k + 5) is a multiple of 3.
k (k + 1) (k + 5) = 3m, where m N ……………….(i)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider, (k + 1) {(k + 1) + 1} {(k + 1) + 5}
= (k + 1) (k + 2) {(k + 5) + 1}
= (k + 1) (k + 2) (k + 5) + (k + 1) (k + 2)
= {k (k + 1) (k + 5) + 2 (k + 1) (k + 5)} + (k + 1) (k + 2)
= 3m + (k + 1) {2 (k + 5) + (k + 2)}
= 3m + (k + 1) {2k + 10 + k + 2}
= 3m + (k + 1) (3k + 12)
= 3m + 3 (k + 1) (k + 4)
= 3 {m + (k + 1) (k + 4)}
= 3 × q, where, q = {m + (k + 1) (k + 4)} is some natural number.
Therefore, (k + 1) {(k + 1) + 1} {(k + 1) + 5} is a multiple of 3.
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.

### Ex 4.1 Class 11 Maths Question 20:

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

102n – 1 + 1 is divisible by 11.

Solution:

### Ex 4.1 Class 11 Maths Question 21:

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

x2n – y2n is divisible by x + y.

Solution:

### Ex 4.1 Class 11 Maths Question 22:

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

32n + 2 – 8n – 9 is divisible by 8.

Solution:

### Ex 4.1 Class 11 Maths Question 23:

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

41n – 14n is a multiple of 27.

Solution:

### Ex 4.1 Class 11 Maths Question 24:

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

(2n + 7) < (n + 3)2

Solution:

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