NCERT Solutions for Class 11 Science Math Chapter 4 Principle Of Mathematical Induction are provided here with simple step-by-step explanations. These solutions for Principle Of Mathematical Induction are extremely popular among Class 11 Science students for Math Principle Of Mathematical Induction Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 11 Science Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 11 Science Math are prepared by experts and are 100% accurate.
Page No 94:
Question 1:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
P(n): 1 + 3 + 32 + …+ 3n–1 =
For n = 1, we have
P(1): 1 =, which is true.
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider
1 + 3 + 32 + … + 3k–1 + 3(k+1) – 1
= (1 + 3 + 32 +… + 3k–1) + 3k
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.
Page No 94:
Question 2:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
P(n):
For n = 1, we have
P(1): 13
= 1 =,
which is true.
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider
13 + 23 + 33 + … + k3 + (k + 1)3
= (13
+ 23 + 33 + …. + k3) +
(k + 1)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.
Page No 94:
Question 3:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
P(n):
For n = 1, we have
P(1): 1 =
which is true.
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider
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.
Page No 94:
Question 4:
Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) =
Answer:
Let the given statement be P(n), i.e.,
P(n): 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) =
For n = 1, we have
P(1): 1.2.3 = 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)
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)
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.
Page No 94:
Question 5:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
P(n)
:
For n = 1, we have
P(1): 1.3
= 3,
which is true.
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider
1.3 + 2.32 + 3.33 + … + k3k+ (k + 1) 3k+1
= (1.3 + 2.32 + 3.33 + …+ k.3k) + (k + 1) 3k+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.
Page No 94:
Question 6:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
P(n):
For n = 1, we have
P(1):
,
which is true.
Let P(k) be true for some positive integer k, i.e.,
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)
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.
Page No 94:
Question 7:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
P(n):
For n = 1, we have
,
which is true.
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider
(1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + {2(k + 1) – 1}{2(k + 1) + 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.
Page No 94:
Question 8:
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
Answer:
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 + … + k.2k = (k – 1) 2k + 1 + 2 … (i)
We shall now prove that P(k + 1) is true.
Consider
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.
Page No 94:
Question 9:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
P(n):
For n = 1, we have
P(1):
,
which is true.
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider
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.
Page No 94:
Question 10:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
P(n):
For n = 1, we have
,
which is true.
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider
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.
Page No 94:
Question 11:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
P(n):
For n = 1, we have
,
which is true.
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider
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.
Page No 95:
Question 12:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
For n = 1, we have
, which is true.
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider
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.
Page No 95:
Question 13:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
For n = 1, we have
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider
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.
Page No 95:
Question 14:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
For n = 1, we have
,
which is true.
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider
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.
Page No 95:
Question 15:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider
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.
Page No 95:
Question 16:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider
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.
Page No 95:
Question 17:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
For n = 1, we have
,
which is true.
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider
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.
Page No 95:
Question 18:
Prove the following by using the principle of mathematical induction for all n ∈ N:
Answer:
Let the given statement be P(n), i.e.,
It can be
noted that P(n) is true for n = 1 since
.
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
Hence,
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.
Page No 95:
Question 19:
Prove the following by using the principle of mathematical induction for all n ∈ N: n (n + 1) (n + 5) is a multiple of 3.
Answer:
Let the given statement be P(n), i.e.,
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, i.e.,
k (k + 1) (k + 5) is a multiple of 3.
∴k (k + 1) (k + 5) = 3m, where m ∈ N … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
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.
Page No 95:
Question 20:
Prove the following by using the principle of mathematical induction for all n ∈ N: 102n – 1 + 1 is divisible by 11.
Answer:
Let the given statement be P(n), i.e.,
P(n): 102n – 1 + 1 is divisible by 11.
It can be observed that P(n) is true for n = 1 since P(1) = 102.1 – 1 + 1 = 11, which is divisible by 11.
Let P(k) be true for some positive integer k, i.e.,
102k – 1 + 1 is divisible by 11.
∴102k – 1 + 1 = 11m, where m ∈ N … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
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.
Page No 95:
Question 21:
Prove the following by using the principle of mathematical induction for all n ∈ N: x2n – y2n is divisible by x + y.
Answer:
Let the given statement be P(n), i.e.,
P(n): x2n – y2n is divisible by x + y.
It can be observed that P(n) is true for n = 1.
This is so because x2 × 1 – y2 × 1 = x2 – y2 = (x + y) (x – y) is divisible by (x + y).
Let P(k) be true for some positive integer k, i.e.,
x2k – y2k is divisible by x + y.
∴x2k – y2k = m (x + y), where m ∈ N … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
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.
Page No 95:
Question 22:
Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n – 9 is divisible by 8.
Answer:
Let the given statement be P(n), i.e.,
P(n): 32n + 2 – 8n – 9 is divisible by 8.
It can be observed that P(n) is true for n = 1 since 32 × 1 + 2 – 8 × 1 – 9 = 64, which is divisible by 8.
Let P(k) be true for some positive integer k, i.e.,
32k + 2 – 8k – 9 is divisible by 8.
∴32k + 2 – 8k – 9 = 8m; where m ∈ N … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
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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Question 23:
Prove the following by using the principle of mathematical induction for all n ∈ N: 41n – 14n is a multiple of 27.
Answer:
Let the given statement be P(n), i.e.,
P(n):41n – 14nis a multiple of 27.
It can be observed that P(n) is true for n = 1 since
,
which is a multiple of 27.
Let P(k) be true for some positive integer k, i.e.,
41k – 14kis a multiple of 27
∴41k – 14k = 27m, where m ∈ N … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
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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Question 24:
Prove the following by using the principle of mathematical induction for all
(2n +7) < (n + 3)2
Answer:
Let the given statement be P(n), i.e.,
P(n): (2n +7) < (n + 3)2
It can be observed that P(n) is true for n = 1 since 2.1 + 7 = 9 < (1 + 3)2 = 16, which is true.
Let P(k) be true for some positive integer k, i.e.,
(2k + 7) < (k + 3)2 … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
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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