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

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

Topics and Sub Topics in Class 11 Maths Chapter 4 Principle of Mathematical Induction:

Section Name	Topic Name
4	Principle of Mathematical Induction
4.1	Introduction
4.2	Motivation
4.3	The Principle of Mathematical Induction

NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1

Ex 4.1 Class 11 Maths Question 1:

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Ex 4.1 Class 11 Maths Question 2:

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

P(n) = 1 ³ + 2 ³ + 3 ³ + …………. + n ³ = \(\left(\frac{n(n+1)}{2}\right)^{2}\).
For n = 1, we have
P(1) : 1 ³ = 1
= \(\left(\frac{1(1+1)}{2}\right)^{2}=\left(\frac{12}{2}\right)^{2}\)
= 1 ² = 1 which is true.
Let P(k) be true for some positive integer k, i.e.,
1 ³ + 2 ³ + 3 ³ + …………. + k ³ = \(\left(\frac{k(k+1)}{2}\right)^{2}\)
We shall now prove that P(k+1) is true.
Consider 1 ³ + 2 ³ + 3 ³ + …………. + k ³ = \(\left(\frac{k(k+1)}{2}\right)^{2}\)

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Ex 4.1 Class 11 Maths Question 3:

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Ex 4.1 Class 11 Maths Question 4:

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Ex 4.1 Class 11 Maths Question 5:

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Ex 4.1 Class 11 Maths Question 6:

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= \(\frac{k(k+1)(k+2)}{3}\) + (k + 1) (k + 2)

= (k + 1) (k + 2) + (\(\frac{k}{3}\) + 1)

= \(\frac{(k+1)(k+2)(k+3)}{3}\)

= \(\frac{(k+1)(k+1+1)(k+1+2)}{3}\)
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:

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Let the given statement be P(n), i.e.,
P(n) : 1 . 3 + 3 . 5 + 5 . 7 + … + (2n – 1) (2n + 1) = \(\frac{n\left(4 n^{2}+6 n-1\right)}{3}\)
For n = 1, we have
P(1) : 1 . 3 = 3
= \(\frac{1\left(4.1^{2}+6.1-1\right)}{3}=\frac{4+6-1}{3}=\frac{9}{3}\) = 3, which is true.

Let P(k) be true for some positive integer k, Le.,
1 . 3 + 3 . 5 + 5 . 7 + ………………. + (2k – 1) (2k + 1) = \(\frac{k\left(4 k^{2}+6 k-1\right)}{3}\) ……………(i)
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.

Ex 4.1 Class 11 Maths Question 8:

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Ex 4.1 Class 11 Maths Question 9:

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

P(n) : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}\)

For n = 1, we have
P(1) : \(\frac{1}{2}\)
= 1 – \(\frac{1}{2^{1}}\)
= \(\frac{1}{2}\), which is true.

Let P(k) be true for some positive integer k, i.e.,
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{k}}=1-\frac{1}{2^{k}}\)

We shall now prove that P(k + 1) is true.
Consider \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{k}}\right)+\frac{1}{2^{k+1}}\)

= \(\left(1-\frac{1}{2^{k}}\right)+\frac{1}{2^{k+1}}\)

= \(1-\frac{1}{2^{k}}+\frac{1}{22^{k}}=1-\frac{1}{2^{k}}\left(1-\frac{1}{2}\right)\)

= \(1-\frac{1}{2^{k}}\left(\frac{1}{2}\right)=1-\frac{1}{2^{k+1}}\)

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.

Ex 4.1 Class 11 Maths Question 10:

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Ex 4.1 Class 11 Maths Question 11:

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Ex 4.1 Class 11 Maths Question 12:

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

P(n) : a + ar + ar ² + …………… + ar ^{n – 1} = \(\frac{a\left(r^{n}-1\right)}{r-1}\)

For n = 1, we have

P(1) : a = \(\frac{a\left(r^{1}-1\right)}{(r-1)}\) = a, which is true.

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

a + ar + ar ² + …………… + ar ^{k – 1} = \(\frac{a\left(r^{k}-1\right)}{r-1}\) …………..(i)

We shall now prove that P(k + 1) is true.
Consider
a + ar + ar ² + …………… + ar ^{k – 1} } + ar ^{(k + 1) – 1} = \(\frac{a\left(r^{k}-1\right)}{r-1}\) + ar ^k [Using eQuestion (i)]

= \(\frac{a\left(r^{k}-1\right)+a r^{k}(k-1)}{r-1}\)

= \(\frac{a\left(r^{k}-1\right)+a r^{k+1}-a r^{k}}{r-1}\)

= \(\frac{a r^{k}-a+a r^{k+1}-a r^{k}}{r-1}=\frac{a r^{k+1}-a}{r-1}\)

= \(\frac{a\left(r^{k+1}-1\right)}{r-1}\)

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.

Ex 4.1 Class 11 Maths Question 13:

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Ex 4.1 Class 11 Maths Question 14:

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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 indcution for all n ∈ N:
1 ² + 3 ² + 5 ² + … + (2n – 1) ² = \(\frac{n(2 n-1)(2 n+1)}{3}\)
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Ex 4.1 Class 11 Maths Question 16:

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Ex 4.1 Class 11 Maths Question 17:

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Ex 4.1 Class 11 Maths Question 18:

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Ex 4.1 Class 11 Maths Question 19:

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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=l 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 ……………….(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 trtie 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:

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Ex 4.1 Class 11 Maths Question 21:

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Ex 4.1 Class 11 Maths Question 22:

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Ex 4.1 Class 11 Maths Question 23:

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Ex 4.1 Class 11 Maths Question 24:

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Prove the following through the principle of mathematical induction for all values of n, where n is a natural number.

1) 1 + 3 + 3² + … . + 3 ^n-1 =\(\frac { { (3 }^{ n }-1) }{ 2 } \)

2: 1³ + 2 ³ + 3 ³ + … … + n ³ = \({ (\frac { n(n+1) }{ 2 } ) }^{ 2 }\)

3: \(1+\frac { 1 }{ 1+2 } +\frac { 1 }{ 1+2+3 } +……+\frac { 1 }{ 1+2+3+…+n } =\frac { 2n }{ n+1 } \)

4: 1.2.3 + 2.3.4 + … + n ( n + 1 ) ( n + 2 ) = \(\frac { n(n+1)(n+2)(n+3) }{ 4 } \)

5: 1.3 + 2. 3 2 + 3.3 3 + … + n . 3 n

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

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