# NCERT solutions for Mathematics Exemplar Class 11 chapter 4 - Principle of Mathematical Induction [Latest edition]

## Chapter 4: Principle of Mathematical Induction

Solved ExamplesExercise
Solved Examples [Pages 61 - 70]

### NCERT solutions for Mathematics Exemplar Class 11 Chapter 4 Principle of Mathematical Induction Solved Examples [Pages 61 - 70]

Solved Examples | Q 1 | Page 61

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

1 + 3 + 5 + ... + (2n – 1) = n2

Solved Examples | Q 2 | Page 62

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

sum_(t = 1)^(n - 1) t(t + 1) = (n(n - 1)(n + 1))/3, for all natural numbers n ≥ 2.

Solved Examples | Q 3 | Page 63

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

(1 - 1/2^2).(1 - 1/3^2)...(1 - 1/n^2) = (n + 1)/(2n), for all natural numbers, n ≥ 2.

Solved Examples | Q 4 | Page 63

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

22n – 1 is divisible by 3.

Solved Examples | Q 5 | Page 64

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

2n + 1 < 2n, for all natual numbers n ≥ 3.

Solved Examples | Q 6.(i) | Page 64

Define the sequence a1, a2, a3 ... as follows:
a1 = 2, an = 5 an–1, for all natural numbers n ≥ 2.

Write the first four terms of the sequence.

Solved Examples | Q 6.(ii) | Page 64

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.

Solved Examples | Q 7 | Page 65

The distributive law from algebra says that for all real numbers c, a1 and a2, we have c(a1 + a2) = ca1 + ca2.

Use this law and mathematical induction to prove that, for all natural numbers, n ≥ 2, if c, a1, a2, ..., an are any real numbers, then c(a1 + a2 + ... + an) = ca1 + ca2 + ... + can.

Solved Examples | Q 8 | Page 65

Prove by induction that for all natural number n sinα + sin(α + β) + sin(α + 2β)+ ... + sin(α + (n – 1)β) = (sin (alpha + (n - 1)/2 beta)sin((nbeta)/2))/(sin(beta/2))

Solved Examples | Q 9 | Page 67

Prove by the Principle of Mathematical Induction that 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! – 1 for all natural numbers n.

Solved Examples | Q 10 | Page 67

Show by the Principle of Mathematical Induction that the sum Sn of the n term of the series 12 + 2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 ... is given by

Sn = {{:((n(n + 1)^2)/2",",  "if n is even"),((n^2(n + 1))/2",",  "if n is odd"):}

#### Objective Type Questions Choose the correct answer in Examples 11 and 12 (M.C.Q.)

Solved Examples | Q 11 | Page 69

Let P(n): “2n < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is ______.

• 1

• 2

• 3

• 4

Solved Examples | Q 12 | Page 69

A student was asked to prove a statement P(n) by induction. He proved that P(k + 1) is true whenever P(k) is true for all k > 5 ∈ N and also that P(5) is true. On the basis of this he could conclude that P(n) is true ______.

• For all n ∈ N

• For all n > 5

• For all n ≥ 5

• For all n < 5

#### Fill in the blanks 13 and 14:

Solved Examples | Q 13 | Page 69

If P(n) : “2.42n+1 + 33n+1 is divisible by λ for all n ∈ N” is true, then the value of λ is ______.

Solved Examples | Q 14 | Page 70

If P(n): “49n + 16n + k is divisible by 64 for n ∈ N” is true, then the least negative integral value of k is ______.

Solved Examples | Q 15 | Page 70

State whether the following proof (by mathematical induction) is true or false for the statement.

P(n): 12 + 22 + ... + n2 = (n(n + 1) (2n + 1))/6

Proof By the Principle of Mathematical induction, P(n) is true for n = 1,

12 = 1 = (1(1 + 1)(2*1 + 1))/6. Again for some k ≥ 1, k2 = (k(k + 1)(2k + 1))/6. Now we prove that

(k + 1)2 = ((k + 1)((k + 1) + 1)(2(k + 1) + 1))/6

• True

• False

Exercise [Pages 70 - 72]

### NCERT solutions for Mathematics Exemplar Class 11 Chapter 4 Principle of Mathematical Induction Exercise [Pages 70 - 72]

Exercise | Q 1 | Page 70

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

Exercise | Q 2 | Page 70

Give an example of a statement P(n) which is true for all n. Justify your answer.

Exercise | Q 3 | Page 70

Prove the statement by using the Principle of Mathematical Induction:

4n – 1 is divisible by 3, for each natural number n.

Exercise | Q 4 | Page 70

Prove the statement by using the Principle of Mathematical Induction:

23n – 1 is divisible by 7, for all natural numbers n.

Exercise | Q 5 | Page 70

Prove the statement by using the Principle of Mathematical Induction:

n3 – 7n + 3 is divisible by 3, for all natural numbers n.

Exercise | Q 6 | Page 70

Prove the statement by using the Principle of Mathematical Induction:

32n – 1 is divisible by 8, for all natural numbers n.

Exercise | Q 7 | Page 71

Prove the statement by using the Principle of Mathematical Induction:

For any natural number n, 7n – 2n is divisible by 5.

Exercise | Q 8 | Page 71

Prove the statement by using the Principle of Mathematical Induction:

For any natural number n, xn – yn is divisible by x – y, where x and y are any integers with x ≠ y.

Exercise | Q 9 | Page 71

Prove the statement by using the Principle of Mathematical Induction:

n3 – n is divisible by 6, for each natural number n ≥ 2.

Exercise | Q 10 | Page 71

Prove the statement by using the Principle of Mathematical Induction:

n(n2 + 5) is divisible by 6, for each natural number n.

Exercise | Q 11 | Page 71

Prove the statement by using the Principle of Mathematical Induction:

n2 < 2n for all natural numbers n ≥ 5.

Exercise | Q 12 | Page 71

Prove the statement by using the Principle of Mathematical Induction:

2n < (n + 2)! for all natural number n.

Exercise | Q 13 | Page 71

Prove the statement by using the Principle of Mathematical Induction:

sqrt(n) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(n), for all natural numbers n ≥ 2.

Exercise | Q 14 | Page 71

Prove the statement by using the Principle of Mathematical Induction:

2 + 4 + 6 + ... + 2n = n2 + n for all natural numbers n.

Exercise | Q 15 | Page 71

Prove the statement by using the Principle of Mathematical Induction:

1 + 2 + 22 + ... + 2n = 2n+1 – 1 for all natural numbers n.

Exercise | Q 16 | Page 71

Prove the statement by using the Principle of Mathematical Induction:

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

#### Long Answer Use the Principle of Mathematical Induction in the following

Exercise | Q 17 | Page 71

A sequence a1, a2, a3 ... is defined by letting a1 = 3 and ak = 7ak – 1 for all natural numbers k ≥ 2. Show that an = 3.7n–1 for all natural numbers.

Exercise | Q 18 | Page 71

A sequence b0, b1, b2 ... is defined by letting b0 = 5 and bk = 4 + bk – 1 for all natural numbers k. Show that bn = 5 + 4n for all natural number n using mathematical induction.

Exercise | Q 19 | Page 71

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

Exercise | Q 20 | Page 71

Prove that for all n ∈ N.
cos α + cos(α + β) + cos(α + 2β) + ... + cos(α + (n – 1)β) = (cos(alpha + ((n - 1)/2)beta)sin((nbeta)/2))/(sin  beta/2).

Exercise | Q 21 | Page 71

Prove that, cosθ cos2θ cos22θ ... cos2n–1θ = (sin 2^n theta)/(2^n sin theta), for all n ∈ N.

Exercise | Q 22 | Page 71

Prove that, sinθ + sin2θ + sin3θ + ... + sinnθ = ((sin ntheta)/2 sin  ((n + 1))/2 theta)/(sin  theta/2), for all n ∈ N.

Exercise | Q 23 | Page 72

Show that n^5/5 + n^3/3 + (7n)/15 is a natural number for all n ∈ N.

Exercise | Q 24 | Page 72

Prove that 1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24, for all natural numbers n > 1.

Exercise | Q 25 | Page 72

Prove that number of subsets of a set containing n distinct elements is 2n, for all n ∈ N.

#### Objective Type Questions from 26 to 30

Exercise | Q 26 | Page 72

If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.

• 5

• 3

• 7

• 1

Exercise | Q 27 | Page 72

For all n ∈ N, 3.52n+1 + 23n+1 is divisible by ______.

• 19

• 17

• 23

• 25

Exercise | Q 28 | Page 72

If xn – 1 is divisible by x – k, then the least positive integral value of k is ______.

• 1

• 2

• 3

• 4

#### Fill in the blanks in the following:

Exercise | Q 29 | Page 72

If P(n): 2n < n!, n ∈ N, then P(n) is true for all n ≥ ______.

Exercise | Q 30 | Page 72

State whether the following statement is true or false. Justify.

Let P(n) be a statement and let P(k) ⇒ P(k + 1), for some natural number k, then P(n) is true for all n ∈ N.

• True

• False

## Chapter 4: Principle of Mathematical Induction

Solved ExamplesExercise

## NCERT solutions for Mathematics Exemplar Class 11 chapter 4 - Principle of Mathematical Induction

NCERT solutions for Mathematics Exemplar Class 11 chapter 4 (Principle of Mathematical Induction) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Mathematics Exemplar Class 11 solutions in a manner that help students grasp basic concepts better and faster.

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Concepts covered in Mathematics Exemplar Class 11 chapter 4 Principle of Mathematical Induction are Motivation, Principle of Mathematical Induction.

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