Math Ncert Exemplar 2019 Solutions for Class 11 Science Maths Chapter 4 - Principle Of Mathematical Induction

Share with your friends

Math Ncert Exemplar 2019 Solutions for Class 11 Science Maths 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 Maths Principle Of Mathematical Induction Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of class 11 Science Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class 11 Science Maths are prepared by experts and are 100% accurate.

Page No 70:

Question 1:

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.

Answer:

Consider the statement

$P (n) : 3 n < n! For n = 1, 3 \times 1 = 3 < 1!, which is not true For n = 2, 3 \times 2 = 6 < 2!, which is not true For n = 3, 3 \times 3 = 9 < 3!, which is not true For n = 4, 3 \times 4 = 12 < 4!, which is true For n = 5, 3 \times 5 = 15 < 5!, which is true$

Page No 70:

Question 2:

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

Answer:

$Consider the statement P (n) : 1 + 2 + 3 + 4 . . . . . . n = \frac{n (n + 1)}{2} When n = 1, P (1) : 1 = \frac{1 (1 + 1)}{2} \Rightarrow 1 = \frac{1 (2)}{2} \Rightarrow LHS = RHS Therefore P (1) is true When n = 2, P (2) : 1 + 2 = \frac{2 (2 + 1)}{2} \Rightarrow 3 = \frac{2 (3)}{2} \Rightarrow LHS = RHS Therefore P (2) is true When n = 3, P (3) : 1 + 2 + 3 = \frac{3 (3 + 1)}{2} \Rightarrow 6 = \frac{3 (4)}{2} \Rightarrow LHS = RHS Therefore P (3) is true Similarly, P (n) will be true for all the other natural numbers .$

Page No 70:

Question 3:

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

Answer:

$P (n) : 4^{n} - 1 is divisible by 3, \forall n \in N P (1) = 4^{1} - 1 = 3, which is divisible by 3 So P (1) is true . Let P (k) be true . P (k) = 4^{k} - 1 = 3 m . . . . . (1) k \in N, m \in Z We shall now prove that P (k + 1) is true whenever P (k) is true . P (k + 1) = 4^{k + 1} - 1 P (k + 1) = 4^{k} . 4 - 1 P (k + 1) = (3 m + 1) . 4 - 1 [using (1)] P (k + 1) = 3 m \times 4 + 4 - 1 P (k + 1) = 12 m + 3 P (k + 1) = 3 (4 m + 1) P (k + 1) is clearly divisible by 3, so it' s true . Hence, by principal of mathematical induction we can say P (n) is true .$

Page No 70:

Question 4:

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

Answer:

$P (n) : 2^{3 n} - 1 is divisible by 7, \forall n \in N P (1) = 2^{3 \times 1} - 1 = 7, which is divisible by 7 So P (1) is true . Let P (k) be true . P (k) = 2^{3 k} - 1 = 7 m . . . . . (1) k \in N, m \in Z We shall now prove that P (k + 1) is true whenever P (k) is true . P (k + 1) = 2^{3 (k + 1)} - 1 P (k + 1) = 2^{3 k + 3} - 1 P (k + 1) = 2^{3 k} . 2^{3} - 1 = 2^{3 k} . 8 - 1 P (k + 1) = (7 m + 1) . 8 - 1 [using (1)] P (k + 1) = 7 m \times 8 + 8 - 1 P (k + 1) = 56 m + 7 P (k + 1) = 7 (8 m + 1) P (k + 1) is clearly divisible by 7, so it' s true . Hence, by principal of mathematical induction we can say P (n) is true .$

Page No 70:

Question 5:

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

Answer:

$P (n) : n^{3} - 7 n + 3 is divisible by 3, \forall n \in N P (1) = 1^{3} - 7 \times 1 + 3 = - 3, which is divisible by 3 So P (1) is true . Let P (k) be true . We shall now prove that P (k + 1) is true whenever P (k) is true . P (k) = k^{3} - 7 k + 3 = 3 m . . . . . (1) k \in N, m \in Z P (k + 1) = {(k + 1)}^{3} - 7 (k + 1) + 3 P (k + 1) = k^{3} + 1 + 3 k^{2} + 3 k - 7 k - 7 + 3 P (k + 1) = (k^{3} - 7 k + 3) + 3 k^{2} + 3 k - 6 P (k + 1) = 3 m + 3 (k^{2} + k - 2) [using (1)] P (k + 1) = 3 (m + k^{2} + k - 2) P (k + 1) is clearly divisible by 3, so it' s true . Hence, by principal of mathematical induction we can say P (n) is true .$

Page No 70:

Question 6:

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

Answer:

$P (n) : 3^{2 n} - 1 is divisible by 8, \forall n \in N P (1) = 3^{2 \times 1} - 1 = 8, which is divisible by 8 So P (1) is true . Let P (k) be true . P (k) = 3^{2 k} - 1 = 8 m . . . . . (1) k \in N, m \in Z We shall now prove that P (k + 1) is true whenever P (k) is true . P (k + 1) = 3^{2 (k + 1)} - 1 P (k + 1) = 3^{2 k + 2} - 1 P (k + 1) = 3^{2 k} . 3^{2} - 1 = 3^{2 k} . 9 - 1 P (k + 1) = (8 m + 1) . 9 - 1 [using (1)] P (k + 1) = 8 m \times 9 + 9 - 1 P (k + 1) = 72 m + 8 P (k + 1) = 8 (9 m + 1) P (k + 1) is clearly divisible by 8, so it' s true . Hence, by principal of mathematical induction we can say P (n) is true .$

Page No 71:

Question 7:

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

Answer:

$P (n) : 7^{n} - 2^{n} is divisible by 5, \forall n \in N P (1) = 7^{1} - 2^{1} = 5, which is divisible by 5 So P (1) is true . Let P (k) be true . P (k) = 7^{k} - 2^{k} = 5 m . . . . . (1) k \in N, m \in Z We shall now prove that P (k + 1) is true whenever P (k) is true . P (k + 1) = 7^{(k + 1)} - 2^{(k + 1)} P (k + 1) = 7^{k} . 7 - 2^{k} . 2 P (k + 1) = (5 m + 2^{k}) . 7 - 2^{k} . 2 [using (1)] P (k + 1) = 35 m + 2^{k} . 7 - 2^{k} . 2 P (k + 1) = 35 m + 5 . 2^{k} P (k + 1) = 5 (7 m + 2^{k}) P (k + 1) is clearly divisible by 5, so it' s true . Hence, by principal of mathematical induction we can say P (n) is true .$

Page No 71:

Question 8:

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

Answer:

$P (n) : x^{n} - y^{n} is divisible by x - y, \forall n \in N P (1) = x^{1} - y^{1} = x - y, which is divisible by x - y So P (1) is true . Let P (k) be true . P (k) = x^{k} - y^{k} = m (x - y) . . . . . (1) k \in N, m \in Z We shall now prove that P (k + 1) is true whenever P (k) is true . P (k + 1) = x^{(k + 1)} - y^{(k + 1)} P (k + 1) = x^{k} . x - y^{k} . y P (k + 1) = \{m (x - y) + y^{k}\} . x - y^{k} . y [using (1)] P (k + 1) = \{m x - m y + y^{k}\} . x - y^{k} . y P (k + 1) = m x^{2} - m x y + x . y^{k} - y^{k} y P (k + 1) = m x (x - y) + y^{k} (x - y) P (k + 1) = (x - y) (m + y^{k}) P (k + 1) is clearly divisible by x - y, so it' s true . Hence, by principal of mathematical induction we can say P (n) is true .$

Page No 71:

Question 9:

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

Answer:

$P (n) : n^{3} - n is divisible by 6, \forall n \in N P (1) = 1^{3} - 1 = 0, which is divisible by 6 So P (1) is true . Let P (k) be true . P (k) = k^{3} - k = 6 m . . . . . (1) k \in N, m \in Z We shall now prove that P (k + 1) is true whenever P (k) is true . P (k + 1) = {(k + 1)}^{3} - (k + 1) P (k + 1) = k^{3} + 1 + 3 k^{2} + 3 k - k - 1 P (k + 1) = (k^{3} - k) + 3 k^{2} + 3 k P (k + 1) = 6 m + 3 (k^{2} + k) [using (1)] P (k + 1) = 6 m + 3 k (k + 1) we can write k (k + 1) = 2 y, y \in Z as it will always be even P (k + 1) = 6 m + 3 (2 y) \Rightarrow 6 (m + y) P (k + 1) is clearly divisible by 6, so it' s true . Hence, by principal of mathematical induction we can say P (n) is true .$

Page No 71:

Question 10:

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

Answer:

$P (n) : n (n^{2} + 5) is divisible by 6, \forall n \in N P (1) = 1 (1^{2} + 5) = 6, which is divisible by 6 So P (1) is true . Let P (k) be true . P (k) = k (k^{2} + 5) = 6 m . . . . . (1) k \in N, m \in Z We shall now prove that P (k + 1) is true whenever P (k) is true . P (k + 1) = (k + 1) \{{(k + 1)}^{2} + 5\} P (k + 1) = (k + 1) (k^{2} + 1 + 2 k + 5) P (k + 1) = k^{3} + 2 k^{2} + 6 k + k^{2} + 2 k + 6 P (k + 1) = k^{3} + 5 k + 3 k^{2} + 3 k + 6 P (k + 1) = 6 m + 3 k (k + 1) + 6 [using (1)] we can write k (k + 1) = 2 y, y \in Z as it will always be even P (k + 1) = 6 m + 3 (2 y) + 6 \Rightarrow 6 (m + y + 1) P (k + 1) is clearly divisible by 6, so it' s true . Hence, by principal of mathematical induction we can say P (n) is true .$

Page No 71:

Question 11:

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

Answer:

$P (n) : n^{2} < 2^{n}, \forall n \geq 5 P (5) : 5^{2} < 2^{5} \Rightarrow 25 < 32 So P (5) is true . Let P (k) be true . Then, P (k) : k^{2} < 2^{k} . . . . . (1) We shall now prove that P (k + 1) is true whenever P (k) is true . {(k + 1)}^{2} = k^{2} + 1 + 2 k < 2^{k} + 2 k + 1 . . . . . (2) [From (1)] Let 2^{k} + 2 k + 1 < 2^{(k + 1)} \Rightarrow 2^{k} + 2 k + 1 < 2^{k} . 2 \Rightarrow 2^{k} + 2 k + 1 < 2^{k} + 2^{k} \Rightarrow 2 k + 1 < 2^{k} \forall k \geq 5 . . . . . (3) So, {(k + 1)}^{2} < 2^{k} + 2^{k} \Rightarrow {(k + 1)}^{2} < 2^{(k + 1)} \Rightarrow P (k + 1) it' s true . Hence, by principal of mathematical induction we can say P (n) is true .$

Page No 71:

Question 12:

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

Answer:

Let P(n) : 2n < (n + 2)! ∀n ∈ N

Check that statement is true for n = 1

P(1): 2(1) < (1+2)!

P(1): 2 < 6

So, P(n) is true for n = 1

Assume P(k) to be true and then prove P(k+1) is true.

Assume that P(k) is true for some natural number k.

⇒ 2k < (k + 2)! ..…(1)

To prove : P(k + 1) is true, we have to show that

2(k + 1) < ((k + 1)+2)!

Or 2(k + 1) < (k + 3)!

∴ 2k < (k + 2)!

Adding 2 both the sides

⇒ 2k + 2 < (k + 2)! + 2

⇒ 2(k + 1) < (k + 2)! + 2 …..(2)

Now, we know that

(k + 3)! = (k + 3)(k + 2)! which is greater than (k + 2)! + 2 as (k + 2)! is (k + 3) times in the former and one time in the later.

⇒ (k + 2)! + 2 < (k + 3)! …..(3)

Using equations (2) & (3), we get

2(k + 1) < (k + 3)!

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

Hence, By Principle of mathematical Induction P(n) is true for all natural numbers n.

Page No 71:

Question 13:

$\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}}$ , for all natural numb n ≥ 2ers.

Answer:

$P (2) is \sqrt{2} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} = 1.414 < 1.70, thus it is true . P (3) i s \sqrt{3} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} = 1.732 < 2.284, thus it is true . Let us consider P (k) = \sqrt{k} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{k}}, is true . Add \sqrt{k + 1} on both side \sqrt{k} + \sqrt{k + 1} - \sqrt{k} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{k}} + \sqrt{k + 1} - \sqrt{k} \Rightarrow \sqrt{k + 1} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k + 1}}$

Hence, by mathematical induction, for each natural number n ≥ 2, P(n) is $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}}$ .

Page No 71:

Question 14:

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

Answer:

Let P(n) :2 + 4 + 6+ …+ 2n = n² + n
P(1): 2 = 12 + 1 = 2, which is true
Hence, P(1) is true.

Let us assume that P(n) is true for some natural number n = k.
∴ P(k): 2 + 4 + 6 + ...+ 2k = k² + k .....(1)

Now, we have to prove that P(k + 1) is true.
P(k + 1): 2 + 4 + 6 + 8 + …+ 2k + 2(k +1)
= k² + k + 2(k+ 1) [Using (1)]
= k² + k + 2k + 2
= k² + 2k + 1 + k + 1
= (k + 1)2 + k + 1
Hence, P(k + 1) is true whenever P(k) is true.

So, by the principle of mathematical induction P(n) is true for any natural number n.

Page No 71:

Question 15:

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

Answer:

Let P(n): 1 + 2 + 22 + … + 2ⁿ = 2^n +1 – 1, for all natural numbers n
P(1): 1 =2^{0 + 1} − 1 = 2 − 1 = 1, which is true.
Hence, P(1) is true.

Let us assume that P(n) is true for some natural number n = k.

P(k): l + 2 + 2² +…+ 2^k = 2^{k +}¹ − 1 .....(1)

Now, we have to prove that P(k + 1) is true.

P(k + 1): 1+2 + 2² +…+ 2^k + 2^{k +}¹

= 2^{k +}¹ − 1 + 2^{k +}¹ [Using (1)]
= 2 × 2^{k +}¹ − 1
= 2⁽^{k +}^{1) + 1} â€‹− 1

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

So, by the principle of mathematical induction P(n) is true for any natural number n.

Page No 71:

Question 16:

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

Answer:

Let P(n): 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers n.
P(1): 1 = 1(2 × 1 – 1) = 1, which is true.
Hence, P(l) is true.

Let us assume that P(n) is true for some natural number n = k.
∴ P(k): l + 5 + 9 +…+ (4k – 3) = k(2k – 1) .....(1)

Now, we have to prove that P(k + 1) is true.
P(k+ 1): 1 + 5 + 9 + … + (4k – 3) + [4(k + 1) – 3]
= 2k² – k + 4k + 4 – 3
= 2k² + 3k + 1
= (k + 1)(2k + 1)
= (k + l)[2(k + l) – l]

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

So, by the principle of mathematical induction P(n) is true for any natural number n.

Page No 71:

Question 17:

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

Answer:

We have a sequence a₁, a₂, a₃ ... is defined by letting a₁ = 3 and a_k = 7a_{k – 1} for all natural numbers k ≥ 2.

Given: a₁ = 3 and a_k = 7a_{k – 1}

$\begin{array}{rcl} a_{2} & = & 7 \times a_{1} \\ = & 7 \times 3 \\ = & 21 \\ = & 3 \cdot 7^{2 - 1} \\ a_{3} & = & 7 \times a_{2} \\ = & 7^{2} \times a_{1} \\ = & 49 \times 3 \\ = & 147 \\ = & 3 \cdot 7^{3 - 1} \\ a_{n} & = & 7 \times a_{n - 1} \\ = & 3 \cdot 7^{n - 1} \end{array}$

Thus,

$\begin{array}{rcl} a_{n + 1} & = & 7 \times a_{n + 1 - 1} \\ = & 7 \times a_{n} \\ = & 7 \times 3 \times 7^{n - 1} \\ = & 3 \times 7^{(n + 1) - 1} \end{array}$

Thus, $a_{n + 1}$ is true for $a_{n}$ is true.

For each natural number n it is true that a_n = 3.7^{n – 1}.

Page No 71:

Question 18:

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

Answer:

We have a sequence b₀, b₁, b₂ ... is defined by letting b₀ = 5 and b_k = 4 + b_{k – 1} for all natural numbers k.

Let P(n): b_n = 5 + 4n, for all natural numbers.

For n = 1, b₁ = 5 + 4 × 1 = 9
Also, b₀ = 5
∴ b₁ = 4 + b₀ = 4 + 5 = 9
Thus, P(1) is true.
Now, let us assume that P(n) is true for some natural number n = m.

∴ P(m): b_m = 5 + 4m

Now, to prove that P(k + 1) is true, we have to show that
P(m + 1): b_{m + 1} = 5 + 4(m + 1)

b_{m + 1} = 4 + b_{(m + 1) − 1}

b_{m + 1} = 4 + b_m = 4 + 5 + 4m = 5 + 4(m + 1)

Hence, P(m + 1) is true whenever P(m) is true.

So, by the principle of mathematical induction P(n) is true for any natural number.

Page No 71:

Question 19:

A sequence d₁, d₂, d₃ ... is defined by letting d₁ = 2 and $d_{k} = \frac{d_{k - 1}}{k}$ for all natural numbers, k ≥ 2. Show that $d_{n} = \frac{2}{n!}$ for all n ∈ N.

Answer:

Given that: d₁ = 2 and $d_{k} = \frac{d_{k - 1}}{k}$

Let P(n): $d_{n} = \frac{2}{n!}$

Now, P(1): $d_{1} = \frac{2}{1!} = 2$ , which is true for P(1).

Let it be true for P(k).

∴ P(k): $d_{k} = \frac{2}{k!}$

Given that: $d_{k} = \frac{d_{k - 1}}{k}$

$∴ d_{k + 1} = \frac{d_{k + 1 - 1}}{k + 1} = \frac{d_{k}}{k + 1} \Rightarrow d_{k + 1} = \frac{1}{k + 1} \cdot d_{k} \Rightarrow d_{k + 1} = \frac{1}{k + 1} \cdot \frac{2}{k!} \Rightarrow d_{k + 1} = \frac{2}{(k + 1)!}$

Which is true for P(k + 1).

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

Page No 71:

Question 20:

Prove that for all n ∈ N
cos α + cos (α + β) + cos (α + 2β) + ... + cos (α + (n –1) β)

$= \frac{\cos (α + (\frac{n - 1}{2}) β) \sin (\frac{n β}{2})}{\sin \frac{β}{2}}$

Answer:

$Let P (n) be the given statement . i . e ., P (n) : \cos α + \cos (α + β) + \cos (α + 2 β) + . . . + \cos (α + (n - 1) β) = \frac{\cos \{α + (\frac{n - 1}{2}) β\} \sin (\frac{n β}{2})}{\sin \frac{β}{2}} We note that P (n) is true for n = 1 since \cos α = \frac{\cos \{α + (\frac{1 - 1}{2}) β\} \sin (\frac{1 \times β}{2})}{\sin \frac{β}{2}} Assume that P (k) is true i . e . P (k) : \cos α + \cos (α + β) + \cos (α + 2 β) + . . . + \cos (α + (k - 1) β) = \frac{\cos \{α + (\frac{k - 1}{2}) β\} \sin (\frac{k β}{2})}{\sin \frac{β}{2}} . . . . . (1) We shall now prove that P (k + 1) is true whenever P (k) is true . We have, P (k + 1) : \cos α + \cos (α + β) + \cos (α + 2 β) + . . . + \cos (α + (k - 1) β) + \cos \{α + (k + 1 - 1) β\} P (k + 1) : \cos α + \cos (α + β) + \cos (α + 2 β) + . . . + \cos \{α + (k - 1) β\} + \cos (α + k β) P (k + 1) : \frac{\cos \{α + (\frac{k - 1}{2}) β\} \sin (\frac{k β}{2})}{\sin \frac{β}{2}} + \cos (α + k β) . . . . . using (1) P (k + 1) : \frac{\cos \{α + (\frac{k - 1}{2}) β\} \sin (\frac{k β}{2}) + \cos (α + k β) \sin \frac{β}{2}}{\sin \frac{β}{2}} P (k + 1) : \frac{2 \cos \{α + (\frac{k - 1}{2}) β\} \sin (\frac{k β}{2}) + 2 \cos (α + k β) \sin \frac{β}{2}}{2 \sin \frac{β}{2}} P (k + 1) : \frac{\sin (α + (\frac{k - 1}{2}) β + \frac{k β}{2}) + \sin (α + (\frac{k - 1}{2}) β - \frac{k β}{2}) + \sin (α + k β + \frac{β}{2}) - \sin (α + k β - \frac{β}{2})}{2 \sin \frac{β}{2}} P (k + 1) : \frac{\sin (α + k β - \frac{β}{2}) - \sin (α - \frac{β}{2}) + \sin (α + k β + \frac{β}{2}) - \sin (α + k β - \frac{β}{2})}{2 \sin \frac{β}{2}} P (k + 1) : \frac{\sin (α + k β + \frac{β}{2}) - \sin (α - \frac{β}{2})}{2 \sin \frac{β}{2}} P (k + 1) : \frac{2 \cos (\frac{α + k β + \frac{β}{2} + α - \frac{β}{2}}{2}) \sin (\frac{α + k β + \frac{β}{2} - α + \frac{β}{2}}{2})}{2 \sin \frac{β}{2}} P (k + 1) : \frac{\cos (\frac{2 α + k β}{2}) \sin (\frac{k β + β}{2})}{\sin \frac{β}{2}} P (k + 1) : \frac{\cos \{α + (\frac{k}{2}) β\} \sin \{(k + 1) β\}}{\sin \frac{β}{2}} Therefore, P (k + 1) is also true whenever P (k) is true . Hence, by mathematical induction P (n) is true for all n \in N .$

Page No 71:

Question 21:

Prove that, cos θ cos 2θ cos2²θ ... cos2^n–1θ = $\frac{\sin 2^{n} θ}{2^{n} \sin θ}$ , for all n ∈ N.

Answer:

Let P(n): cos θ cos 2θ cos2²θ ... cos2^n–1θ = $\frac{\sin 2^{n} θ}{2^{n} \sin θ}$

$P (1) : \cos θ = \frac{\sin 2^{1} θ}{2^{1} \sin θ} = \frac{\sin 2 θ}{2 \sin θ} = \frac{2 \sin θ \cos θ}{2 \sin θ} = \cos θ, which is true .$

Hence, P(1) is true.

Now, let us assume that P(n) is true for some natural number n = k.

$P (k) : \cos θ \cos 2^{2} θ . . . \cos 2^{k - 1} θ = \frac{\sin 2^{k} θ}{2^{k} \sin θ} . . . . . (1)$

To prove that P(k + 1) is true, we have to show that

$P (k + 1) : \cos θ \cos 2^{2} θ . . . \cos 2^{k - 1} θ \cos 2^{k} θ = \frac{\sin 2^{k + 1} θ}{2^{k + 1} \sin θ}$

$\begin{array}{rcl} Now, \cos θ \cos 2^{2} θ . . . \cos 2^{k - 1} θ \cos 2^{k} θ & = & \frac{\sin 2^{k} θ}{2^{k} \sin θ} \cdot \cos 2^{k} θ \\ = & \frac{2 \cdot \sin 2^{k} θ \cos 2^{k} θ}{2 \cdot 2^{k} \sin θ} \\ = & \frac{\sin 2 \cdot 2^{k} θ}{2^{k + 1} \sin θ} \\ = & \frac{\sin 2^{k + 1} θ}{2^{k + 1} \sin θ} \end{array}$

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

Page No 71:

Question 22:

Prove that, sin θ + sin 2θ + sin 3 θ + ... + sin nθ $= \frac{\frac{\sin n θ}{2} \sin \frac{(n + 1)}{2} θ}{\sin \frac{θ}{2}}$ , for all n ∈ N.

Answer:

Given that: sin θ + sin 2θ + sin 3 θ + ... + sin nθ $= \frac{\frac{\sin n θ}{2} \sin \frac{(n + 1)}{2} θ}{\sin \frac{θ}{2}}$ , for all n ∈ N

Now, for n = 1
$P (1) : \sin θ = \frac{\frac{\sin θ}{2} \sin \frac{(1 + 1)}{2} θ}{\sin \frac{θ}{2}} P (1) : \sin θ = \frac{\frac{\sin θ}{2} \sin θ}{\sin \frac{θ}{2}} P (1) : \sin θ = \sin θ$

Hence, P(1) is true.

Assume that P(n) is true for a natural number n = k.

P(k) : sin θ + sin 2θ + sin 3 θ + ... + sin kθ $= \frac{\frac{\sin k θ}{2} \sin \frac{(k + 1)}{2} θ}{\sin \frac{θ}{2}}$ ....(1)

Now, to prove that P(k + 1) is true, we have to show that:

P(k + 1) : sin θ + sin 2θ + sin 3 θ + ... + sin kθ + sin (k + 1)θ $= \frac{\frac{\sin (k + 1) θ}{2} \sin \frac{(k + 1 + 1)}{2} θ}{\sin \frac{θ}{2}}$

From (1), we have

P(k + 1) : sin θ + sin 2θ + sin 3 θ + ... + sin kθ + sin (k + 1)θ $= \frac{\frac{\sin k θ}{2} \sin \frac{(k + 1)}{2} θ}{\sin \frac{θ}{2}} + \sin (k + 1) θ$

$= \frac{\frac{\sin k θ}{2} \sin \frac{(k + 1)}{2} θ + \sin (k + 1) θ \sin \frac{θ}{2}}{\sin \frac{θ}{2}} = \frac{\cos [\frac{k θ}{2} - (\frac{k + 1}{2}) θ] - \cos [\frac{k θ}{2} + (\frac{k + 1}{2}) θ] + \cos [(k + 1) θ - \frac{θ}{2}]}{2 \sin \frac{θ}{2}} = \frac{\cos \frac{θ}{2} - \cos (k θ + \frac{θ}{2}) + \cos (k θ + \frac{θ}{2}) - \cos (k θ + \frac{3 θ}{2})}{2 \sin \frac{θ}{2}} = \frac{\cos \frac{θ}{2} - \cos (k θ + \frac{3 θ}{2})}{2 \sin \frac{θ}{2}} = \frac{2 \sin \frac{1}{2} (k θ + \frac{3 θ}{2} + \frac{θ}{2}) \cdot \sin \frac{1}{2} (k θ + \frac{3 θ}{2} - \frac{θ}{2})}{2 \sin \frac{θ}{2}} = \frac{2 \sin (\frac{k θ + 2 θ}{2}) \cdot \sin (\frac{k θ + θ}{2})}{2 \sin \frac{θ}{2}} = \frac{\frac{\sin (k + 1) θ}{2} \sin \frac{(k + 1 + 1)}{2} θ}{\sin \frac{θ}{2}}$ \

Hence, P(k + 1) is true when P(k) is true.

So, by the principle of mathematical induction, P(n) is true for any natural number n.

Page No 72:

Question 23:

Show that $\frac{n^{5}}{5} + \frac{n^{3}}{3} + \frac{7 n}{15}$ is a natural number for all n ∈ N.

Answer:

Given that, $\frac{n^{5}}{5} + \frac{n^{3}}{3} + \frac{7 n}{15}$ is a natural number for all n ∈ N.

For n = 1

$P (1) : \frac{1^{5}}{5} + \frac{1^{3}}{3} + \frac{7 (1)}{15} = \frac{15}{15} = 1$

Hence, P(1) is true.

Assume that P(n) is true for a natural number n = k.

$P (k) : \frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{7 k}{15}$ is a natural number. .....(1)

To prove: $P (k + 1) = \frac{{(k + 1)}^{5}}{5} + \frac{{(k + 1)}^{3}}{3} + \frac{7 (k + 1)}{15}$

$\begin{array}{rcl} P (k + 1) & = & \frac{k^{5} + 5 k^{4} + 10 k^{3} + 10 k^{2} + 5 k + 1}{5} + \frac{k^{3} + 3 k^{2} + 3 k + 1}{3} + \frac{7 k + 7}{15} \\ = & \frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{7 k}{15} + k^{4} + 3 k^{3} + 2 k^{2} + k + \frac{1}{5} + k^{2} + k + \frac{1}{3} + \frac{7}{15} \\ = & \frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{7 k}{15} + k^{4} + 3 k^{3} + 3 k^{2} + 2 k + 1 \end{array}$

Page No 72:

Question 24:

Prove that $\frac{1}{n + 1} + \frac{1}{n + 2} + . . . + \frac{1}{2 n} > \frac{13}{24}$ , for all natural numbers n > 1.

Answer:

Let $P (n) : \frac{1}{n + 1} + \frac{1}{n + 2} + . . . + \frac{1}{2 n} > \frac{13}{24} for all natural number n > 1 .$

$P (2) : \frac{1}{2 + 1} + \frac{1}{2 + 2} > \frac{13}{24} \Rightarrow \frac{7}{12} > \frac{13}{24}, which is true . Thus, P (2) is true . Let assume P (n) is true for some natural number n = k . ∴ P (k) : \frac{1}{k + 1} + \frac{1}{k + 2} + . . . + \frac{1}{2 k} > \frac{13}{24} Now, n = k + 1 P (k + 1) : \frac{1}{k + 1 + 1} + \frac{1}{k + 1 + 2} + . . . + \frac{1}{2 (k + 1)} > \frac{13}{24} \Rightarrow \frac{1}{k + 1} + \frac{1}{k + 2} + \frac{1}{k + 3} + . . . + \frac{1}{2 k} + \frac{1}{2 k + 1} + \frac{1}{2 (k + 1)} - \frac{1}{k + 1} > \frac{13}{24} \Rightarrow \frac{1}{k + 1} + \frac{1}{k + 2} + \frac{1}{k + 3} + . . . + \frac{1}{2 k} + \frac{1}{2 (2 k + 1) (k + 1)} > \frac{13}{24} Thus, P (k + 1) is ture . Hence, by principle of mathematical induction P (n) : \frac{1}{n + 1} + \frac{1}{n + 2} + . . . + \frac{1}{2 n} > \frac{13}{24} is true for all natural number n > 1 .$

Page No 72:

Question 25:

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

Answer:

Let P(n): Number of subset of a set containing n distinct elements is 2ⁿ, for all n ∈ N.
For n = 1, consider set A = {1}.
So, set of subsets is {{1}, ∅}, which contains 21 elements.
So, P(1) is true.
Let us assume that P(n) is true, for some natural number n = k.
P(k): Number of subsets of a set containing k distinct elements is 2k To prove that P(k + 1) is true,
we have to show that P(k + 1): Number of subsets of a set containing (k + 1) distinct elements is 2k+1
We know that, with the addition of one element in the set, the number of subsets become double.
Number of subsets of a set containing (k+ 1) distinct elements = 2×2k = 2k+1
So, P(k + 1) is true.
Hence, P(n) is true.

Page No 72:

Question 26:

Choose the correct answer from the given four options:
If 10ⁿ + 3.4ⁿ⁺²+ k is divisible by 9 for all n ∈ N, then the least positive integral value of k is
(A) 5
(B) 3
(C) 7
(D) 1

Answer:

Let P(n): 10ⁿ + 3.4ⁿ⁺²+ k is divisible by 9 for all n ∈ N.

For n = 1, the given statement is also true 10¹ + 3.4¹⁺²+ k is divisible by 9.

⇒ 10 + 192 + k is divisible by 9

⇒ (202 + k) is divisible by 9

If (202 + kâ€‹) is divisible by 9, then the least value of k must be 5.

⇒ 207 is divisible by 9

Thus, the least value of k is 5.

Hence, the correct answer is option (A).

Page No 72:

Question 27:

Choose the correct answer from the given four options:
For all n ∈ N, 3.5²ⁿ⁺¹ + 2³ⁿ⁺¹is divisible by
(A) 19
(B) 17
(C) 23
(D) 25

Answer:

Given that: 3.5²ⁿ⁺¹ + 2³ⁿ⁺¹

For n = 1, 3.5²⁺¹ + 2³⁺¹

= 3 × 125 + 16

= 391 = 17 × 23

Which is divisible by both 17 and 23.

Hence, the correct answer is option (B) and (C).

Page No 72:

Question 28:

Choose the correct answer from the given four options:
If xⁿ – 1 is divisible by x – k, then the least positive integral value of k is
(A) 1
(B) 2
(C) 3
(D) 4

Answer:

Let P(n): xⁿ – 1 is divisible by (x – k).

For n = 1, x¹ – 1 is divisible by (x – 1).

Since, if x – 1 is divisible by (x – k), then the least possible integral value of k is 1.

Hence, the correct answer is option (A).

Page No 72:

Question 29:

Fill in the blanks in the following :
If P(n) : 2n < n!, n ∈ N, then P(n) is true for all n ≥ ________.

Answer:

Given that: P(n) : 2n < n!, n ∈ N

For, n = 1, 2 < 1! ⇒ 2 < 1 [false]
For, n = 2, 4â€‹ < 2! ⇒ 2 < 2 [false]
For, n = 3, 6 â€‹< 3! ⇒ 6 < 6 [false]
For, n = 4, 8 â€‹< 4! ⇒ 8 < 24 [true]
For, n = 5, 10 â€‹< 5! ⇒ 10 < 120 [true]

Hence, P(n) is true for all n ≥ 4.

Page No 72:

Question 30:

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.

Answer:

False;

The given statement is false because P(1) is true has not been proved.

View NCERT Solutions for all chapters of Class 11