Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 4- Methods of Induction and Binomial Theorem

Class 11: Maths Chapter 4 solutions. Complete Class 11 Maths Chapter 4 Notes.

Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 4- Methods of Induction and Binomial Theorem

Maharashtra Board 11th Maths Chapter 4, Class 11 Maths Chapter 4 solutions

Ex 4.1

Prove by the method of induction, for all n ∈ N.

Question 1.
2 + 4 + 6 + …… + 2n = n(n + 1)
Solution:
Let P(n) = 2 + 4 + 6 + …… + 2n = n(n + 1), for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 2
R.H.S. = 1(1 + 1) = 2
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 2 + 4 + 6 + ….. + 2k = k(k + 1) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
2 + 4 + 6 + …… + 2(k + 1) = (k + 1) (k + 2)
L.H.S. = 2 + 4 + 6 + …+ 2(k + 1)
= 2 + 4 + 6+ ….. + 2k + 2(k + 1)
= k(k + 1) + 2(k + 1) …..[From (i)]
= (k + 1).(k + 2)
= R.H.S.
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 2 + 4 + 6 + …… + 2n = n(n + 1) for all n ∈ N.

Question 2.
3 + 7 + 11 + ……… to n terms = n(2n + 1)
Solution:
Let P(n) = 3 + 7 + 11 + ……… to n terms = n(2n +1), for all n ∈ N.
But 3, 7, 11, …. are in A.P.
∴ a = 3 and d = 4
Let t_n be the nth term.
∴ t_n = a + (n – 1)d = 3 + (n – 1)4 = 4n – 1
∴ P(n) = 3 + 7 + 11 + ……. + (4n – 1) = n(2n + 1)

Step I:
Put n = 1
L.H.S. = 3
R.H.S. = 1[2(1)+ 1] = 3
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 3 + 7 + 11 + ….. + (4k – 1) = k(2k + 1) …..(i)

Sept III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
3 + 7 + 11 + …+ [4(k + 1) – 1] = (k + 1)(2k + 3)
L.H.S. = 3 + 7 + 11 + …… + [4(k + 1) – 1]
= 3 + 7 + 11 + ….. + (4k – 1) + [4(k+ 1) – 1]
= k(2k + 1) + (4k + 4 – 1) …..[From (i)]
= 2k² + k + 4k + 3
= 2k² + 2k + 3k + 3
= 2k(k + 1) + 3(k + 1)
= (k + 1) (2k + 3)
= R.H.S.
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 3 + 7 + 11 + ….. to n terms = n(2n + 1) for all n ∈ N.

∴ P(n) is true for n = k + 1.

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

i.e., to prove that

∴ P(n) is true for n = k + 1.

i.e., 2^3k – 1 is a multiple of 7.
∴ 2^3k – 1 = 7a, where a ∈ N
∴ 2^3k = 7a + 1 ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
2^3(k+1) – 1 = 7b, where b ∈ N.
∴ P(k + 1) = 2^3(k+1) – 1
= 2^3k+3 – 1
= 2^3k . (2³) – 1
= (7a + 1)8 – 1 …..[From (i)]
= 56a + 8 – 1
= 56a + 7
= 7(8a + 1)
7b, where b = (8a + 1) ∈ N
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ (2⁴ⁿ – 1) is divisible by 7, for all n ∈ N.

Question 11.
(2⁴ⁿ – 1) is divisible by 15.
Solution:
(2⁴ⁿ – 1) is divisible by 15 if and only if (2⁴ⁿ – 1) is a multiple of 15.
Let P(n) ≡ (2⁴ⁿ – 1) = 15m, where m ∈ N.

Step I:
Put n = 1
∴ 2⁴⁽¹⁾ – 1 = 16 – 1 = 15
∴ (2⁴ⁿ – 1) is a multiple of 15.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 2^4k – 1 = 15a, where a ∈ N
∴ 2^4k = 15a + 1 …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
∴ 2^4(k+1) – 1 = 15b, where b ∈ N
∴ P(k + 1) = 2^4(k+1) – 1 = 2^4k+4 – 1
= 2^4k . 2⁴ – 1
= 16 . (2^4k) – 1
= 16(15a + 1) – 1 …..[From (i)]
= 240a + 16 – 1
= 240a + 15
= 15(16a + 1)
= 15b, where b = (16a + 1) ∈ N
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ (2⁴ⁿ – 1) is divisible by 15, for all n ∈ N.

Question 12.
3ⁿ – 2n – 1 is divisible by 4.
Solution:
(3ⁿ – 2n – 1) is divisible by 4 if and only if (3ⁿ – 2n – 1) is a multiple of 4.
Let P(n) ≡ (3ⁿ – 2n – 1) = 4m, where m ∈ N.

Step I:
Put n = 1
∴ (3ⁿ – 2n – 1) = 3⁽¹⁾ – 2(1) – 1 = 0 = 4(0)
∴ (3ⁿ – 2n – 1) is a multiple of 4.
∴ P(n) is tme for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 3^k – 2k – 1 = 4a, where a ∈ N
∴ 3^k = 4a + 2k + 1 ….(i)

Step III:
We have to prove that P(n) is tme for n = k + 1,
i.e., to prove that
3^(k+1) – 2(k + 1) – 1 = 4b, where b ∈ N
P(k + 1) = 3^k+1 – 2(k + 1) – 1
= 3^k . 3 – 2k – 2 – 1
= (4a + 2k + 1) . 3 – 2k – 3 …….[From (i)]
= 12a + 6k + 3 – 2k – 3
= 12a + 4k
= 4(3a + k)
= 4b, where b = (3a + k) ∈ N
∴ P(n) is tme for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is tme for all n ∈ N.
∴ 3ⁿ – 2n – 1 is divisible by 4, for all n ∈ N.

R.H.S. = cos[(1)θ] + i sin[(1)θ] = cos θ + i sin θ
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ (cos θ + i sin θ)^k = cos kθ + i sin kθ …….(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
(cos θ + i sin θ)^k+1 = cos (k + 1)θ + i sin (k + 1)θ
L.H.S. = (cos θ + i sin θ)^k+1
= (cos θ + i sin θ)^k . (cos θ + i sin θ)
= (cos kθ + i sin kθ) . (cos θ + i sin θ) ……[From (i)]
= cos kθ cos θ + i sin θ cos kθ + i sin kθ cosθ – sin kθ sin θ ……[∵ i² = -1]
= (cos kθ cos θ – sin k θ sin θ) + i(sin kθ cos θ + cos kθ sin θ)
= cos(kθ + θ) + i sin(kθ + θ)
= cos(k + 1) θ + i sin (k + 1) θ
= R.H.S.
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ (cos θ + i sin θ)ⁿ = cos (nθ) + i sin (nθ), for all n ∈ N.

Question 15.
Given that t_n+1 = 5 t_n+4, t₁ = 4, prove by method of induction that t_n = 5ⁿ – 1.
Solution:
Let the statement P(n) has L.H.S. a recurrence relation t_n+1 = 5 t_n+4, t₁ = 4 and R.H.S. a general statement t_n = 5ⁿ – 1.
Step I:
Put n = 1
L.H.S. = 4
R.H.S. = 5¹ – 1 = 4
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.
Put n = 2
L.H.S. = t₂ = 5t₁ + 4 = 24
R.H.S. = t₂ = 5² – 1 = 24
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 2.

Step II:
Let us assume that P(n) is true for n = k.
∴ t_k+1 = 5 t_k+4 and t_k = 5^k – 1

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that t_k+1 = 5^k+1 – 1
Since t_k+1 = 5 t_k+4 and t_k = 5^k – 1 …..[From Step II]
t_k+1 = 5 (5^k – 1) + 4 = 5^k+1 – 1
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ t_n = 5ⁿ – 1, for all n ∈ N.

Ex 4.2

Question 1.
Expand:
(i) (√3 + √2)⁴
Solution:
Here, a = √3, b = √2 and n = 4.
Using binomial theorem,

∴ (√3 + √2)⁴ = 1(9) (1) + 4(3√3) (√2) + 6(3)(2) + 4(√3) (2√2) + 1(1)(4)
= 9 + 12√6 + 36 + 8√6 + 4
= 49 + 20√6

(ii) (√5 – √2)⁵
Solution:
Here, a = √5, b = √2 and n = 5.
Using binomial theorem,

Question 2.
Expand:
(i) (2x² + 3)⁴
Solution:
Here, a = 2x², b = 3 and n = 4.
Using binomial theorem,

Question 3.
Find the value of
(i) (√3 + 1)⁴ – (√3 – 1)⁴
Solution:

(ii) (2 + √5)⁵ + (2 – √5)⁵
Solution:

Adding (i) and (ii), we get
∴ (2 + √5 )⁵ + (2 – √5)⁵ = (32 + 80√5 + 400 + 200√5 + 250 + 25√5) + (32 – 80√5 + 400 – 200√5+ 250 – 25√5 )
= 64 + 800 + 500
= 1364

Question 4.
Prove that:
(i) (√3 + √2)⁶ + (√3 – √2)⁶ = 970
Solution:

(ii) (√5 + 1)⁵ – (√5 – 1)⁵ = 352
Solution:

Question 5.
Using binomial theorem, find the value of
(i) (102)⁴
Solution:

(ii) (1.1)⁵
Solution:

Question 6.
Using binomial theorem, find the value of
(i) (9.9)³
Solution:

(ii) (0.9)⁴
Solution:

Question 7.
Without expanding, find the value of
(i) (x + 1)⁴ – 4(x + 1)³ (x – 1) + 6(x + 1)² (x – 1)² – 4(x + 1) (x – 1)³ + (x – 1)⁴
Solution:

(ii) (2x – 1)⁴ + 4(2x – 1)³ (3 – 2x) + 6(2x – 1)² (3 – 2x)² + 4(2x – 1)¹ (3 – 2x)³ + (3 – 2x)⁴
Solution:

Question 8.
Find the value of (1.02)⁶, correct upto four places of decimals.
Solution:

Question 9.
Find the value of (1.01)⁵, correct upto three places of decimals.
Solution:

Question 10.
Find the value of (0.9)⁶, correct upto four places of decimals.
Solution:

Ex 4.3

Ex 4.4

Question 1.
State, by writing the first four terms, the expansion of the following, where |x| < 1.
(i) (1 + x)^-4
Solution:

(ii) (1 – x)^1/3
Solution:

(iii) (1 – x²)^-3
Solution:

(iv) (1 + x)^-1/5
Solution:

(v) (1 + x²)^-1
Solution:

Question 2.
State by writing first four terms, the expansion of the following, where |b| < |a|.
(i) (a – b)^-3
Solution:
(a – b)^-3 = [a(1−ba)]−3

(ii) (a + b)^-4
Solution:

(iii) (a + b)^1/4
Solution:

(iv) (a – b)^-1/4
Solution:
(a – b)^-1/4 = [a(1−ba)]−14

(v) (a + b)^-1/3
Solution:

Question 3.
Simplify the first three terms in the expansion of the following:
(i) (1 + 2x)^-4
Solution:

(ii) (1 + 3x)^-1/2
Solution:

(iii) (2 – 3x)^1/3
Solution:

(iv) (5 + 4x)^-1/2
Solution:

(v) (5 – 3x)^-1/3
Solution:

Question 4.
Use the binomial theorem to evaluate the following upto four places of decimals.
(i) √99
Solution:

= 10 [1 – 0.005 – 0.0000125 – ……]
= 10(0.9949875)
= 9.94987 5
= 9.9499

(ii) 126−−−√3
Solution:

(iii) 16.08−−−−√4
Solution:

(iv) (1.02)^-5
Solution:

(v) (0.98)^-3
Solution:

Ex 4.5

Question 1.
Show that C₀ + C₁ + C₂ + ….. + C₈ = 256
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 8, we get
C₀ + C₁ + C₂ + ….. + C₈ = 2⁸
∴ C₀ + C₁ + C₂ + ….. + C₈ = 256

Question 2.
Show that C₀ + C₁ + C₂ + …… + C₉ = 512
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 9, we get
C₀ + C₁ + C₂ + ….. + C₉ = 2⁹
∴ C₀ + C₁ + C₂ + …… + C₉ = 512

Question 3.
Show that C₁ + C₂ + C₃ + ….. + C₇ = 127
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 7, we get
C₀ + C₁ + C₂ + ….. + C₇ = 2⁷
∴ C₀ + C₁ + C₂ +….. + C₇ = 128
But, C₀ = 1
∴ 1 + C₁ + C₂ + ….. + C₇ = 128
∴ C₁ + C₂ + ….. + C₇ = 128 – 1 = 127

Question 4.
Show that C₁ + C₂ + C₃ + ….. + C₆ = 63
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 6, we get
C₀ + C₁ + C₂ + ….. + C₆ = 2⁶
∴ C₀ + C₁ + C₂ + …… + C₆ = 64
But, C₀ = 1
∴ 1 + C₁ + C₂ + ….. + C₆ = 64
∴ C₁ + C₂ + ….. + C₆ = 64 – 1 = 63

Question 5.
Show that C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇ = 128
Solution:
Since C₀ + C₁ + C₂ + C₃ + …… + C_n = 2ⁿ
Putting n = 8, we get
C₀ + C₁ + C₂ + C₃ + …… + C₈ = 2⁸
But, sum of even coefficients = sum of odd coefficients
∴ C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇
Let C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇ = k
Now, C₀ + C₁ + C₂ + C₃ + C₄ + C₅ + C₆ + C₇ + C₈ = 256
∴ (C₀ + C₂ + C₄ + C₆ + C₈) + (C₁ + C₃ + C₅ + C₇) = 256
∴ k + k = 256
∴ 2k = 256
∴ k = 128
∴ C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇ = 128

Question 6.
Show that C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ – 1
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
But, C₀ = 1
∴ 1 + C₁ + C₂ + C₃ + …… + C_n = 2ⁿ
∴ C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ – 1

Question 7.
Show that C₀ + 2C₁ + 3C₂ + 4C₃ + ….. + (n + 1)C_n = (n + 2) 2^n-1
Solution:

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Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 4- Methods of Induction and Binomial Theorem

Download PDF: Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 4- Methods of Induction and Binomial Theorem PDF

Chapterwise Maharashtra Board Solutions Class 11 Arts & Science Maths (Part 2) :

• Chapter 1- Complex Numbers
• Chapter 2- Sequences and Series
• Chapter 3- Permutations and Combination
• Chapter 4- Methods of Induction and Binomial Theorem
• Chapter 5- Sets and Relations
• Chapter 6- Functions
• Chapter 7- Limits
• Chapter 8- Continuity
• Chapter 9- Differentiation

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