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Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board chapter 4 - Methods of Induction and Binomial Theorem [Latest edition]

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Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board chapter 4 - Methods of Induction and Binomial Theorem - Shaalaa.com
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Solutions for Chapter 4: Methods of Induction and Binomial Theorem

Below listed, you can find solutions for Chapter 4 of Maharashtra State Board Balbharati for Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board.


Exercise 4.1Exercise 4.2Exercise 4.3Exercise 4.4Exercise 4.5Miscellaneous Exercise 4.1Miscellaneous Exercise 4.2
Exercise 4.1 [Pages 73 - 74]

Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board 4 Methods of Induction and Binomial Theorem Exercise 4.1 [Pages 73 - 74]

1Page 73

Prove by method of induction, for all n ∈ N:

2 + 4 + 6 + ..... + 2n = n (n+1)

2Page 73

Prove by method of induction, for all n ∈ N:

3 + 7 + 11 + ..... + to n terms = n(2n+1)

3Page 73

Prove by method of induction, for all n ∈ N:

12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`

4Page 73

Prove by method of induction, for all n ∈ N:

12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`

5Page 73

Prove by method of induction, for all n ∈ N:

13 + 33 + 53 + .... to n terms = n2(2n2 − 1)

6Page 73

Prove by method of induction, for all n ∈ N:

1.2 + 2.3 + 3.4 + ..... + n(n + 1) = `"n"/3 ("n" + 1)("n" + 2)`

7Page 73

Prove by method of induction, for all n ∈ N:

1.3 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)`

8Page 73

Prove by method of induction, for all n ∈ N:

`1/(1.3) + 1/(3.5) + 1/(5.7) + ... + 1/((2"n" - 1)(2"n" + 1)) = "n"/(2"n" + 1)`

9Page 74

Prove by method of induction, for all n ∈ N:

`1/(3.5) + 1/(5.7) + 1/(7.9) + ...` to n terms = `"n"/(3(2"n" + 3))`

10Page 74

Prove by method of induction, for all n ∈ N:

(23n − 1) is divisible by 7

11Page 74

Prove by method of induction, for all n ∈ N:

(24n−1) is divisible by 15

12Page 74

Prove by method of induction, for all n ∈ N:

3n − 2n − 1 is divisible by 4

13Page 74

Prove by method of induction, for all n ∈ N:

5 + 52 + 53 + .... + 5n = `5/4(5^"n" - 1)`

14Page 74

Prove by method of induction, for all n ∈ N:

(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)

15Page 74

Prove by method of induction, for all n ∈ N:

Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 1

16Page 74

Prove by method of induction, for all n ∈ N:

`[(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)]` ∀ n ∈ N

Exercise 4.2 [Page 77]

Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board 4 Methods of Induction and Binomial Theorem Exercise 4.2 [Page 77]

1. (i)Page 77

Expand: `(sqrt(3) + sqrt(2))^4`

1. (ii)Page 77

Expand: `(sqrt(5) - sqrt(2))^5`

2. (i)Page 77

Expand: (2x2 + 3)4 

2. (ii)Page 77

Expand: `(2x - 1/x)^6`

3. (i)Page 77

Find the value of `(sqrt(3) + 1)^4- (sqrt(3) - 1)^4`.

3. (ii)Page 77

Find the value of `(2 + sqrt(5))^5 + (2 - sqrt(5))^5`

4. (i)Page 77

Prove that `(sqrt(3) + sqrt(2))^6 + (sqrt(3) - sqrt(2))^6` = 970

4. (ii)Page 77

Prove that `(sqrt(5) + 1)^5 - (sqrt(5) - 1)^5` = 352

5. (i)Page 77

Using binomial theorem, find the value of (102)4 

5. (ii)Page 77

Using binomial theorem, find the value of (1.1)5 

6. (i)Page 77

Using binomial theorem, find the value of (9.9)3 

6. (ii)Page 77

Using binomial theorem, find the value of (0.9)4 

7. (i)Page 77

Without expanding, find the value of (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

7. (ii)Page 77

Without expanding, find the value of (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 

8Page 77

Find the value of (1.02)6, correct upto four places of decimal

9Page 77

Find the value of (1.01)5, correct up to three places of decimals.

10Page 77

Find the value of (0.9)6, correct upto four places of decimal

Exercise 4.3 [Page 80]

Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board 4 Methods of Induction and Binomial Theorem Exercise 4.3 [Page 80]

1. (i)Page 80

In the following expansion, find the indicated term:

`(2x^2 + 3/(2x))^8`, 3rd term.

1. (ii)Page 80

In the following expansion, find the indicated term.

`(x^2 - 4/(x^3))^11`, 5th term

1. (iii)Page 80

In the following expansion, find the indicated term.

`((4x)/5 - 5/(2x))^9`, 7th term

1. (iv)Page 80

In the following expansion, find the indicated term.

`(1/3 + "a"^2)^12`, 9th term

1. (v)Page 80

In the following expansion, find the indicated term.

`(3"a" + 4/"a")^13`, 10th term

2. (i)Page 80

In the following expansion, find the indicated coefficient.

x3 in `(x^2 + (3sqrt(2))/x)^9`

2. (ii)Page 80

In the following expansion, find the indicated coefficient.

x8 in `(2x^5 - 5/x^3)^8`

2. (iii)Page 80

In the following expansion, find the indicated coefficient.

x9 in `(1/x + x^2)^18`

2. (iv)Page 80

In the following expansion, find the indicated coefficient.

x–3 in `(x - 1/(2x))^5`

2. (v)Page 80

In the following expansion, find the indicated coefficient.

x–20 in `(x^3 - 1/(2x^2))^15`

3. (i)Page 80

Find the constant term (term independent of x) in the expansion of `(2x + 1/(3x^2))^9`

3. (ii)Page 80

Find the constant term (term independent of x) in the expansion of `(x - 2/x^2)^15`

3. (iii)Page 80

Find the constant term (term independent of x) in the expansion of `(sqrt(x) - 3/x^2)^10`

3. (iv)Page 80

Find the constant term (term independent of x) in the expansion of `(x^2 - 1/x)^9`

3. (v)Page 80

Find the constant term (term independent of x) in the expansion of `(2x^2 - 5/x)^9`

4. (i)Page 80

Find the middle term in the expansion of `(x/y + y/x)^12`

4. (ii)Page 80

Find the middle terms in the expansion of `(x^2 + 1/x)^7`

4. (iii)Page 80

Find the middle term in the expansion of `(x^2 - 2/x)^8`

4. (iv)Page 80

Find the middle term in the expansion of `(x/"a" - "a"/x)^10`

4. (v)Page 80

Find the middle terms in the expansion of `(x^4 - 1/x^3)^11`

5Page 80

In the expansion of (k + x)8, the coefficient of x5 is 10 times the coefficient of x6. Find the value of k.

6Page 80

Find the term containing x6 in the expansion of (2 − x) (3x + 1)9

7Page 80

The coefficient of x2 in the expansion of (1 + 2x)m is 112. Find m

Exercise 4.4 [Page 82]

Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board 4 Methods of Induction and Binomial Theorem Exercise 4.4 [Page 82]

1. (i)Page 82

State, by writing first four terms, the expansion of the following, where |x| < 1

(1 + x)−4

1. (ii)Page 82

State, by writing first four terms, the expansion of the following, where |x| < 1

`(1 - x)^(1/3)`

1. (iii)Page 82

State, by writing first four terms, the expansion of the following, where |x| < 1

(1 – x2)–3

1. (iv)Page 82

State, by writing first four terms, the expansion of the following, where |x| < 1

`(1 + x)^(-1/5)`

1. (v)Page 82

State, by writing first four terms, the expansion of the following, where |x| < 1

(1 + x2)–1

2. (i)Page 82

State, by writing first four terms, the expansion of the following, where |b| < |a|

(a − b)−3 

2. (ii)Page 82

State, by writing first four terms, the expansion of the following, where |b| < |a| 

(a + b)−4 

2. (iii)Page 82

State, by writing first four terms, the expansion of the following, where |b| < |a| 

`("a" + "b")^(1/4)`

2. (iv)Page 82

State, by writing first four terms, the expansion of the following, where |b| < |a| 

`("a" - "b")^(-1/4)`

2. (v)Page 82

State, by writing first four terms, the expansion of the following, where |b| < |a| 

`("a" + "b")^(-1/3)`

3. (i)Page 82

Simplify first three terms in the expansion of the following

(1 + 2x)–4 

3. (ii)Page 82

Simplify first three terms in the expansion of the following

`(1 + 3x)^(-1/2)`

3. (iii)Page 82

Simplify first three terms in the expansion of the following

`(2 - 3x)^(1/3)`

3. (iv)Page 82

Simplify first three terms in the expansion of the following

`(5 + 4x)^(-1/2)`

3. (v)Page 82

Simplify first three terms in the expansion of the following

`(5 - 3x)^(-1/3)`

4. (i)Page 82

Use binomial theorem to evaluate the following upto four places of decimal

`sqrt(99)`

4. (ii)Page 82

Use binomial theorem to evaluate the following upto four places of decimal

`root(3)(126)`

4. (iii)Page 82

Use binomial theorem to evaluate the following upto four places of decimal

`root(4)(16.08)`

4. (iv)Page 82

Use binomial theorem to evaluate the following upto four places of decimal

(1.02)–5 

4. (v)Page 82

Use binomial theorem to evaluate the following upto four places of decimal

(0.98)–3 

Exercise 4.5 [Page 84]

Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board 4 Methods of Induction and Binomial Theorem Exercise 4.5 [Page 84]

1Page 84

Show That C0 + C1 + C2 + .... C8 = 256

2Page 84

Show That C0 + C1 + C2 + .... C9 = 512

3Page 84

Show That C1 + C2 + C3 + .... C7 = 127

4Page 84

Show That C1 + C2 + C3 + .... C6 = 63

5Page 84

Show That C0 + C2 + C4 + C6 + C8 = C1 + C3 + C5 + C7 = 128

6Page 84

Show That C1 + C2 + C3 + .... Cn = 2n − 1

7Page 84

Show That C0 + 2C1 + 3C2 + 4C3 + ... + (n + 1)Cn = (n + 2)2n−1

Miscellaneous Exercise 4.1 [Page 85]

Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board 4 Methods of Induction and Binomial Theorem Miscellaneous Exercise 4.1 [Page 85]

I. (1)Page 85

Select the correct answer from the given alternatives.

The total number of terms in the expression of (x + y)100 + (x − y)100 after simplification is:

  • 50

  • 51

  • 100

  • 202

I. (2)Page 85

Select the correct answer from the given alternatives.

The middle term in the expansion of (1 + x)2n will be :

  • (n − 1)th 

  • nth 

  • (n + 1)th 

  • (n + 2)th 

I. (3)Page 85

Select the correct answer from the given alternatives.

In the expansion of (x2 − 2x)10, the coefficient of x16 is

  • −1680

  • 1680

  • 3360

  • 6720

I. (4)Page 85

Select the correct answer from the given alternatives.

The term not containing x in expansion of `(1 - x)^2 (x + 1/x)^10` is 

  • 11C

  • 10C5 

  • 10C

  • 10C

I. (5)Page 85

Select the correct answer from the given alternatives.

The number of terms in expansion of (4y + x)8 − (4y − x)8 

  • 4

  • 5

  • 8

  • 9

I. (6)Page 85

Select the correct answer from the given alternatives.

The value 14C1 + 14C3 + 14C5 + ..... + 14C11 is

  • 214 − 1

  • 214 − 14

  • 212 

  • 213 − 14 

I. (7)Page 85

Select the correct answer from the given alternatives.

The value 11C2 + 11C4 + 11C6 + 11C8 is equal to

  • 210 − 1

  • 210 − 11

  • 210 + 12

  • 210 − 12 

I. (8)Page 85

Select the correct answer from the given alternatives.

In the expansion of (3x + 2)4, the coefficient of the middle term is

  • 36

  • 54

  • 81

  • 216

I. (9)Page 85

Select the correct answer from the given alternatives.

The coefficient of the 8th term in the expansion of (1 + x)10 is:

  • 7

  • 120

  • 10C8 

  • 210

I. (10)Page 85

Select the correct answer from the given alternatives.

If the coefficient of x2 and x3 in the expansion of (3 + ax)9 are the same, then the value of a is

  • `-7/9`

  • `-9/7`

  • `7/9`

  • `9/7`

Miscellaneous Exercise 4.2 [Pages 85 - 86]

Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board 4 Methods of Induction and Binomial Theorem Miscellaneous Exercise 4.2 [Pages 85 - 86]

II. (1) (i)Page 85

Answer the following:

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

8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`

II. (1) (ii)Page 85

Answer the following:

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

12 + 42 + 72 + ... + (3n − 2)2 = `"n"/2 (6"n"^2 - 3"n" - 1)`

II. (1) (iii)Page 85

Answer the following:

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

2 + 3.2 + 4.22 + ... + (n + 1)2n–1 = n.2n 

II. (1) (iv)Page 85

Answer the following:

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

`1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "n"/(("n" + 2)("n" + 3)("n" + 4)) = ("n"("n" + 1))/(6("n" + 3)("n" + 4))`

II. (2)Page 85

Answer the following:

Given that tn+1 = 5tn − 8, t1 = 3, prove by method of induction that tn = 5n−1 + 2

II. (3)Page 85

Answer the following:

Prove by method of induction

`[(3, -4),(1, -1)]^"n" = [(2"n" + 1, -4"n"),("n", -2"n" + 1)], ∀  "n" ∈ "N"`

II. (4)Page 85

Expand (3x2 + 2y)5 

II. (5)Page 85

Answer the following:

Expand `((2x)/3 - 3/(2x))^4`

II. (6)Page 85

Answer the following:

Find third term in the expansion of `(9x^2 - y^3/6)^4`

II. (7)Page 85

Answer the following:

Find tenth term in the expansion of `(2x^2 + 1/x)^12`

II. (8) (i)Page 85

Answer the following:

Find the middle term (s) in the expansion of `((2"a")/3 - 3/(2"a"))^6`

II. (8) (ii)Page 85

Answer the following:

Find the middle term (s) in the expansion of `(x - 1/(2y))^10`

II. (8) (iii)Page 85

Answer the following:

Find the middle term (s) in the expansion of (x2+ 2y2)

II. (8) (iv)Page 85

Answer the following:

Find the middle term (s) in the expansion of `((3x^2)/2 - 1/(3x))^9`

II. (9) (i)Page 86

Find the coefficients of x6 in the expansion of `(3x^2 - 1/(3x))^9`.

II. (9) (ii)Page 86

Find the coefficients of x60 in the expansion of `(1/x^2 + x^4)^18`

II. (10) (i)Page 86

Answer the following:

Find the constant term in the expansion of `((4x^2)/3 + 3/(2x))^9`

II. (10) (ii)Page 86

Answer the following

Find the constant term in the expansion of `(2x^2 - 1/x)^12`

II. (11) (i)Page 86

Answer the following:

Prove by method of induction loga xn = n logax, x > 0, n ∈ N

II. (11) (ii)Page 86

Answer the following:

Prove by method of induction 152n–1 + 1 is divisible by 16, for all n ∈ N.

II. (11) (iii)Page 86

Answer the following:

Prove by method of induction 52n − 22n is divisible by 3, for all n ∈ N

II. (12)Page 86

Answer the following:

If the coefficient of x16 in the expansion of (x2 + ax)10 is 3360, find a

II. (13)Page 86

Answer the following:

If the middle term in the expansion of `(x + "b"/x)^6` is 160, find b

II. (14)Page 86

Answer the following:

If the coefficient of x2 and x3 in the expansion of (3 + kx)9 are equal, find k

II. (15)Page 86

Answer the following:

If the constant term in the expansion of `(x^3 + "k"/x^8)^11` is 1320, find k

II. (16)Page 86

Answer the following:

Show that there is no term containing x6 in the expansion of `(x^2 - 3/x)^11`

II. (17)Page 86

Answer the following:

Show that there is no constant term in the expansion of `(2x - x^2/4)^9`

II. (18)Page 86

Answer the following:

State, first four terms in the expansion of `(1 - (2x)/3)^(-1/2)`

II. (19)Page 86

Answer the following:

State, first four terms in the expansion of `(1 - x)^(-1/4)`

II. (20)Page 86

Answer the following:

State, first three terms in the expansion of `(5 + 4x) ^(-1/2)`

II. (21)Page 86

Answer the following:

Using binomial theorem, find the value of `root(3)(995)` upto four places of decimals

II. (22)Page 86

Answer the following:

Find approximate value of `1/4.08` upto four places of decimals

II. (23)Page 86

Answer the following:

Find the term independent of x in the in expansion of `(1 - x^2) (x + 2/x)^6`

II. (24)Page 86

Answer the following:

(a + bx) (1 − x)6 = 3 − 20x + cx2 + ..... then find a, b, c

II. (25)Page 86

Answer the following:

The 3rd term of (1 + x)n is 36x2. Find 5th term

II. (26)Page 86

Answer the following:

Suppose (1 + kx)n = 1 − 12x + 60x2 − .... find k and n.

Solutions for 4: Methods of Induction and Binomial Theorem

Exercise 4.1Exercise 4.2Exercise 4.3Exercise 4.4Exercise 4.5Miscellaneous Exercise 4.1Miscellaneous Exercise 4.2
Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board chapter 4 - Methods of Induction and Binomial Theorem - Shaalaa.com

Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board chapter 4 - Methods of Induction and Binomial Theorem

Shaalaa.com has the Maharashtra State Board Mathematics Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board Maharashtra State Board solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. Balbharati solutions for Mathematics Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board Maharashtra State Board 4 (Methods of Induction and Binomial Theorem) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.

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Concepts covered in Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board chapter 4 Methods of Induction and Binomial Theorem are Principle of Mathematical Induction, Binomial Theorem for Positive Integral Index, General Term in Expansion of (a + b)n, Middle term(s) in the expansion of (a + b)n, Binomial Theorem for Negative Index Or Fraction, Binomial Coefficients.

Using Balbharati Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board solutions Methods of Induction and Binomial Theorem exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in Balbharati Solutions are essential questions that can be asked in the final exam. Maximum Maharashtra State Board Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board students prefer Balbharati Textbook Solutions to score more in exams.

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