RD Sharma Solutions for Class 11 Chapter 18 - Binomial Theorem Exercise 18.1

This exercise mainly deals with binomial theorem for positive integral index and some important conclusions from the binomial theorem. Students are advised to go through RD Sharma solutions thoroughly before the board exams to score well and improve their problem-solving abilities. These solutions are formulated by experts in Maths by using shortcut techniques to solve the exercise-wise problems in a precise manner. By practising regularly, students will be able to achieve their desired scores. Students can access RD Sharma Class 11 Maths Solutions from the below-mentioned links.

Download the pdf of RD Sharma Solutions for Class 11 Maths Exercise 18.1 Chapter 18 – Binomial Theorem

 

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Exercise 18.2 Solutions

Access answers to RD Sharma Solutions for Class 11 Maths Exercise 18.1 Chapter 18 – Binomial Theorem

1. Using binomial theorem, write down the expressions of the following:

RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem image - 1

Solution:

(i) (2x + 3y) 5

Let us solve the given expression:

(2x + 3y) 5 = 5C0 (2x)5 (3y)0 + 5C1 (2x)4 (3y)1 + 5C2 (2x)3 (3y)2 + 5C3 (2x)2 (3y)3 + 5C4 (2x)1 (3y)4 + 5C5 (2x)0 (3y)5

= 32x5 + 5 (16x4) (3y) + 10 (8x3) (9y)2 + 10 (4x)2 (27y)3 + 5 (2x) (81y4) + 243y5

= 32x5 + 240x4y + 720x3y2 + 1080x2y3 + 810xy4 + 243y5

(ii) (2x – 3y) 4

Let us solve the given expression:

(2x – 3y) 4 = 4C0 (2x)4 (3y)04C1 (2x)3 (3y)1 + 4C2 (2x)2 (3y)24C3 (2x)1 (3y)3 + 4C4 (2x)0 (3y)4

= 16x4 – 4 (8x3) (3y) + 6 (4x2) (9y2) – 4 (2x) (27y3) + 81y4

= 16x4 – 96x3y + 216x2y2 – 216xy3 + 81y4

RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem image - 2

(iv) (1 – 3x) 7

Let us solve the given expression:

(1 – 3x) 7 = 7C0 (3x)07C1 (3x)1 + 7C2 (3x)27C3 (3x)3 + 7C4 (3x)47C6 (3x)67C7 (3x)7

= 1 – 7 (3x) + 21 (9x)2 – 35 (27x3) + 35 (81x4) – 21 (243x5) + 7 (729x6) – 2187(x7)

= 1 – 21x + 189x2 – 945x3 + 2835x4 – 5103x5 + 5103x6 – 2187x7

RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem image - 3

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(viii) (1 + 2x – 3x2)5

Let us solve the given expression:

Let us consider (1 + 2x) and 3x2 as two different entities and apply the binomial theorem.

(1 + 2x – 3x2)5 = 5C0 (1 + 2x)5 (3x2)05C1 (1 + 2x)4 (3x2)1 + 5C2 (1 + 2x)3 (3x2)25C3 (1 + 2x)2 (3x2)3 + 5C4 (1 + 2x)1 (3x2)45C5 (1 + 2x)0 (3x2)5

= (1 + 2x)5 – 5(1 + 2x)4 (3x2) + 10 (1 + 2x)3 (9x4) – 10 (1 + 2x)2 (27x6) + 5 (1 + 2x) (81x8) – 243x10

= 5C0 (2x)0 + 5C1 (2x)1 + 5C2 (2x)2 + 5C3 (2x)3 + 5C4 (2x)4 + 5C5 (2x)5 – 15x2 [4C0 (2x)0 + 4C1 (2x)1 + 4C2 (2x)2 + 4C3 (2x)3 + 4C4 (2x)4] + 90x4 [1 + 8x3 + 6x + 12x2] – 270x6(1 + 4x2 + 4x) + 405x8 + 810x9 – 243x10

= 1 + 10x + 40x2 + 80x3 + 80x4 + 32x5 – 15x2 – 120x3 – 3604 – 480x5 – 240x6 + 90x4 + 720x7 + 540x5 + 1080x6 – 270x6 – 1080x8 – 1080x7 + 405x8 + 810x9 – 243x10

= 1 + 10x + 25x2 – 40x3 – 190x4 + 92x5 + 570x6 – 360x7 – 675x8 + 810x9 – 243x10

RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem image - 6

(x) (1 – 2x + 3x2)3

Let us solve the given expression:

RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem image - 7

2. Evaluate the following:

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Solution:

RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem image - 10

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RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem image - 13

Let us solve the given expression:

= 2 [5C0 (2x)0 + 5C2 (2x)2 + 5C4 (2x)4]

= 2 [1 + 10 (4x) + 5 (16x2)]

= 2 [1 + 40x + 80x2]

RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem image - 14

Let us solve the given expression:

= 2 [6C0 (2)6 + 6C2 (2)4 + 6C4 (2)2 + 6C6 (2)0]

= 2 [8 + 15 (4) + 15 (2) + 1]

= 2 [99]

= 198

RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem image - 15

Let us solve the given expression:

= 2 [5C1 (34) (2)1 + 5C3 (32) (2)3 + 5C5 (30) (2)5]

= 2 [5 (81) (2) + 10 (9) (22) + 42]

= 22 (405 + 180 + 4)

= 11782

RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem image - 16

Let us solve the given expression:

= 2 [7C0 (27) (3)0 + 7C2 (25) (3)2 + 7C4 (23) (3)4 + 7C6 (21) (3)6]

= 2 [128 + 21 (32)(3) + 35(8)(9) + 7(2)(27)]

= 2 [128 + 2016 + 2520 + 378]

= 2 [5042]

= 10084

RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem image - 17

Let us solve the given expression:

= 2 [5C1 (3)4 + 5C3 (3)2 + 5C5 (3)0]

= 2 [5 (9) + 10 (3) + 1]

= 2 [76]

= 152

RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem image - 18

Let us solve the given expression:

= (1 – 0.01)5 + (1 + 0.01)5

= 2 [5C0 (0.01)0 + 5C2 (0.01)2 + 5C4 (0.01)4]

= 2 [1 + 10 (0.0001) + 5 (0.00000001)]

= 2 [1.00100005]

= 2.0020001

RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem image - 19

Let us solve the given expression:

= 2 [6C1 (3)5 (2)1 + 6C3 (3)3 (2)3 + 6C5 (3)1 (2)5]

= 2 [6 (93) (2) + 20 (33) (22) + 6 (3) (42)]

= 2 [6 (54 + 120 + 24)]

= 396 6

RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem image - 20

= 2 [a8 + 6a6 – 6a4 + a4 + 1 – 2a2]

= 2a8 + 12a6 – 10a4 – 4a2 + 2

3. Find (a + b) 4 – (a – b) 4. Hence, evaluate (√3 + √2)4 – (√3 – √2)4.

Solution:

Firstly, let us solve the given expression:

(a + b) 4 – (a – b) 4

The above expression can be expressed as,

(a + b) 4 – (a – b) 4 = 2 [4C1 a3b1 + 4C3 a1b3]

= 2 [4a3b + 4ab3]

= 8 (a3b + ab3)

Now,

Let us evaluate the expression:

(√3 + √2)4 – (√3 -√2)4

So consider, a = √3 and b = √2 we get,

(√3 + √2)4 – (√3 -√2)4 = 8 (a3b + ab3)

= 8 [(√3)3 (√2) + (√3) (√2)3]

= 8 [(3√6) + (2√6)]

= 8 (5√6)

= 40√6

4. Find (x + 1) 6 + (x – 1) 6. Hence, or otherwise evaluate (√2 + 1)6 + (√2 – 1)6.

Solution:

Firstly, let us solve the given expression:

(x + 1) 6 + (x – 1) 6

The above expression can be expressed as,

(x + 1) 6 + (x – 1) 6 = 2 [6C0 x6 + 6C2 x4 + 6C4 x2 + 6C6 x0]

= 2 [x6 + 15x4 + 15x2 + 1]

Now,

Let us evaluate the expression:

(√2 + 1)6 + (√2 – 1)6

So consider, x = √2 then we get,

(√2 + 1)6 + (√2 – 1)6 = 2 [x6 + 15x4 + 15x2 + 1]

= 2 [(√2)6 + 15 (√2)4 + 15 (√2)2 + 1]

= 2 [8 + 15 (4) + 15 (2) + 1]

= 2 [8 + 60 + 30 + 1]

= 198

5. Using binomial theorem evaluate each of the following:

(i) (96)3

(ii) (102)5

(iii) (101)4

(iv) (98)5

Solution:

(i) (96)3

We have,

(96)3

Let us express the given expression as two different entities and apply the binomial theorem.

(96)3 = (100 – 4)3

= 3C0 (100)3 (4)03C1 (100)2 (4)1 + 3C2 (100)1 (4)23C3 (100)0 (4)3

= 1000000 – 120000 + 4800 – 64

= 884736

(ii) (102)5

We have,

(102)5

Let us express the given expression as two different entities and apply the binomial theorem.

(102)5 = (100 + 2)5

= 5C0 (100)5 (2)0 + 5C1 (100)4 (2)1 + 5C2 (100)3 (2)2 + 5C3 (100)2 (2)3 + 5C4 (100)1 (2)4 + 5C5 (100)0 (2)5

= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32

= 11040808032

(iii) (101)4

We have,

(101)4

Let us express the given expression as two different entities and apply the binomial theorem.

(101)4 = (100 + 1)4

= 4C0 (100)4 + 4C1 (100)3 + 4C2 (100)2 + 4C3 (100)1 + 4C4 (100)0

= 100000000 + 4000000 + 60000 + 400 + 1

= 104060401

(iv) (98)5

We have,

(98)5

Let us express the given expression as two different entities and apply the binomial theorem.

(98)5 = (100 – 2)5

= 5C0 (100)5 (2)05C1 (100)4 (2)1 + 5C2 (100)3 (2)25C3 (100)2 (2)3 + 5C4 (100)1 (2)45C5 (100)0 (2)5

= 10000000000 – 1000000000 + 40000000 – 800000 + 8000 – 32

= 9039207968

6. Using binomial theorem, prove that 23n – 7n – 1 is divisible by 49, where n ∈ N.

Solution:

Given:

23n – 7n – 1

So, 23n – 7n – 1 = 8n – 7n – 1

Now,

8n – 7n – 1

8n = 7n + 1

= (1 + 7) n

= nC0 + nC1 (7)1 + nC2 (7)2 + nC3 (7)3 + nC4 (7)2 + nC5 (7)1 + … + nCn (7) n

8n = 1 + 7n + 49 [nC2 + nC3 (71) + nC4 (72) + … + nCn (7) n-2]

8n – 1 – 7n = 49 (integer)

So now,

8n – 1 – 7n is divisible by 49

Or

23n – 1 – 7n is divisible by 49.

Hence proved.

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