RD Sharma Solutions Class 11 Maths Chapter 18 – Binomial Theorem: It is RD Sharma Solutions Class 11 Maths Chapter 18 Binomial Theorem in which all the parts of the blog are associated with the fundamental aspects of Binomial Theorem like the benefits of following RD Sharma Solutions Class 11 Maths Chapter 18 Maths Solutions along with exercise-wise explanations.
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RD Sharma Solutions Class 11 Maths Chapter 18
Exercise-Wise RD Sharma Solutions Class 11 Maths Chapter 18
| RD Sharma Solutions Exercise 18.1 |
| RD Sharma Solutions Exercise 18.2 |
Access RD Sharma Solutions Class 11 Maths Chapter 18
1. Using the binomial theorem, write down the expressions of the following:
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)0 – 4C1 (2x)3 (3y)1 + 4C2 (2x)2 (3y)2 – 4C3 (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
(iv) (1 – 3x) 7
Let us solve the given expression
(1 – 3x) 7 = 7C0 (3x)0 – 7C1 (3x)1 + 7C2 (3x)2 – 7C3 (3x)3 + 7C4 (3x)4 – 7C5 (3x)5 + 7C6 (3x)6 – 7C7 (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
(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)0 – 5C1 (1 + 2x)4 (3x2)1 + 5C2 (1 + 2x)3 (3x2)2 – 5C3 (1 + 2x)2 (3x2)3 + 5C4 (1 + 2x)1 (3x2)4 – 5C5 (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
(x) (1 – 2x + 3x2)3
Let us solve the given expression
2. Evaluate the following:
Solution:
Let us solve the given expression
= 2 [5C0 (2√x)0 + 5C2 (2√x)2 + 5C4 (2√x)4]
= 2 [1 + 10 (4x) + 5 (16x2)]
= 2 [1 + 40x + 80x2]
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
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) (2√2) + 4√2]
= 2√2 (405 + 180 + 4)
= 1178√2
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
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
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
Let us solve the given expression
= 2 [6C1 (√3)5 (√2)1 + 6C3 (√3)3 (√2)3 + 6C5 (√3)1 (√2)5]
= 2 [6 (9√3) (√2) + 20 (3√3) (2√2) + 6 (√3) (4√2)]
= 2 [√6 (54 + 120 + 24)]
= 396 √6
= 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 the 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)0 – 3C1 (100)2 (4)1 + 3C2 (100)1 (4)2 – 3C3 (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)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
= 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.
EXERCISE 18.2 PAGE NO: 18.37
1. Find the 11th term from the beginning and the 11th term from the end in the expansion of (2x – 1/x2)25.
Solution:
Given:
(2x – 1/x2)25
The given expression contains 26 terms.
So, the 11th term from the end is the (26 − 11 + 1) th term from the beginning.
In other words, the 11th term from the end is the 16th term from the beginning.
Then,
T16 = T15+1 = 25C15 (2x)25-15 (-1/x2)15
= 25C15 (210) (x)10 (-1/x30)
= – 25C15 (210 / x20)
Now, we shall find the 11th term from the beginning.
T11 = T10+1 = 25C10 (2x)25-10 (-1/x2)10
= 25C10 (215) (x)15 (1/x20)
= 25C10 (215 / x5)
2. Find the 7th term in the expansion of (3x2 – 1/x3)10.
Solution:
Given:
(3x2 – 1/x3)10
Let us consider the 7th term as T7
So,
T7 = T6+1
= 10C6 (3x2)10-6 (-1/x3)6
= 10C6 (3)4 (x)8 (1/x18)
= [10×9×8×7×81] / [4×3×2×x10]
= 17010 / x10
∴ The 7th term of the expression (3x2 – 1/x3)10 is 17010 / x10.
3. Find the 5th term in the expansion of (3x – 1/x2)10.
Solution:
Given:
(3x – 1/x2)10
The 5th term from the end is the (11 – 5 + 1)th, is., 7th term from the beginning.
So,
T7 = T6+1
= 10C6 (3x)10-6 (-1/x2)6
= 10C6 (3)4 (x)4 (1/x12)
= [10×9×8×7×81] / [4×3×2×x8]
= 17010 / x8
∴ The 5th term of the expression (3x – 1/x2)10 is 17010 / x8.
4. Find the 8th term in the expansion of (x3/2 y1/2 – x1/2 y3/2)10.
Solution:
Given:
(x3/2 y1/2 – x1/2 y3/2)10
Let us consider the 8th term as T8
So,
T8 = T7+1
= 10C7 (x3/2 y1/2)10-7 (-x1/2 y3/2)7
= -[10×9×8]/[3×2] x9/2 y3/2 (x7/2 y21/2)
= -120 x8y12
∴ The 8th term of the expression (x3/2 y1/2 – x1/2 y3/2)10 is -120 x8y12.
5. Find the 7th term in the expansion of (4x/5 + 5/2x) 8.
Solution:
Given:
(4x/5 + 5/2x) 8
Let us consider the 7th term as T7
So,
T7 = T6+1
∴ The 7th term of the expression (4x/5 + 5/2x) 8 is 4375/x4.
6. Find the 4th term from the beginning and 4th term from the end in the expansion of (x + 2/x) 9.
Solution:
Given:
(x + 2/x) 9
Let Tr+1 be the 4th term from the end.
Then, Tr+1 is (10 − 4 + 1)th, i.e., 7th, the term from the beginning.
7. Find the 4th term from the end in the expansion of (4x/5 – 5/2x) 9.
Solution:
Given:
(4x/5 – 5/2x) 9
Let Tr+1 be the 4th term from the end of the given expression.
Then, Tr+1 is (10 − 4 + 1)th term, i.e., the 7th term, from the beginning.
T7 = T6+1
∴ The 4th term from the end is 10500/x3.
Detailed Exercise-wise Explanation with Listing of Important Topics in the Exercise
RD Sharma Class 11 Chapter 18 Exercise 18A
In Chapter 18A RD Sharma Class 11, the students will be able to examine their performances by attempting objective-type questions that are associated with the binomial theorem for positive integral index and some important conclusions from the binomial theorem.
After going through the solutions of RD Sharma Class 11 Chapter 18A the students will be able to understand the applications of the Binomial Theorem. The preparation style of the students will undergo a lot of changes which will be beneficial for their performances in the final Maths Exam. Hence, you must buy a copy for yourself right now.
RD Sharma Class 11 Chapter 18 Exercise 18B
In the RD Sharma Class 11 Chapter 18 Exercise 18B Solutions you will be exposed to all kinds of tricky problems from the Binomial Theorem. Try to attempt all the questions from RD Sharma Class 11 Chapter 18B without seeing the answers from Chapter 18 solutions. Analyze your performance and make a note of it.
Important concepts from RD Sharma Solutions Class 11 Maths Chapter 18
- Binomial theorem for positive integral index.
- Some important conclusions from the binomial theorem.
- General terms and middle terms in a binomial expansion.
RD Sharma Solutions Class 11 Maths Chapter 18
RD Sharma’s solution can play an important role in handling difficult problems from the Binomial Theorem. Collecting a copy of the RD Sharma Chapter 18 Solutions can help you to clear all the basic concepts of the Binomial Theorem. All the things will be easy to understand once you examine the relevant parts.
Some of the parts need special attention and the students will be able to keep track of it by referring to the solutions. There are different kinds of benefits for referring to RD Sharma Chapter 18 Class 11 Solutions and now we will talk about those solutions. You must collect it after you finish your syllabus.
- The students will be able to prepare themselves in a better way when they go through the contents of Maths Chapter 18.
- The students will be able to cross all the hurdles about the problems that are involved with Binomial Theorems.
- All the parts in connection with the basics of Binomial Theorems have been explained with proper presentations and if you examine the contents carefully you will be left with zero queries.
- You will get new ideas to improve your methods of preparation and that is the beauty of following CBSE Chapter 18 Binomial Theorems.
- Shaping study materials becomes easier for you when you are ready with solutions.
Access Other Important Chapters of RD Sharma Solutions Class 11 Maths
- Chapter 1 – Sets
- Chapter 2 – Relations
- Chapter 3 – Functions
- Chapter 4 – Measurement of Angles
- Chapter 5 – Trigonometric Functions
- Chapter 6 – Graphs of Trigonometric Functions
- Chapter 7 – Trigonometric Ratios of Compound Angles
- Chapter 8 – Transformation Formulae
- Chapter 9 – Trigonometric Ratios of Multiple and Sub-Multiple Angles
- Chapter 10 – Sine and Cosine Formulae and Their Applications
- Chapter 11 – Trigonometric Equations
- Chapter 12 – Mathematical Induction
- Chapter 13 – Complex Numbers
- Chapter 14 – Quadratic Equations
- Chapter 15 – Linear Inequations
- Chapter 16 – Permutations
- Chapter 17 – Combinations
- Chapter 19 – Arithmetic Progressions
- Chapter 20 – Geometric Progressions
- Chapter 21 – Some Special Series
- Chapter 22 – Brief Review of Cartesian System of Rectangular Coordinates
- Chapter 23 – The Straight Lines
- Chapter 24 – The Circle
- Chapter 25 – Parabola
- Chapter 26 – Ellipse
- Chapter 27 – Hyperbola
- Chapter 28 – Introduction To 3D Coordinate Geometry
- Chapter 29 – Limits
- Chapter 30 – Derivatives
- Chapter 31 – Mathematical Reasoning
- Chapter 32 – Statistics
- Chapter 33 – Probability
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