NCERT Solutions Class 11 Maths Chapter 8

Class 11 is an important year academically for every student. Overall performance and marks during this year will help students to build the knowledge they need for Grade 12. It acts like a trial run for your board examinations because most of the syllabus and concepts covered in the NCERT syllabus will prepare the students for Class 12 examinations.

Mathematics as a subject in Class 11 is very important. To make it easier for students to understand, Extramarks provides the NCERT Solutions Class 11 Mathematics Chapter 8. It is an important study material and is recommended by all teachers and professors. The solutions cover all important information such as theorems, derivations, formulas and more. The NCERT Solutions Class 11 Mathematics Chapter 8 talks about the Binomial theorem. This chapter is easy to understand. Regular practice of all solved examples and problems will help students to gain better score.

Extramarks is an online learning platform which provides high-quality study materials such as NCERT Solutions Class 8, NCERT Solutions Class 9 and NCERT Solutions Class 10. These academic notes enhance students’ preparation and enable them to attain high scores in the exams.

Key Topics Covered In NCERT Solutions Class 11 Mathematics Chapter 8

Students can completely depend on the NCERT Solutions for Class 11 Mathematics Chapter 8 provided by Extramarks, as all answers are explained in a detailed and well-structured manner.

The following key topics are covered in the Class 11 Mathematics Chapter 8 Solutions:

Exercise	Topic
5.1	Introduction
5.2	Binomial Theorem for Positive Integral Indices
5.3	General and Middle Terms

8.1 Introduction

The Chapter on binomial theorem helps students to evaluate numerical values of bigger numbers using the formulas of squares and cubes of a + b and a – b. The Binomial theorem overcomes the difficulty of using repeated multiplication. Using the NCERT Solutions Class 11 Mathematics Chapter 8, students will learn to expand (a + b)n, where n is either a rational number or an integer.

8.2 Binomial Theorem for Positive Integral Indices

The following formulas are included in this section:

(a + b)0 = 1, where a + b ≠ 0
(a + b)1 = a + b
(a + b)2= a2+ 2ab + b2
(a + b)3= a3+ 3a2b + 3ab2 + b3
(a + b)4= a4+ 4a3b + 6a2b2 + 4ab3+ b4

Here students will also learn to distinguish index and coefficients and arrange them in the form of a triangle. Using these patterns, they will be able to write the expansion of any n in (a + b)n.

Refer to the NCERT Solutions Class 11 Mathematics Chapter 8 to practice several sums based on this chapter.

Pascal’s Triangle

The array of numbers arranged in the form of a triangle is called Pascal’s triangle. It gives the coefficients for the expanded binomial of the form (a + b)n. The concepts of combination are used in this section. We know Crn= n!r! (n-r)!, 0rn and C0n=Cnn=1. With the help of the pascal triangle, students can visualise the expansion of a binomial with respect to any positive integral index n in the form of combinations.

Binomial theorem

In the NCERT Solutions Class 11 Mathematics Chapter 8, students will learn the formula for the expansion of (a + b)n using the binomial theorem, which is given by,

(a + b)n= C0nan+ C1nan-1bn+C2nan-2b2+….+ Cn-1nabn-1+bn.

In this section, with the help of the principle of mathematical induction, the proof of the above formula is explained in a stepwise manner. It is important that students understand and practice this theorem well. Several observations given in this part are mentioned below:

k=0nCknan-kbk is used to denote C0nan+ C1nan-1bn+C2nan-2b2+….+ Cn-1nabn-1+bn. Therefore we can say that (a + b)n= k=0nCknan-kbk
The coefficients of Crn are called binomial coefficients.
The terms in the binomial expansion are always one more than the index. If the index is n, then there will be (n+1) terms in the expansion.

Also, many special cases of the binomial theorem are explained in detail in the NCERT Solutions Class 11 Mathematics Chapter 8. Students can also practice the solved examples and exercise problems to understand the concept thoroughly.

8.3 General and Middle terms

This section include a detailed explanation of the general terms, denoted by Tr+1. Some rules are mentioned below:

If n is even, the middle term will be (n2+1)th term.
If n is odd, the two middle terms will be (n+12)th term and (n+12+1)th term.
In the expansion of (x+1x)2n, the middle term will be (n+1)th term.

The NCERT Solutions Class 11 Mathematics Chapter 8 includes unlimited practice problems for students to solve and become experts in this chapter.

NCERT Solutions Class 11 Mathematics Chapter 8: Exercise & Solutions

Students can visit the Extramarks’ website to gain access to NCERT Solutions Class 11 Mathematics Chapter 8. Students can attain high scores in the annual examinations by practising the questions present in the solutions . In case of difficulty, they can refer to the step-by-step solutions to understand logic and know their mistakes. The solutions will help students to develop a better comprehension and deeper knowledge of the concepts through detailed answers to the textual questions. By using the NCERT Solutions Class 11 Mathematics Chapter 8, students will be able to tackle all questions as they will be well-versed with the method used to solve the equations in exams.

The links mentioned below will provide the best study material, such as the NCERT Solutions Class 11 Mathematics Chapter 8- Binomial theorem.

Extramarks also provides other study materials such as the NCERT Solutions Class 1 and NCERT Solutions Class 2. Several sample papers, questions papers, practice sheets and mock tests are available for students to analyse themselves based on their preparation level. Students appearing for grade 12 board exams must refer to the detailed NCERT Solutions Class 12, which adheres to the latest guidelines of the CBSE board.

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NCERT Exemplar Class 11 Mathematics

The NCERT Exemplar provides a variety of quality problems of various difficulty levels. These problems are described in various formats for students to practice. It is beneficial for students to use the NCERT Solutions Class 11 Mathematics Chapter 8 and NCERT Exemplar notes for effective preparation. It covers the CBSE syllabus and prepares the aspirants for board exams as well as several competitive examinations such as JEE Main and JEE advanced. Several properties of the binomial theorem of positive integers can be studied with the help of the all-inclusive NCERT Solutions Class 11 Mathematics Chapter 8.

Key Features of NCERT Solutions Class 11 Mathematics Chapter 8

The notes follow the NCERT textbook, marking system, weightage of each concept and the latest guideline issued by the board.
The NCERT Solutions Class 11 Mathematics Chapter 8 provides conceptual clarity with its well-explained and detailed solutions.
The notes are prepared by expert teachers and professors after reviewing and analysing the question papers of past years’.
It covers all the topics and sub-topics mentioned in the NCERT textbook. The solutions contain answers that are detailed, well-structured and explained in an easy language.
The NCERT Solutions Class 11 Mathematics Chapter 8 enables students to establish a strong command of the Chapter.

Q.1 Using binomial theorem, evaluate the following: (99)⁵

Ans

$\begin{array}{l} Since, {(a - b)}^{n} =^{n} C_{0} a^{n} -^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} - . .. + {(- 1)}^{n}^{n} C_{n} b^{n} \\ ∵ {(99)}^{5} = {(100 - 1)}^{5} \\ So, {(100 - 1)}^{5} =^{5} C_{0} {(100)}^{5} -^{5} C_{1} {(100)}^{4} (1) +^{5} C_{2} {(100)}^{3} {(1)}^{2} \\ -^{5} C_{3} {(100)}^{2} {(1)}^{3} +^{5} C_{4} {(100)}^{1} {(1)}^{4} -^{5} C_{5} {(1)}^{5} \\ = (1) (10000000000) - (5) (100000000) (1) \\ + (10) (1000000) (1) - (10) (10000) (1) \\ + (5) (100) (1) - (1) (1) \\ = 10000000000 - 500000000 + 10000000 \\ - 100000 + 500 - 1 \\ = 9509900499 \end{array}$

Q.2 Expand the expression: (1 – 2x)⁵.

Ans

$\begin{array}{l} Since, {(1 - x)}^{n} =^{n} C_{0} -^{n} C_{1} x +^{n} C_{2} x^{2} - . .. + {(- 1)}^{n}^{n} C_{n} x^{n} \\ So, {(1 - 2x)}^{5} =^{5} C_{0} -^{5} C_{1} (2 x) +^{5} C_{2} {(2 x)}^{2} -^{5} C_{3} {(2 x)}^{3} +^{5} C_{4} {(2 x)}^{4} -^{5} C_{5} {(2 x)}^{5} \\ = 1 - 10 x + 40 x^{2} - 80 x^{3} + 80 x^{4} - 32 x^{5} \end{array}$

Q.3

$Expand the expression: {(\frac{2}{x} – \frac{x}{2})}^{5} .$

Ans

$\begin{array}{l} Since, {(a - b)}^{n} =^{n} C_{0} a^{n} -^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} - . .. + {(- 1)}^{n}^{n} C_{n} b^{n} \\ So, {(\frac{2}{x} - \frac{x}{2})}^{5} =^{5} C_{0} {(\frac{2}{x})}^{5} -^{5} C_{1} {(\frac{2}{x})}^{4} (\frac{x}{2}) +^{5} C_{2} {(\frac{2}{x})}^{3} {(\frac{x}{2})}^{2} \\ -^{5} C_{3} {(\frac{2}{x})}^{2} {(\frac{x}{2})}^{3} +^{5} C_{4} (\frac{2}{x}) {(\frac{x}{2})}^{4} -^{5} C_{5} {(\frac{x}{2})}^{5} \\ = (1) (\frac{32}{x^{5}}) - (5) (\frac{16}{x^{4}}) (\frac{x}{2}) + (10) (\frac{8}{x^{3}}) (\frac{x^{2}}{4}) \\ - (10) (\frac{4}{x^{2}}) (\frac{x^{3}}{8}) + (5) (\frac{2}{x}) (\frac{x^{4}}{16}) - (1) (\frac{x^{5}}{32}) \\ = \frac{32}{x^{5}} - \frac{40}{x^{3}} + \frac{20}{x} - 5 x + \frac{5}{8} x^{3} - \frac{x^{5}}{32} \end{array}$

Q.4 Expand the expression: (2x – 3)⁶.

Ans

$\begin{array}{l} Since, {(a - b)}^{n} =^{n} C_{0} a^{n} -^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} - . .. + {(- 1)}^{n}^{n} C_{n} b^{n} \\ So, {(2x - 3)}^{6} =^{6} C_{0} {(2 x)}^{6} -^{6} C_{1} {(2 x)}^{5} (3) +^{6} C_{2} {(2 x)}^{4} {(3)}^{2} -^{6} C_{3} {(2 x)}^{3} {(3)}^{3} \\ +^{6} C_{4} {(2 x)}^{2} {(3)}^{4} -^{6} C_{5} {(2 x)}^{1} {(3)}^{5} +^{6} C_{6} {(2 x)}^{0} {(3)}^{6} \\ = (1) 2^{6} x^{6} - 6 \times 2^{5} x^{5} \times 3 + 15 \times 2^{4} x^{4} \times 3^{2} - 20 \times 2^{3} x^{3} \times 3^{3} \\ + 15 \times 2^{2} x^{2} \times 3^{4} - 6 \times 2^{1} x^{1} \times 3^{5} + (1) 3^{6} \\ = 64 x^{6} - 576 x^{5} + 2160 x^{4} - 4320 x^{3} + 4860 x^{2} \\ - 2916 x + 729 \end{array}$

Q.5

$Expand the expression: {(\frac{x}{3} + \frac{1}{x})}^{5} .$

Ans

$\begin{array}{l} Since, {(a + b)}^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} + . .. +^{n} C_{n} b^{n} \\ So, {(\frac{x}{3} + \frac{1}{x})}^{5} =^{5} C_{0} {(\frac{x}{3})}^{5} +^{5} C_{1} {(\frac{x}{3})}^{4} (\frac{1}{x}) +^{5} C_{2} {(\frac{x}{3})}^{3} {(\frac{1}{x})}^{2} \\ +^{5} C_{3} {(\frac{x}{3})}^{2} {(\frac{1}{x})}^{3} +^{5} C_{4} (\frac{x}{3}) {(\frac{1}{x})}^{4} +^{5} C_{5} {(\frac{1}{x})}^{5} \\ = (1) (\frac{x^{5}}{243}) + (5) (\frac{x^{4}}{81}) (\frac{1}{x}) + (10) (\frac{x^{3}}{27}) (\frac{1}{x^{2}}) \\ + (10) (\frac{x^{2}}{9}) (\frac{1}{x^{3}}) + (5) (\frac{x}{3}) (\frac{1}{x^{4}}) + (1) (\frac{1}{x^{5}}) \\ = \frac{x^{5}}{243} + \frac{5 x^{3}}{81} + \frac{10 x}{27} + \frac{10}{9 x} + \frac{5}{3 x^{3}} + \frac{1}{x^{5}} \end{array}$

Q.6

$Expand the expression: {(x+ \frac{1}{x})}^{6} .$

Ans

$\begin{array}{l} Since, {(a + b)}^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} + . .. +^{n}^{n} C_{n} b^{n} \\ So, {(x + \frac{1}{x})}^{6} =^{6} C_{0} {(x)}^{6} +^{6} C_{1} {(x)}^{5} (\frac{1}{x}) +^{6} C_{2} {(x)}^{4} {(\frac{1}{x})}^{2} +^{6} C_{3} {(x)}^{3} {(\frac{1}{x})}^{3} \\ +^{6} C_{4} {(x)}^{2} {(\frac{1}{x})}^{4} +^{6} C_{5} {(x)}^{1} {(\frac{1}{x})}^{5} +^{6} C_{6} {(x)}^{0} {(\frac{1}{x})}^{6} \\ = (1) x^{6} + 6 \times x^{4} + 15 \times x^{2} + 20 + 15 \times \frac{1}{x^{2}} + 6 \times \frac{1}{x^{4}} + (1) \frac{1}{x^{6}} \\ = x^{6} + 6 x^{4} + 15 x^{2} + 20 + \frac{15}{x^{2}} + \frac{6}{x^{4}} + \frac{1}{x^{6}} \end{array}$

Q.7 Using binomial theorem, evaluate the following: (96)³

Ans

$\begin{array}{l} Since, \\ {(a - b)}^{n} =^{n} C_{0} a^{n} -^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} - . .. + {(- 1)}^{n}^{n} C_{n} b^{n} \\ ∴ {(96)}^{3} = {(100 - 4)}^{3} \\ =^{3} C_{0} {(100)}^{3} -^{3} C_{1} {(100)}^{3 - 1} (4) +^{3} C_{2} {(100)}^{3 - 2} {(4)}^{2} \\ -^{3} C_{3} {(100)}^{3 - 3} {(4)}^{3} \\ = 1000000 - 3 \times 10000 \times 4 + 3 \times 100 \times 16 - 1 \times 1 \times 64 \\ = 1000000 - 120000 + 4800 - 64 \\ = 884736 \end{array}$

Q.8 Using binomial theorem, evaluate the following: (102)⁵.

Ans

$\begin{array}{l} Since, {(a + b)}^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} + . .. +^{n} C_{n} b^{n} \\ ∵ {(102)}^{5} = {(100 + 2)}^{5} \\ So, {(100 + 2)}^{5} =^{5} C_{0} {(100)}^{5} +^{5} C_{1} {(100)}^{4} (2) +^{5} C_{2} {(100)}^{3} {(2)}^{2} \\ +^{5} C_{3} {(100)}^{2} {(2)}^{3} +^{5} C_{4} (100) {(2)}^{4} +^{5} C_{5} {(2)}^{5} \\ = (1) (10000000000) + (5) (100000000) (2) \\ + (10) (1000000) (4) + (10) (10000) (8) \\ + (5) (100) (16) + (1) (32) \\ = 10000000000 + 1000000000 + 40000000 + 800000 \\ + 8000 + 32 \\ = 11040808032 \end{array}$

Q.9 Using binomial theorem, evaluate the following: (101)⁴

Ans

$\begin{array}{l} Since, {(a + b)}^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} + . .. +^{n} C_{n} b^{n} \\ ∵ {(101)}^{4} = {(100 + 2)}^{4} \\ So, {(100 + 1)}^{4} =^{4} C_{0} {(100)}^{4} +^{4} C_{1} {(100)}^{3} (1) +^{4} C_{2} {(100)}^{2} {(1)}^{2} \\ +^{4} C_{3} {(100)}^{1} {(1)}^{3} +^{4} C_{4} {(1)}^{4} \\ = (1) (100000000) + (4) (1000000) (1) \\ + (6) (10000) (1) + (4) (100) (1) + (1) (1) \\ = 100000000 + 4000000 + 60000 + 400 + 1 \\ = 104060401 \end{array}$

Q.10 Using Binomial Theorem, indicate which number is larger (1.1)¹⁰⁰⁰⁰ or 1000.

Ans

$\begin{array}{l} {(1.1)}^{10000} = {(1+0.1)}^{10000} \\ =1+10000 (0.1) + higher powers . .. \\ =1+1000+ higher powers . .. \\ >1001 (Leaving higher power terms) \\ ∵ 1001>1000 \\ So, {(1.1)}^{10000} >1000. \end{array}$

Q.11

${Find (a + b)}^{4} - {(a - b)}^{4} . Hence, evaluate {(\sqrt{3} + \sqrt{2})}^{4} - {(\sqrt{3} - \sqrt{2})}^{4} .$

Ans

$\begin{array}{l} Since, {(a + b)}^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} + . .. +^{n} C_{n} b^{n} \\ {(a + b)}^{4} - {(a – b)}^{4} = (^{4} C_{0} a^{4} +^{4} C_{1} a^{3} b +^{4} C_{2} a^{2} b^{2} +^{4} C_{3} a^{1} b^{3} +^{4} C_{4} b^{4}) \\ - (^{4} C_{0} a^{4} -^{4} C_{1} a^{3} b +^{4} C_{2} a^{2} b^{2} -^{4} C_{3} a^{1} b^{3} +^{4} C_{4} b^{4}) \\ =^{4} C_{0} a^{4} +^{4} C_{1} a^{3} b +^{4} C_{2} a^{2} b^{2} +^{4} C_{3} a^{1} b^{3} +^{4} C_{4} b^{4} \\ -^{4} C_{0} a^{4} +^{4} C_{1} a^{3} b -^{4} C_{2} a^{2} b^{2} +^{4} C_{3} a^{1} b^{3} -^{4} C_{4} b^{4} \\ = 2 (^{4} C_{1} a^{3} b +^{4} C_{3} a^{1} b^{3}) \\ = 2 (4 a^{3} b + 4 {ab}^{3}) \\ = 8 (a^{3} b + {ab}^{3}) \\ Putting a = \sqrt{3} and b = \sqrt{2}, we get \\ {(\sqrt{3} + \sqrt{2})}^{4} - {(\sqrt{3} - \sqrt{2})}^{4} \\ = 8 {{(\sqrt{3})}^{3} (\sqrt{2}) + (\sqrt{3}) {(\sqrt{2})}^{3}} \\ = 8 {3 \sqrt{6} + 2 \sqrt{6}} \\ = 8 (5 \sqrt{6}) \\ = 40 \sqrt{6} \end{array}$

Q.12

$\begin{array}{l} {Find (x + 1)}^{6} +(x - {1)}^{6} . Hence or otherwise evaluate \\ {(\sqrt{2} +1)}^{6} - {(\sqrt{2} - 1)}^{6} . \end{array}$

Ans

$\begin{array}{l} Since, {(a + b)}^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} + . .. +^{n}^{n} C_{n} b^{n} \\ So, {(x + 1)}^{6} =^{6} C_{0} {(x)}^{6} +^{6} C_{1} {(x)}^{5} +^{6} C_{2} {(x)}^{4} +^{6} C_{3} {(x)}^{3} +^{6} C_{4} {(x)}^{2} \\ +^{6} C_{5} {(x)}^{1} +^{6} C_{6} {(x)}^{0} \\ = (1) x^{6} + 6 x^{5} + 15 x^{4} + 20 x^{3} + 15 x^{2} + 6 x + 1 \\ {(x - 1)}^{6} =^{6} C_{0} {(x)}^{6} -^{6} C_{1} {(x)}^{5} +^{6} C_{2} {(x)}^{4} -^{6} C_{3} {(x)}^{3} +^{6} C_{4} {(x)}^{2} \\ -^{6} C_{5} {(x)}^{1} +^{6} C_{6} {(x)}^{0} \\ = (1) x^{6} - 6 x^{5} + 15 x^{4} - 20 x^{3} + 15 x^{2} - 6 x + 1 \\ {(x + 1)}^{6} {+(x – 1)}^{6} \\ = 2 x^{6} + 30 x^{4} + 30 x^{2} + 2 \\ = 2 (x^{6} + 15 x^{4} + 15 x^{2} + 1) \\ Putting x = \sqrt{2}, we get \\ {(\sqrt{2} + 1)}^{6} - {(\sqrt{2} - 1)}^{6} \\ = 2 {{(\sqrt{2})}^{6} + 15 {(\sqrt{2})}^{4} + 15 {(\sqrt{2})}^{2} + 1} \\ = 2 (8 + 60 + 30 + 1) \\ = 2 (99) \\ = 198 \end{array}$

Q.13 Show that 9ⁿ⁺¹ – 8n – 9 is divisible by 64, whenever n is a positive integer.

Ans

$\begin{array}{l} 9^{n + 1} = {(1 + 8)}^{n + 1} \\ = 1 + (n + 1) 8 + \frac{(n + 1) n}{2!} 8^{2} + \frac{(n + 1) n (n - 1)}{3!} 8^{3} + . .. \\ \Rightarrow 9^{n + 1} - 1 - 8 n - 8 = \frac{(n + 1) n}{2!} 8^{2} + \frac{(n + 1) n (n - 1)}{3!} 8^{3} + . .. \\ 9^{n + 1} - 8 n - 9 = 8^{2} \{\frac{(n + 1) n}{2!} + \frac{(n + 1) n (n - 1)}{3!} 8 + . ..\} \\ 9^{n + 1} - 8 n - 9 = 64 \{\frac{(n + 1) n}{2!} + \frac{(n + 1) n (n - 1)}{3!} 8 + . ..\} \\ 9^{n + 1} - 8 n - 9 = multiple of 64, because n is an integer. \\ \Rightarrow 9^{n + 1} - 8 n - 9 is divisible by 64. \end{array}$

Q.14

$Prove that \sum_{r=0}^{n} 3^{r}^{n} C_{r} {=4}^{n} .$

Ans

$\begin{array}{l} L . H . S . = \sum_{r = 0}^{n} 3^{r}^{n} C_{r} = 3^{0}^{n} C_{0} + 3^{1}^{n} C_{1} + 3^{2}^{n} C_{2} + 3^{3}^{n} C_{3} + 3^{4}^{n} C_{4} + . .. \\ =^{n} C_{0} . 3^{0} +^{n} C_{1} . 3^{1} +^{n} C_{2} . 3^{2} +^{n} C_{3} . 3^{3} + . .. \\ = {(1 + 3)}^{n} [\begin{array}{l} ∵ {(1 + x)}^{n} =^{n} C_{0} . x^{0} +^{n} C_{1} . x^{1} +^{n} C_{2} . x^{2} \\ +^{n} C_{3} . x^{3} + . .. \end{array}] \\ = 4^{n} = R . H . S . \end{array}$

Q.15 Find the coefficient of x⁵ in (x + 3)⁸.

Ans

$\begin{array}{l} {The (r + 1)}^{th} {term of the expansion (x + y)}^{n} is given by \\ T_{r +1} =^{n} C_{r} {x^{n}}^{- r} y^{r} . \\ In the expansion of {(x + 3)}^{8}, \\ T_{r +1} =^{8} C_{r} {x^{8}}^{- r} 3^{r} \\ For {the coefficient of x}^{5}, \\ let 8 - r = 5 \\ \Rightarrow r = 8 - 5 = 3 \\ So, {the coefficient of x}^{5} \\ =^{8} C_{3} 3^{3} \\ = 27 \times \frac{8!}{3! 5!} \\ = 27 \times \frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times 5!} \\ = 1512 \end{array}$

Q.16 Find the coefficient of a⁵b⁷in (a – 2b)¹².

Ans

$\begin{array}{l} {The (r + 1)}^{th} {term of the expansion (x + y)}^{n} is given by \\ T_{r +1} =^{n} C_{r} {x^{n}}^{- r} y^{r} . \\ In the expansion of {(a - 2b)}^{12}, \\ T_{r +1} =^{12} C_{r} {a^{12}}^{- r} {(- 2b)}^{r} \\ For {the coefficient of a}^{5} b^{7}, \\ let 12 - r = 5 \\ \Rightarrow r = 12 - 5 \\ \begin{array}{l} = 7 \\ So, {the coefficient of a}^{5} b^{7} \\ =^{12} C_{7} {(- 2)}^{7} \\ = - 128 \times \frac{12!}{7! 5!} \\ = - 128 \times \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{7! \times 5 \times 4 \times 3 \times 2 \times 1} \\ = - 101376 \end{array} \end{array}$

Q.17 Write the general term in the expansion of (x² – y)⁶.

Ans

$\begin{array}{l} The general term of {(x^{2} - y)}^{6} is \\ T_{r + 1} =^{6} C_{r} {(x^{2})}^{6 - r} {(- y)}^{r} \\ = {(- 1)}^{r}^{6} C_{r} {(x)}^{12 - 2 r} . y^{r} \end{array}$

Q.18 Find the 4^th term in the expansion of (x – 2y)¹².

Ans

$\begin{array}{l} Let { 4}^{th} term in the expression of {(x - 2y)}^{12} {be T}_{4}, then \\ T_{4} = T_{3 + 1} \\ =^{12} C_{3} x^{12 - 3} {(- 2 y)}^{3} \\ =^{12} C_{3} x^{9} {(- 2)}^{3} y^{3} \\ = - 8 \times \frac{12!}{3! 9!} x^{9} y^{3} \\ = - 8 \times \frac{12 \times 11 \times 10 \times 9!}{3 \times 2 \times 1 \times 9!} x^{9} y^{3} \\ = - 1760 x^{9} y^{3} \end{array}$

Q.19

${Find the 13}^{th} term in the expansion of {(9x - \frac{1}{3 \sqrt{x}})}^{18}, x \neq 0.$

Ans

$\begin{array}{l} {Let 13}^{th} term in the expression of {(9 x - \frac{1}{3 \sqrt{x}})}^{18} {be T}_{13}, then \\ T_{13} = T_{12 + 1} \\ =^{18} C_{12} {(9 x)}^{18 - 12} {(- \frac{1}{3 \sqrt{x}})}^{12} \\ =^{18} C_{12} (9^{6} x^{6}) {(- \frac{1}{3})}^{12} (\frac{1}{x^{6}}) \\ = \frac{9^{6}}{3^{12}} \times \frac{18!}{12! 6!} x^{6} (\frac{1}{x^{6}}) \\ = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2} \times \frac{x^{6}}{x^{6}} \\ = 18564 \end{array}$

Q.20

$Write {the general term in the expansion of (x}^{2} - yx)^{12}, x \neq 0 .$

Ans

$\begin{array}{l} The general term of {(x^{2} - yx)}^{12} is \\ T_{r + 1} =^{12} C_{r} {(x^{2})}^{12 - r} {(- yx)}^{r} \\ = {(- 1)}^{r}^{12} C_{r} {(x)}^{24 - 2 r} . x^{r} . y^{r} \\ = {(- 1)}^{r}^{12} C_{r} {(x)}^{24 - r} . y^{r} \end{array}$

Q.21

$Find the middle term in the expansion of {(3 - \frac{x^{3}}{6})}^{7} .$

Ans

$\begin{array}{l} {(3 - \frac{x^{3}}{6})}^{7} \\ Here, n = 7 (odd) \\ So, middle terms in expansion \\ = {(\frac{7 + 1}{2})}^{th} term and {(\frac{7 + 1}{2} + 1)}^{th} term \\ = 4^{th} { term and 5}^{th} term \\ {Then, T}_{4} = T_{3 + 1} \\ =^{7} C_{3} . 3^{7 - 3} . {(- \frac{x^{3}}{6})}^{3} \\ =^{7} C_{3} . 3^{4} . {(- \frac{1}{6})}^{3} x^{9} \\ = - \frac{35 \times 27 \times 3}{36 \times 6} x^{9} \\ = - \frac{105}{8} x^{9} \\ and T_{5} = T_{4 + 1} \end{array}$ $\begin{array}{l} =^{7} C_{4} . 3^{7 - 4} . {(- \frac{x^{3}}{6})}^{4} \\ =^{7} C_{4} . 3^{3} {(- \frac{1}{6})}^{4} x^{12} \\ = \frac{35 \times 27}{36 \times 6 \times 6} x^{12} \\ = \frac{35}{48} x^{12} \\ Thus, the middle terms of the expansion are - \frac{105}{8} x^{9} and \frac{35}{48} x^{12} . \end{array}$

Q.22

$Find the middle term in the expansion of {(\frac{x}{3} +9y)}^{10} .$

Ans

$\begin{array}{l} Given : {(\frac{x}{3} + 9 y)}^{10} \\ Here, n = 10 (even) \\ The middle term in the expansion = {(\frac{10}{2} + 1)}^{th} term \\ = 6^{th} term \\ {Then, T}_{6} = T_{5 + 1} \\ =^{10} C_{5} {(\frac{x}{3})}^{10 - 5} {(9 y)}^{5} \\ = \frac{10!}{5! (10 - 5)!} \times \frac{x^{5}}{3^{5}} \times 9^{5} y^{5} \\ = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5 \times 4 \times 3 \times 2 \times 1 \times 5!} \times \frac{x^{5}}{3^{5}} \times 3^{10} y^{5} \\ = 6 \times 42 \times 9 \times 27 x^{5} y^{4} \\ = 61236 x^{5} y^{5} \\ Thus, the middle term of given expression is 61236 x^{5} y^{5} . \end{array}$

Q.23 In the expansion of (1 + a)^m+n, prove that coefficients of a^m and aⁿ are equal.

Ans

$\begin{array}{l} The given expression is {(1 + a)}^{m + n} . \\ The general term (T_{r + 1}) =^{m + n} C_{r} a^{r} . .. (i) \\ Putting r = m in equation (i), we get \\ Coefficient {of a}^{m} =^{m + n} C_{m} \\ Putting r = n in equation (i), we get \\ Coefficient {of a}^{n} =^{m + n} C_{n} \\ =^{m + n} C_{m + n - n} [∵^{n} C_{r} =^{n} C_{n - r}] \\ =^{m + n} C_{m} \\ Thus, \\ {the coefficient of a}^{m} = {the coefficient of a}^{n} . \end{array}$

Q.24 The coefficients of the (r – 1)^th, r^th and (r + 1)^th terms in the expansion of (x + 1)ⁿare in the ratio 1 : 3 : 5. Find n and r.

Ans

$\begin{array}{l} The coefficients of the {(r - 1)}^{th}, r^{th} and {(r + 1)}^{th} terms in the \\ expansion of {(x + 1)}^{n} are in the ratio 1 : 3 : 5.Then, \\ T_{r - 1} = T_{(r - 2) +1} \\ =^{n} C_{r - 2} x^{n - (r - 2)} \\ = \frac{n!}{(r - 2)! (n - r + 2)!} x^{(n - r + 2)} \\ T_{r} = T_{(r - 1) +1} \\ =^{n} C_{(r - 1)} x^{n - (r - 1)} \\ = \frac{n!}{(r - 1)! (n - r + 1)!} x^{(n - r + 1)} \\ T_{r + 1} = T_{r} \\ =^{n} C_{r} x^{n - r} \\ = \frac{n!}{r! (n - r)!} x^{(n - r)} \\ Since, \\ \frac{Coefficient of T_{r - 1}}{Coefficient of T_{r}} = \frac{\frac{n!}{(r - 2)! (n - r + 2)!}}{\frac{n!}{(r - 1)! (n - r + 1)!}} \\ \frac{1}{3} = \frac{(r - 1)! (n - r + 1)!}{(r - 2)! (n - r + 2)!} \\ \frac{1}{3} = \frac{(r - 1) . (r - 2)! (n - r + 1)!}{(r - 2)! (n - r + 2) . (n - r + 1)!} \end{array}$ $\begin{array}{l} \frac{1}{3} = \frac{(r - 1)}{(n - r + 2)} \\ n - r + 2 = 3 r - 3 \\ n - 4 r = - 5 . .. (i) \\ \frac{Coefficient of T_{r}}{Coefficient of T_{r + 1}} = \frac{\frac{n!}{(r - 1)! (n - r + 1)!}}{\frac{n!}{r! (n - r)!}} \\ \frac{3}{5} = \frac{r . (r - 1)! (n - r)!}{(r - 1)! (n - r + 1) . (n - r)!} \\ \frac{3}{5} = \frac{r}{(n - r + 1)} \\ 3 n - 3 r + 3 = 5 r \\ 3 n - 8 r = - 3 . .. (ii) \\ On solving equation (i) and equation (ii), we get \\ n = 7 and r = 3. \end{array}$

Q.25 Prove that the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1 + x)^{2n – 1}.

Ans

$\begin{array}{l} Given expressions are: {(1 + x)}^{2n} and {(1 + x)}^{2n-1} \\ For {(1 + x)}^{2n} : \\ Since, T_{r+1} =^{2 n} C_{r} x^{r} \\ Putting r = n, we get \\ {coefficient of x}^{n} =^{2 n} C_{n} \end{array}$ $\begin{array}{l} = \frac{2 n!}{n! (2 n - n)!} \\ = \frac{2 n!}{n! n!} \\ = \frac{2 n (2 n - 1)!}{n (n - 1)! n!} \\ = 2 \times \frac{(2 n - 1)!}{(n - 1)! n!} \\ For {(1 + x)}^{2n-1} : \\ Since, T_{r + 1} =^{2 n - 1} C_{r} x^{r} \\ Putting r = n, we get \\ {coefficient of x}^{n} =^{2 n - 1} C_{n} \\ = \frac{(2 n - 1)!}{n! (2 n - 1 - n)!} \\ = \frac{(2 n - 1)!}{n! (n - 1)!} \\ Thus, {coefficient of x}^{n} in {(1 + x)}^{n} is double of the coefficient \\ {of x}^{n} in {(1 + x)}^{2 n - 1} . \end{array}$

Q.26 Find a positive value of m for which the coefficient of x² in the expansion (1 + x)^m is 6.

Ans

$\begin{array}{l} {Given: Coefficient of x}^{2} in the expansion of {(1 + x)}^{m} is 6 . \\ {The (r + 1)}^{th} term of the expansion {(1 + x)}^{m} is given by \\ T_{r+1} =^{m} C_{r} x^{r} \\ Putting r = 2, we get \\ {coefficient of x}^{2} =^{m} C_{2} \\ 6 = \frac{m!}{2! (m - 2)!} \\ \Rightarrow 6 = \frac{m (m - 1) (m - 2)!}{2 (m - 2)!} \\ \Rightarrow 12 = m (m - 1) \\ \Rightarrow 4 \times 3 = m (m - 1) \\ \Rightarrow 4 \times (4 - 1) = m (m - 1) \\ \Rightarrow m = 4 \end{array}$

Q.27 Find a, b and n in the expansion of (a + b)ⁿ if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Ans

$\begin{array}{l} Three terms in the expansion of {(a + b)}^{n} are 729, 7290 and \\ 30375 respectively. \\ T_{1} = 729 \\ \Rightarrow T_{0 + 1} =^{n} C_{0} a^{n} b^{0} \\ \Rightarrow^{n} C_{0} a^{n} = 729 \\ \Rightarrow a^{n} = 729 . .. (i) \\ \Rightarrow T_{2} = 7290 \\ \Rightarrow T_{1 + 1} =^{n} C_{1} a^{n - 1} b^{1} \\ \Rightarrow {na}^{n - 1} b = 7290 . .. (ii) \end{array}$ $\begin{array}{l} \Rightarrow \frac{{na}^{n} b}{a} = 7290 \\ \Rightarrow \frac{n (729) b}{a} = 7290 [From eqution (i)] \\ \Rightarrow \frac{nb}{a} = \frac{7290}{729} \\ \Rightarrow \frac{nb}{a} = 10 . .. (iii) \\ T_{3} = 30375 \\ \Rightarrow T_{2 + 1} = 30375 \\ \Rightarrow^{n} C_{2} a^{n - 2} b^{2} = 30375 \\ \Rightarrow \frac{n!}{2! (n - 2)!} a^{n - 2} b^{2} = 30375 \\ \Rightarrow \frac{n (n - 1)}{2} a^{n - 2} b^{2} = 30375 . .. (iv) \\ Dividing equation (iv) by equation (ii), we get \\ \frac{\frac{n (n - 1)}{2} a^{n - 2} b^{2}}{{na}^{n - 1} b} = \frac{30375}{7290} \\ \Rightarrow \frac{(n - 1) b}{a} = \frac{2 \times 30375}{7290} \\ \Rightarrow \frac{nb}{a} - \frac{b}{a} = \frac{25}{3} \\ \Rightarrow 10 - \frac{b}{a} = \frac{25}{3} [From equation (ii)] \\ \Rightarrow \frac{b}{a} = 10 - \frac{25}{3} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \Rightarrow \frac{b}{a} = \frac{5}{3} \\ From equation (iii), we get \\ n (\frac{5}{3}) = 10 \\ \Rightarrow n = 10 \times \frac{3}{5} \\ = 6 \\ From equation (i), we have \\ a^{6} = 729 \\ a^{6} = 3^{6} \\ \Rightarrow a = 3 \\ Putting the values of n and a in equation (iii), we get \\ \frac{6 b}{3} = 10 \\ \Rightarrow b = \frac{10}{2} \\ = 5 \\ Thus, a = 3, b = 5 and n = 6. \end{array}$

Q.28 Find a, if the coefficients of x² and x³ in the expansion of (3 + ax)⁹ are equal.

Ans

$\begin{array}{l} {(r+1)}^{th} term in the expansion of {(3 + ax)}^{9} is \\ T_{r+1} =^{9} C_{r} 3^{9 - r} {(ax)}^{r} . .. (i) \\ Putting r = 2 in equation (i), we get \end{array}$ $\begin{array}{l} T_{2+1} =^{9} C_{2} 3^{9 - 2} {(ax)}^{2} \\ \Rightarrow coefficient {of x}^{2} =^{9} C_{2} 3^{7} a^{2} \\ = \frac{9! \times 3^{7}}{2! (9 - 2)!} a^{2} \\ Putting r = 3 in equation (i), we get \\ T_{3+1} =^{9} C_{3} 3^{9 - 3} {(ax)}^{3} \\ \Rightarrow coefficient {of x}^{3} =^{9} C_{3} 3^{6} a^{3} \\ = \frac{9! \times 3^{6}}{3! (9 - 3)!} a^{3} \\ Since, \\ Coefficient {of x}^{3} = {coefficient of x}^{2} \\ \frac{9! \times 3^{6}}{3! (9 - 3)!} a^{3} = \frac{9! \times 3^{7}}{2! (9 - 2)!} a^{2} \\ \Rightarrow \frac{a}{3} = \frac{3}{7} \\ \Rightarrow a = \frac{9}{7} \\ Thus, the required value of a is \frac{9}{7} . \end{array}$

Q.29 Find the coefficient of x⁵ in the product (1 + 2x)⁶ (1 – x)⁷ using binomial theorem.

Ans

$\begin{array}{l} Using binomial theorem, we have \\ {(1 + 2x)}^{6} = 1 + 6 (2 x) + \frac{6 .5}{2!} {(2 x)}^{2} + \frac{6 .5.4}{3!} {(2 x)}^{3} + \frac{6 .5.4.3}{4!} {(2 x)}^{4} \end{array}$ $\begin{array}{l} + \frac{6 .5.4.3.2}{5!} {(2 x)}^{5} + \frac{6 .5.4.3.2.1}{6!} {(2 x)}^{6} \\ = 1 + 12 x + 60 x^{2} + 160 x^{3} + 240 x^{4} + 192 x^{5} + 64 x^{6} \\ {(1 - x)}^{7} = 1 - 7 x + \frac{7 .6}{2!} x^{2} - \frac{7 .6.5}{3!} x^{3} + \frac{7 .6.5.4}{4!} x^{4} - \frac{7 .6.5.4.3}{5!} x^{5} \\ + \frac{7 .6.5.4.3.2}{6!} x^{6} - \frac{7 .6.5.4.3.2.1}{7!} x^{7} \\ = 1 - 7 x + 21 x^{2} - 35 x^{3} + 35 x^{4} - 21 x^{5} + 7 x^{6} - x^{7} \\ Now, \\ {(1 + 2x)}^{6} {(1 - x)}^{7} \\ = (1 + 12 x + 60 x^{2} + 160 x^{3} + 240 x^{4} + 192 x^{5} + 64 x^{6}) \\ \times (1 - 7 x + 21 x^{2} - 35 x^{3} + 35 x^{4} - 21 x^{5} + 7 x^{6} - x^{7}) \\ Sum of the coefficients {of x}^{5} \\ = - 21 + 12 \times 35 - 60 \times 35 + 160 \times 21 + 240 \times - 7 + 192 \times 1 \\ = - 21 + 420 - 2100 + 3360 - 1680 + 192 \\ = 171 \end{array}$

Q.30 If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer.

Ans

$\begin{array}{l} \begin{array}{l} Since, a = a - b + b \\ So, a^{n} = {(a - b + b)}^{n} \\ = {\{(a - b) + b\}}^{n} \\ =^{n} C_{0} {(a - b)}^{n} +^{n} C_{1} {(a - b)}^{n - 1} b +^{n} C_{2} {(a - b)}^{n - 2} b^{2} + . .. + \\ ^{n} C_{n - 1} (a - b) b^{n - 1} +^{n} C_{n} b^{n} \end{array} \\ = {(a - b)}^{n} +^{n} C_{1} {(a - b)}^{n - 1} b +^{n} C_{2} {(a - b)}^{n - 2} b^{2} + . .. + \\ ^{n} C_{n - 1} (a - b) b^{n - 1} + b^{n} \\ a^{n} - b^{n} = (a - b) \{\begin{array}{l} {(a - b)}^{n - 1} +^{n} C_{1} {(a - b)}^{n - 2} b +^{n} C_{2} {(a - b)}^{n - 3} b^{2} + . .. \\ +^{n} C_{n - 1} b^{n - 1} \end{array}\} \\ = (a - b) λ, where λ is an integer. \\ Thus, (a - b) is a factor of (a^{n} - b^{n}) . \end{array}$

Q.31

$Evaluate : {(\sqrt{3} + \sqrt{2})}^{6} - {(\sqrt{3} - \sqrt{2})}^{6} .$

Ans

$\begin{array}{l} Since, {(a + b)}^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} + . .. +^{n}^{n} C_{n} b^{n} \\ Putting a = \sqrt{3}, b = \sqrt{2} and n = 6, we get \\ {(\sqrt{3} + \sqrt{2})}^{6} =^{6} C_{0} {(\sqrt{3})}^{6} +^{6} C_{1} {(\sqrt{3})}^{6 - 1} (\sqrt{2}) +^{6} C_{2} {(\sqrt{3})}^{6 - 2} {(\sqrt{2})}^{2} \\ +^{6} C_{3} {(\sqrt{3})}^{6 - 3} {(\sqrt{2})}^{3} +^{6} C_{4} {(\sqrt{3})}^{6 - 4} {(\sqrt{2})}^{4} +^{6} C_{5} {(\sqrt{3})}^{6 - 5} {(\sqrt{2})}^{5} \\ +^{6} C_{6} {(\sqrt{3})}^{6 - 6} {(\sqrt{2})}^{6} \\ = 27 + 6 {(\sqrt{3})}^{5} (\sqrt{2}) + 15 {(\sqrt{3})}^{4} {(\sqrt{2})}^{2} + 20 {(\sqrt{3})}^{3} {(\sqrt{2})}^{3} \\ + 15 {(\sqrt{3})}^{2} {(\sqrt{2})}^{4} + 6 (\sqrt{3}) {(\sqrt{2})}^{5} + {(\sqrt{2})}^{6} \\ {(\sqrt{3} - \sqrt{2})}^{6} =^{6} C_{0} {(\sqrt{3})}^{6} -^{6} C_{1} {(\sqrt{3})}^{6 - 1} (\sqrt{2}) +^{6} C_{2} {(\sqrt{3})}^{6 - 2} {(\sqrt{2})}^{2} \\ -^{6} C_{3} {(\sqrt{3})}^{6 - 3} {(\sqrt{2})}^{3} +^{6} C_{4} {(\sqrt{3})}^{6 - 4} {(\sqrt{2})}^{4} -^{6} C_{5} {(\sqrt{3})}^{6 - 5} {(\sqrt{2})}^{5} \\ +^{6} C_{6} {(\sqrt{3})}^{6 - 6} {(\sqrt{2})}^{6} \end{array}$ $\begin{array}{l} = 27 - 6 {(\sqrt{3})}^{5} (\sqrt{2}) + 15 {(\sqrt{3})}^{4} {(\sqrt{2})}^{2} - 20 {(\sqrt{3})}^{3} {(\sqrt{2})}^{3} \\ + 15 {(\sqrt{3})}^{2} {(\sqrt{2})}^{4} - 6 (\sqrt{3}) {(\sqrt{2})}^{5} + {(\sqrt{2})}^{6} \\ ∴ {(\sqrt{3} + \sqrt{2})}^{6} - {(\sqrt{3} - \sqrt{2})}^{6} \\ = 2 {6 {(\sqrt{3})}^{5} (\sqrt{2}) + 20 {(\sqrt{3})}^{3} {(\sqrt{2})}^{3} + 6 (\sqrt{3}) {(\sqrt{2})}^{5}} \\ = 2 {54 \sqrt{6} + 120 \sqrt{6} + 24 \sqrt{6}} \\ = 2 (198 \sqrt{6}) \\ = 396 \sqrt{6} \\ Thus, \\ {(\sqrt{3} + \sqrt{2})}^{6} - {(\sqrt{3} - \sqrt{2})}^{6} = 396 \sqrt{6} . \end{array}$

Q.32

$Find the value of {(a^{2} + \sqrt{a^{2} - 1})}^{4} + {(a^{2} - \sqrt{a^{2} - 1})}^{4} .$

Ans

$\begin{array}{l} Since, {(a + b)}^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} + . .. +^{n}^{n} C_{n} b^{n} \\ {(a^{2} + \sqrt{a^{2} - 1})}^{4} =^{4} C_{0} {(a^{2})}^{4} +^{4} C_{1} {(a^{2})}^{4 - 1} (\sqrt{a^{2} - 1}) +^{4} C_{2} {(a^{2})}^{4 - 2} {(\sqrt{a^{2} - 1})}^{2} \\ +^{4} C_{3} {(a^{2})}^{4 - 3} {(\sqrt{a^{2} - 1})}^{3} +^{4} C_{4} {(a^{2})}^{4 - 4} {(\sqrt{a^{2} - 1})}^{4} \\ = {(a^{2})}^{4} +^{4} C_{1} {(a^{2})}^{3} (\sqrt{a^{2} - 1}) +^{4} C_{2} {(a^{2})}^{2} {(\sqrt{a^{2} - 1})}^{2} \\ +^{4} C_{3} (a^{2}) {(\sqrt{a^{2} - 1})}^{3} + {(\sqrt{a^{2} - 1})}^{4} \\ {(a^{2} - \sqrt{a^{2} - 1})}^{4} =^{4} C_{0} {(a^{2})}^{4} -^{4} C_{1} {(a^{2})}^{4 - 1} (\sqrt{a^{2} - 1}) +^{4} C_{2} {(a^{2})}^{4 - 2} {(\sqrt{a^{2} - 1})}^{2} \end{array}$ $\begin{array}{l} -^{4} C_{3} {(a^{2})}^{4 - 3} {(\sqrt{a^{2} - 1})}^{3} +^{4} C_{4} {(a^{2})}^{4 - 4} {(\sqrt{a^{2} - 1})}^{4} \\ = {(a^{2})}^{4} -^{4} C_{1} {(a^{2})}^{3} (\sqrt{a^{2} - 1}) +^{4} C_{2} {(a^{2})}^{2} {(\sqrt{a^{2} - 1})}^{2} \\ -^{4} C_{3} (a^{2}) {(\sqrt{a^{2} - 1})}^{3} + {(\sqrt{a^{2} - 1})}^{4} \\ ∴ {(a^{2} + \sqrt{a^{2} - 1})}^{4} + {(a^{2} - \sqrt{a^{2} - 1})}^{4} \\ = 2 {{(a^{2})}^{4} +^{4} C_{2} {(a^{2})}^{2} {(\sqrt{a^{2} - 1})}^{2} + {(\sqrt{a^{2} - 1})}^{4}} \\ = 2 {a^{8} + 6 a^{4} (a^{2} - 1) + {(a^{2} - 1)}^{2}} \\ = 2 {a^{8} + 6 a^{6} - 6 a^{4} + a^{4} - 2 a^{2} + 1} \\ = 2 a^{8} + 12 a^{6} - 10 a^{4} - 4 a^{2} + 2 \end{array}$

Q.33 Find an approximation of (0.99)⁵ using the first three terms of its expansion.

Ans

$\begin{array}{l} {(0.99)}^{5} = {(1 - 0 .01)}^{5} \\ = 1 -^{5} C_{1} (0 .01) +^{5} C_{2} {(0 .01)}^{2} \\ = 1 - 5 (0 .01) + 10 (0 .0001) \\ = 1 - 0 .05 + 0 .001 \\ = 1 .001 - 0 .05 \\ = 0 .951 \\ Thus, an approximation of {(0 .99)}^{5} using the first three terms is 0.951. \end{array}$

Q.34

$\begin{array}{l} Find n, if the ratio of the fifth term from the beginning to the fifth term from the end \\ in the expansion of {(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})}^{n} is \sqrt{6} :1. \end{array}$

Ans

$\begin{array}{l} We are given that: {(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})}^{n} \\ T_{r + 1} =^{n} C_{r} {(\sqrt[4]{2})}^{n - r} {(\frac{1}{\sqrt[4]{3}})}^{r} \\ {For 5}^{th} term from starting: \\ T_{5} = T_{4 + 1} \\ =^{n} C_{4} {(\sqrt[4]{2})}^{n - 4} {(\frac{1}{\sqrt[4]{3}})}^{4} \\ =^{n} C_{4} {(\sqrt[4]{2})}^{n - 4} (\frac{1}{3}) \\ {For 5}^{th} term from end: \\ {Since, r}^{th} term from end = (n + 1) - (r - 1) \\ {So, 5}^{th} term from end = (n + 1) - (5 - 1) \\ T_{n+1- (5-1)} = T_{(n - 4) + 1} \\ =^{n} C_{n - 4} {(\sqrt[4]{2})}^{4} {(\frac{1}{\sqrt[4]{3}})}^{n - 4} \\ =^{n} C_{n - n + 4} {(\sqrt[4]{2})}^{4} {(\frac{1}{\sqrt[4]{3}})}^{n - 4} [∵^{n} C_{r} =^{n} C_{n - r}] \\ =^{n} C_{4} (2) {(\frac{1}{\sqrt[4]{3}})}^{n - 4} \\ =^{n} C_{4} (2) {(\frac{1}{\sqrt[4]{3}})}^{n - 4} \\ According to given conditions: \\ \frac{T_{5}}{T_{(n + 1) - (5 - 1)}} = \frac{\sqrt{6}}{1} \\ \frac{^{n} C_{4} {(\sqrt[4]{2})}^{n - 4} (\frac{1}{3})}{^{n} C_{4} (2) {(\frac{1}{\sqrt[4]{3}})}^{n - 4}} = \frac{\sqrt{6}}{1} \\ \frac{{(\sqrt[4]{2})}^{n - 4} (\frac{1}{3})}{(2) {(\frac{1}{\sqrt[4]{3}})}^{n - 4}} = \frac{\sqrt{6}}{1} \\ {(\sqrt[4]{6})}^{n - 4} = \frac{6 \sqrt{6}}{1} \\ 6^{\frac{n - 4}{4}} = 6^{\frac{3}{2}} \\ \Rightarrow \frac{n - 4}{4} = \frac{3}{2} \\ \Rightarrow 2 n - 8 = 12 \\ \Rightarrow n = \frac{20}{2} \\ \Rightarrow n = 10 \end{array}$

Q.35

$Expand using Binomial Theorem {(1+ \frac{x}{2} - \frac{2}{x})}^{4}, x \neq 0.$

Ans

$\begin{array}{l} Using Binomial Theorem, \\ {(1 + \frac{x}{2} - \frac{2}{x})}^{4} =^{4} C_{0} {(1 + \frac{x}{2})}^{4} -^{4} C_{1} {(1 + \frac{x}{2})}^{3} {(\frac{2}{x})}^{1} +^{4} C_{2} {(1 + \frac{x}{2})}^{2} {(\frac{2}{x})}^{2} \\ -^{4} C_{3} {(1 + \frac{x}{2})}^{1} {(\frac{2}{x})}^{3} +^{4} C_{4} {(\frac{2}{x})}^{4} \\ = {(1 + \frac{x}{2})}^{4} - 4 {(1 + \frac{x}{2})}^{3} (\frac{2}{x}) + 6 {(1 + \frac{x}{2})}^{2} (\frac{4}{x^{2}}) \\ - 4 (1 + \frac{x}{2}) (\frac{8}{x^{3}}) + (\frac{16}{x^{4}}) \\ = (1 + 4 . \frac{x}{2} + 6 . \frac{x^{2}}{4} + 4 . \frac{x^{3}}{8} + \frac{x^{4}}{16}) - (1 + 3 . \frac{x}{2} + 3 . \frac{x^{2}}{4} + \frac{x^{3}}{8}) (\frac{8}{x}) \\ + (1 + 2 . \frac{x}{2} + \frac{x^{2}}{4}) (\frac{24}{x^{2}}) - (1 + \frac{x}{2}) (\frac{32}{x^{3}}) + (\frac{16}{x^{4}}) \\ = 1 + 4 . \frac{x}{2} + 6 . \frac{x^{2}}{4} + 4 . \frac{x^{3}}{8} + \frac{x^{4}}{16} - \frac{8}{x} - 12 - 6 x - x^{2} + \frac{24}{x^{2}} \\ + \frac{24}{x} + 6 - \frac{32}{x^{3}} - \frac{16}{x^{2}} + \frac{16}{x^{4}} \\ = - 5 + 2 x - 6 x - x^{2} + \frac{3}{2} x^{2} + \frac{x^{3}}{2} + \frac{x^{4}}{16} - \frac{8}{x} + \frac{24}{x} + \frac{24}{x^{2}} \\ - \frac{16}{x^{2}} - \frac{32}{x^{3}} + \frac{16}{x^{4}} \\ = - 5 - 4 x + \frac{x^{2}}{2} + \frac{x^{3}}{2} + \frac{x^{4}}{16} + \frac{16}{x} + \frac{8}{x^{2}} - \frac{32}{x^{3}} + \frac{16}{x^{4}} \\ = \frac{16}{x} + \frac{8}{x^{2}} - \frac{32}{x^{3}} + \frac{16}{x^{4}} - 4 x + \frac{x^{2}}{2} + \frac{x^{3}}{2} + \frac{x^{4}}{16} - 5 \end{array}$

Q.36 Find the expansion of (3x² – 2ax + 3a²)³ using binomial theorem.

Ans

$\begin{array}{l} Using binomial theorem, we can expand \\ {({3x}^{2} - 2ax + {3a}^{2})}^{3} = {\{({3x}^{2} - 2ax) + {3a}^{2}\}}^{3} \\ =^{3} C_{0} {({3x}^{2} - 2ax)}^{3} +^{3} C_{1} {({3x}^{2} - 2ax)}^{2} {({3a}^{2})}^{1} \\ +^{3} C_{2} {({3x}^{2} - 2ax)}^{1} {({3a}^{2})}^{2} +^{3} C_{3} {({3a}^{2})}^{3} \\ = {({3x}^{2} - 2ax)}^{3} + 3 {({3x}^{2} - 2ax)}^{2} ({3a}^{2}) \\ + 3 {({3x}^{2} - 2ax)}^{1} ({9a}^{4}) + {27a}^{6} \\ =^{3} C_{0} {({3x}^{2})}^{3} -^{3} C_{1} {({3x}^{2})}^{2} (2ax) +^{3} C_{2} {({3x}^{2})}^{1} {(2ax)}^{2} \\ -^{3} C_{3} {(2ax)}^{3} + \{9 x^{4} - 12 {ax}^{3} + 4 a^{2} x^{2}\} (9 a^{2}) \\ + 81 a^{4} x^{2} - 54 a^{5} x + 27 a^{6} \\ = {27x}^{6} - 3 ({9x}^{4}) (2ax) + 3 ({3x}^{2}) ({4a}^{2} x^{2}) - {8a}^{3} x^{3} \\ + 81 a^{2} x^{4} - 108 a^{3} x^{3} + 36 a^{4} x^{2} + 81 a^{4} x^{2} - 54 a^{5} x \\ + 27 a^{6} \\ = {27x}^{6} - 54 {ax}^{5} + 36 a^{2} x^{4} - {8a}^{3} x^{3} + 81 a^{2} x^{4} \\ - 108 a^{3} x^{3} + 36 a^{4} x^{2} + 81 a^{4} x^{2} - 54 a^{5} x + 27 a^{6} \\ = {27x}^{6} - 54 {ax}^{5} + 117 a^{2} x^{4} - {116a}^{3} x^{3} + 117 a^{4} x^{2} \\ - 54 a^{5} x + 27 a^{6} \end{array}$

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FAQs (Frequently Asked Questions)

1. Will the NCERT Solutions Class 11 Mathematics Chapter 8 help students understand important concepts for the exam?

To understand the method of expansion, students are advised to refer to the solved examples present in the Class 11 Mathematics Chapter 8 Solutions before solving the exercise. Each solution is solved step-by-step to help students understand the concepts clearly. Practising regularly with the help of the NCERT Solutions Class 11 Mathematics Chapter 8 will enable the students to attain good scores in the exams.

2. What are the important sub-topics included in the NCERT Solutions Class 11 Mathematics Chapter 8?

The important topics that are included in the NCERT Solutions Class 11 Mathematics Chapter 8- Binomial Theorem are mentioned below

8.1 Introduction

8.2 Binomial Theorem for Positive Integral Indices

Pascal’s Triangle
Binomial theorem
Special cases

8.3 General and Middle Terms

Several solved examples and exercise questions are solved thoroughly in the NCERT Solutions Class 11 Mathematics Chapter 8 to help students complete their preparation for the examinations.

NCERT Solutions Class 11 Maths Chapter 8

NCERT Solutions Class 11 Mathematics Chapter 8 – Binomial theorem

Key Topics Covered In NCERT Solutions Class 11 Mathematics Chapter 8

NCERT Solutions Class 11 Mathematics Chapter 8: Exercise & Solutions

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Key Features of NCERT Solutions Class 11 Mathematics Chapter 8

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2. What are the important sub-topics included in the NCERT Solutions Class 11 Mathematics Chapter 8?

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NCERT Solutions Class 11 Maths Chapter 8

NCERT Solutions Class 11 Mathematics Chapter 8 – Binomial theorem

Key Topics Covered In NCERT Solutions Class 11 Mathematics Chapter 8

NCERT Solutions Class 11 Mathematics Chapter 8: Exercise & Solutions

NCERT Exemplar Class 11 Mathematics

Key Features of NCERT Solutions Class 11 Mathematics Chapter 8

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1. Will the NCERT Solutions Class 11 Mathematics Chapter 8 help students understand important concepts for the exam?

2. What are the important sub-topics included in the NCERT Solutions Class 11 Mathematics Chapter 8?

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