Binomial Theorem

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Binomial Theorem Class 11 Maths Chapter 8 deals with the study of essential topics such as Positive Integral Indices, Pascal’s Triangle, the Binomial theorem for any positive integer and some special cases. Class 11 Maths Chapter 8 explains the key concepts related to Linear Inequalities with examples. The concepts taught in Class 11 Chapter 8 Binomial Theorem will also help students to gain a thorough understanding of this topic and its practical applications. The expert faculty of Toppr produced these solutions to assist students with their first-term exam preparations.

Binomial Theorem Class 11 Chapter 8 also covers topics such as Using Binomial Theorem and Pascal’s Triangle to expand expressions and finding the general term as well as the middle term in an exponential expression. Students will quickly learn about these concepts if they practice Chapter 8 Maths Class 11 NCERT Solutions of Binomial Theorem. All of these solutions have been created with the new CBSE pattern in mind, ensuring that students have a thorough understanding of their exams. It goes over all major concepts in depth, allowing students to better understand the ideas.

Chapter 8 Maths Class 11 Questions and Answers can help you get good grades in tests and properly prepare for all of the important concepts. Several examples provided in the solutions will assist students with a better understanding of the binomial theorem. Binomial Theorem Class 11 NCERT Solutions Chapter 8 are available below in pdf format, and a few solutions are also included in the exercises. These solutions explain the topics covered with examples so that students can easily relate to the notion being discussed.

Table of Content

Exercise 8.1

Question 1

Expand $(1−2x)_{5}$

Solution

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Using Binomial Theorem, the expression $(1−2x)_{5}$ can be expanded as

$(1−2x)_{5}=_{5}C_{0}(1)_{5}−_{5}C_{1}(1)_{4}(2x)+_{5}C_{2}(1)_{3}(2x)_{2}−_{5}C_{3}(1)_{2}(2x)_{3}+_{5}C_{4}(1)_{1}(2x)_{4}−_{5}C_{5}(2x)_{5}$

$=1−5(2x)+10(4x_{2})−10(8x_{3})+5(16x_{4})−(32x_{5})$

$=1−10x+40x_{2}−80x_{3}+80x_{4}−32x_{5}$

$(1−2x)_{5}=_{5}C_{0}(1)_{5}−_{5}C_{1}(1)_{4}(2x)+_{5}C_{2}(1)_{3}(2x)_{2}−_{5}C_{3}(1)_{2}(2x)_{3}+_{5}C_{4}(1)_{1}(2x)_{4}−_{5}C_{5}(2x)_{5}$

$=1−5(2x)+10(4x_{2})−10(8x_{3})+5(16x_{4})−(32x_{5})$

$=1−10x+40x_{2}−80x_{3}+80x_{4}−32x_{5}$

Question 2

Expand the expression $(2x−3)_{6}$

Solution

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Using Binomial theorem,

$(2x−3)_{6}=_{6}C_{0}(2x)_{6}−_{6}C_{1}(2x)_{5}(3)+_{6}C_{2}(2x)_{4}(3)_{2}−_{6}C_{3}(2x)_{3}(3)_{3}+_{6}C_{4}(2x)_{2}(3)_{4}−_{6}C_{5}(2x)(3)_{5}+_{6}C_{6}(3)_{6}$

$=64x_{6}−6(32x_{5})(3)+15(16x_{4})(9)−20(8x_{3})(27)+15(4x_{2})(81)−6(2x)(243)+729$

$=64x_{6}−576x_{5}+2160x_{4}−4320x_{3}+4860x_{2}−2916x+729$

$(2x−3)_{6}=_{6}C_{0}(2x)_{6}−_{6}C_{1}(2x)_{5}(3)+_{6}C_{2}(2x)_{4}(3)_{2}−_{6}C_{3}(2x)_{3}(3)_{3}+_{6}C_{4}(2x)_{2}(3)_{4}−_{6}C_{5}(2x)(3)_{5}+_{6}C_{6}(3)_{6}$

$=64x_{6}−6(32x_{5})(3)+15(16x_{4})(9)−20(8x_{3})(27)+15(4x_{2})(81)−6(2x)(243)+729$

$=64x_{6}−576x_{5}+2160x_{4}−4320x_{3}+4860x_{2}−2916x+729$

Question 3

Expand $(3x +x1 )_{5}$

Solution

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Using Binomial theorem,

$(3x +x1 )_{5}$

$=_{5}C_{0}(3x )_{5}+_{5}C_{1}(3x )_{4}(x1 )+_{5}C_{2}(3x )_{3}(x1 )_{2}$

$(3x +x1 )_{5}$

$=_{5}C_{0}(3x )_{5}+_{5}C_{1}(3x )_{4}(x1 )+_{5}C_{2}(3x )_{3}(x1 )_{2}$

$+_{5}C_{3}(3x )_{2}(x1 )_{3}+_{5}C_{4}(3x )(x1 )_{4}+_{5}C_{5}(x1 )_{5}$

$=243x_{5} +5(81x_{4} )(x1 )+10(27x_{3} )(x_{2}1 )+10(9x_{2} )(x_{3}1 )+5(3x )(x_{4}1 )+x_{5}1 $

$=243x_{3} +815x_{3} +2710x +9x10 +3x_{3}5 +x_{5}1 $

$=243x_{5} +5(81x_{4} )(x1 )+10(27x_{3} )(x_{2}1 )+10(9x_{2} )(x_{3}1 )+5(3x )(x_{4}1 )+x_{5}1 $

$=243x_{3} +815x_{3} +2710x +9x10 +3x_{3}5 +x_{5}1 $

Question 4

Expand $(x+x1 )_{6}$

Solution

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$(x+x1 )_{6}$

$=_{6}C+0(x)_{6}+_{1}(x)_{5}(x1 )+_{6}C_{2}(x)_{4}(x1 )_{2}+_{6}C_{3}(x)_{3}(x1 )_{3}+$

$=_{6}C+0(x)_{6}+_{1}(x)_{5}(x1 )+_{6}C_{2}(x)_{4}(x1 )_{2}+_{6}C_{3}(x)_{3}(x1 )_{3}+$

$_{6}C_{4}(x)_{2}(x1 )_{4}+_{6}C_{5}(x)(x1 )_{5}+_{6}C_{6}(x1 )_{6}$

$=x_{6}+6(x)_{5}(x1 )+15(x)_{4}(x_{2}1 )+20(x)_{3}(x_{3}1 )+15(x)_{2}(x_{4}1 )+6(x)(x_{5}1 )+x_{6}1 $

$=x_{6}+6x_{4}+15x_{2}+20+x_{2}15 +x_{4}6 +x_{6}1 $

$=x_{6}+6(x)_{5}(x1 )+15(x)_{4}(x_{2}1 )+20(x)_{3}(x_{3}1 )+15(x)_{2}(x_{4}1 )+6(x)(x_{5}1 )+x_{6}1 $

$=x_{6}+6x_{4}+15x_{2}+20+x_{2}15 +x_{4}6 +x_{6}1 $

Question 5

Evaluate $(96)_{3}$

Solution

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$(96)_{3}=(100−4)_{3}$

$=_{3}C_{0}(100)_{3}−_{3}C_{1}(100)_{2}(4)+_{3}C_{2}(100)(4)_{2}−_{3}C_{3}(4)_{3}$

$=(100)_{3}−3(100)_{2}(4)+3(100)(4)_{2}−(4)_{3}$

$=1000000−120000+4800−64$

$=_{3}C_{0}(100)_{3}−_{3}C_{1}(100)_{2}(4)+_{3}C_{2}(100)(4)_{2}−_{3}C_{3}(4)_{3}$

$=(100)_{3}−3(100)_{2}(4)+3(100)(4)_{2}−(4)_{3}$

$=1000000−120000+4800−64$

$=884736$

Question 6

Using Binomial Theorem, evaluate $(102)_{5}$

Solution

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$(102)_{5}=(100+2)_{5}$

$=_{5}C_{0}(100)_{5}+_{5}C_{1}(100)_{4}(2)+_{5}C_{2}(100)_{3}(2)_{2}+_{5}C_{3}(100)_{2}(2)_{3}+_{5}C_{4}(100)(2)_{4}+_{5}C_{5}(2)_{5}$

$=(100)_{5}+5(100)_{4}(2)+10(100)_{3}(2)_{2}+10(100)_{2}(2)_{3}+5(100)(2)_{4}+(2)_{5}$

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

$=_{5}C_{0}(100)_{5}+_{5}C_{1}(100)_{4}(2)+_{5}C_{2}(100)_{3}(2)_{2}+_{5}C_{3}(100)_{2}(2)_{3}+_{5}C_{4}(100)(2)_{4}+_{5}C_{5}(2)_{5}$

$=(100)_{5}+5(100)_{4}(2)+10(100)_{3}(2)_{2}+10(100)_{2}(2)_{3}+5(100)(2)_{4}+(2)_{5}$

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

$=11040808032$

Question 7

Using Binomial Theorem, evaluate $(101)_{4}$

Solution

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$(101)_{4}=(100+1)_{4}$

$=_{4}C_{0}(100)_{4}+_{4}C_{1}(100)_{3}(1)+_{4}C_{2}(100)_{2}(1)_{2}+_{4}C_{3}(100)(1)_{3}+_{4}C_{4}(1)_{4}$

$=(100)_{4}+4(100)_{3}+6(100)_{2}+4(100)+(1)_{4}$

$=100000000+4000000+60000+400+1$

$=_{4}C_{0}(100)_{4}+_{4}C_{1}(100)_{3}(1)+_{4}C_{2}(100)_{2}(1)_{2}+_{4}C_{3}(100)(1)_{3}+_{4}C_{4}(1)_{4}$

$=(100)_{4}+4(100)_{3}+6(100)_{2}+4(100)+(1)_{4}$

$=100000000+4000000+60000+400+1$

$=104060401$

Question 8

Using Binomial theorem, evaluate $(99)_{5}$

Solution

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$(99)_{5}=(100−1)_{5}$

$=_{5}C_{0}(100)_{5}−_{5}C_{1}(100)_{4}(1)+_{5}C_{2}(100)_{3}(1)_{2}−_{5}C_{3}(100)_{2}(1)_{3}+_{5}C_{4}(100)(1)_{4}−_{5}C_{5}(1)_{5}$

$=(100)_{5}−5(100)_{4}+10(100)_{3}−10(100)_{2}+5(100)−1$

$=10000000000−500000000+10000000−100000+500−1$

$=10010000500−500100001$

$=_{5}C_{0}(100)_{5}−_{5}C_{1}(100)_{4}(1)+_{5}C_{2}(100)_{3}(1)_{2}−_{5}C_{3}(100)_{2}(1)_{3}+_{5}C_{4}(100)(1)_{4}−_{5}C_{5}(1)_{5}$

$=(100)_{5}−5(100)_{4}+10(100)_{3}−10(100)_{2}+5(100)−1$

$=10000000000−500000000+10000000−100000+500−1$

$=10010000500−500100001$

$=9509900499$

Question 9

Find $(a+b)_{4}−(a−b)_{4}$. Hence find $s$ if $(3 +2 )_{4}−(3 −2 )_{4}=40s $

Solution

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Using Binomial theorem, the expressions $(a+b)_{4}$ and $(a−b)_{4}$ can be expanded as,

$(a+b)_{4}=_{4}C_{0}a_{4}+_{4}C_{1}a_{3}b+_{4}C_{2}a_{2}b_{2}+_{4}C_{3}ab_{3}+_{4}C_{4}b_{4})$

$(a−b)_{4}=_{4}C_{0}a_{4}−_{4}C_{1}a_{3}b−_{4}C_{2}a_{2}b_{2}−_{4}C_{3}ab_{3}+_{4}C_{4}b_{4})$

$∴(a+b)_{4}−(a−b)_{4}$ =$_{4}C_{0}a_{4}+_{4}C_{1}a_{3}b+_{4}C_{2}a_{2}b_{2}+_{4}C_{3}ab_{3}+_{4}C_{4}b_{4}−[_{4}C_{0}a_{4}−_{4}C_{1}a_{3}b−_{4}C_{2}a_{2}b_{2}−_{4}C_{3}ab_{3}+_{4}C_{4}b_{4}]$

$=$ $2(_{4}C_{1}a_{3}b+_{4}C_{3}ab_{3})=2(4a_{3}b+4ab_{3})$

$=$ $8ab(a_{2}+b_{2})$

Now by putting $a=3 $ and $b=2 $, we obtain

$(3 +2 )_{4}−(3 −2 )_{4}=8(3 )(2 ){(3 )_{2}+(2 )_{2}}=8(6 ){3+2}=406 $

$(a+b)_{4}=_{4}C_{0}a_{4}+_{4}C_{1}a_{3}b+_{4}C_{2}a_{2}b_{2}+_{4}C_{3}ab_{3}+_{4}C_{4}b_{4})$

$(a−b)_{4}=_{4}C_{0}a_{4}−_{4}C_{1}a_{3}b−_{4}C_{2}a_{2}b_{2}−_{4}C_{3}ab_{3}+_{4}C_{4}b_{4})$

$∴(a+b)_{4}−(a−b)_{4}$ =$_{4}C_{0}a_{4}+_{4}C_{1}a_{3}b+_{4}C_{2}a_{2}b_{2}+_{4}C_{3}ab_{3}+_{4}C_{4}b_{4}−[_{4}C_{0}a_{4}−_{4}C_{1}a_{3}b−_{4}C_{2}a_{2}b_{2}−_{4}C_{3}ab_{3}+_{4}C_{4}b_{4}]$

$=$ $2(_{4}C_{1}a_{3}b+_{4}C_{3}ab_{3})=2(4a_{3}b+4ab_{3})$

$=$ $8ab(a_{2}+b_{2})$

Now by putting $a=3 $ and $b=2 $, we obtain

$(3 +2 )_{4}−(3 −2 )_{4}=8(3 )(2 ){(3 )_{2}+(2 )_{2}}=8(6 ){3+2}=406 $

Question 10

Using binomial theorem, indicate which is larger $(1.1)_{10000}$ or $1000$.

Solution

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$(1.1)_{10000}=(1+0.1)_{10000}$

$=10000_{C_{0}}+10000_{C_{1}}(1.1)$ + other positive terms

$=1+10000×1.1$ + other positive terms

$=1+11000$ + other positive terms

$>1000$

Hence, $(1.1)_{10000}>1000$.

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**Binomial Theorem**

The mathematical technique of algebraically expanding the power of sums of two or more binomials is defined as the *binomial theorem*. The Binomial Theorem is a simplified method for expanding a number in the traditional form of *(a + b) ^{n}*, where

*Binomial theorem for any positive integer n –*

The expansion of a binomial for any positive integer *n *is given by the Binomial Theorem, which is,

*(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-1}a.b^{n-1} + ^{n}C_{n}b^{n}*

**Pascal’s Triangle**

The coefficients of the expansions are organized in an array. Pascal's triangle is the name given to this array. The French mathematician Blaise Pascal introduced the concept of Pascal's triangle. The higher powers of a binomial can also be expanded using Pascal's triangle.

**General and Middle Terms**

The middle term in the expansion (a + b)^{n}:

- If n is even, middle term = [(n/2) + 1]
^{th}term - If n is odd, middle term = [(n+1)/2]
^{th}term, [{(n+1)/2} + 1]^{th}term

Related Chapters

- Chapter 1 : SetsChapter 2 : Relations and FunctionsChapter 3 : Trigonometric FunctionsChapter 4 : Principle of Mathematical InductionChapter 5 : Complex Numbers and Quadratic EquationsChapter 6 : Linear InequalitiesChapter 7 : Permutations and CombinationsChapter 9 : Sequences and SeriesChapter 10 : Straight LinesChapter 11 : Conic SectionsChapter 12 : Introduction to Three Dimensional GeometryChapter 13 : Limits and DerivativesChapter 14 : Mathematical ReasoningChapter 15 : StatisticsChapter 16 : Probability

**Q1. What exactly is the Binomial Theorem?**

**Answer:** The technique of algebraically expanding the power of sums of two or more binomials is defined as the binomial theorem. Binomial coefficients are the coefficients of binomial terms during the expansion process. The first sections of these chapters contain appropriate formulations of various facets of the binomial theorem.

**Q2. ****How many exercises are there in the Binomial Theorem Class 11 NCERT Solutions****?**

**Answer:** Chapter 8 Class 11 Maths Binomial Theorem comprises of 36 questions divided into 14 simple formula-based sums, 6 intermediate issues, and 16 difficult problems. The miscellaneous exercise is made up of high-order thinking issues that demand learners to use their critical thinking skills. The solutions to the exercise-specific problems are also accessible in PDF format, with the goal of assisting students in performing well in the yearly exam.

**Q3. What are the key subtopics in Binomial Theorem Class 11 Maths Chapter 8 that could be tested?**

**Answer:** NCERT Solutions Class 11 Maths Chapter 8 begins with a brief overview of the Binomial Theorem. The NCERT Solutions Class 11 Maths Chapter 8 focuses on two important topics: expression expansion using the binomial theorem and locating a term within an exponential expression. Students can now study and stay up to date on the latest CBSE syllabus by using the NCERT Solutions, which are available in PDF format.

**Q4. Is it necessary to practice all of the questions in Binomial Theorem Class 11 NCERT Solutions?**

**Answer:** Binomial Theorem Class 11 NCERT Solutions are well-designed to help students study Binomial Theorem, Positive Integral Indices, and Pascal’s Triangle easily. It thoroughly covers all essential principles, allowing students to grasp the concepts. These solutions use examples to clarify the subjects presented so that students may easily relate to the concepts being addressed.

Related Chapters

- Chapter 1 : SetsChapter 2 : Relations and FunctionsChapter 3 : Trigonometric FunctionsChapter 4 : Principle of Mathematical InductionChapter 5 : Complex Numbers and Quadratic EquationsChapter 6 : Linear InequalitiesChapter 7 : Permutations and CombinationsChapter 9 : Sequences and SeriesChapter 10 : Straight LinesChapter 11 : Conic SectionsChapter 12 : Introduction to Three Dimensional GeometryChapter 13 : Limits and DerivativesChapter 14 : Mathematical ReasoningChapter 15 : StatisticsChapter 16 : Probability