**Solutions of class 11 NCERT maths exercise 8.1 of chapter 8 Binomial Theorem**

**Solutions of class 11 NCERT maths exercise 8.1 of chapter 8 Binomial Theorem** are created here for the purpose of helping the students in their preparation for the exam and in completing their worksheets. All these **NCERT solutions** **of class 11 maths** are prepared by an expert in **maths** who has huge experience in the field of **maths** teaching. These **NCERT solutions** **of class 11 maths** are the most valuable study material for all CBSE students, therefore they have to be thorough with the knowledge about each **exercise** of each **chapter** with proper understanding which is needed them to solve unsolved questions in sample papers, previous year question papers, and then in CBSE board exam.

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**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**Class 11 -Physics**

**Solutions of class 11 NCERT maths exercise 8.1 of chapter 8 Binomial Theorem**

**Q1.Expand the expression (1-2x) ^{5}**

Ans. We are given the expression (1-2x)^{5}

The expansion of the expression (1-a)^{n }is given as following

(1-a)^{n} = 1+^{n}C_{1}a + ^{n}C_{2}a^{2} +^{ n}C_{3}a^{3} +…..^{ n}C_{n}a^{n}

∴(1-2x)^{5 }= 1-^{5}C_{1} 2x+ ^{5}C_{2}(2x)^{2} –^{ 5}C_{3}(2x)^{3} + ^{5}C_{4}(2x)^{4 }– ^{5}C_{5}(2x)^{5}

= 1 – 10x +40x²-80x³ +80x^{4 }-32x^{5}

**Q2.Expand the expression**

(2/x – x/2)^{5}

Ans. The given expression is (2/x – x/2)^{5}

The expansion of the expression (a – b)^{n} is given as

(a – b)^{n} = ^{n}C_{0}a^{n}b^{0} + ^{n}C_{1}a^{n-1}b^{1}+ ^{n}C_{2}a^{n-2}b^{2}+…. ^{n}C_{n}a^{0}b^{n}

(2/x – x/2)^{5 } = ^{5}C_{0}(2/x)^{5}(x/2)^{0} – ^{5}C_{1}(2/x)^{4}(x/2)^{1}+^{5}C_{2}(2/x)^{3}(x/2)^{2}– ^{5}C_{3}(2/x)^{2}(x/2)^{3}+^{5}C_{4}(2/x)^{1}(x/2)^{4}–^{5}C_{5}(2/x)^{0}(x/2)^{5}

**Q3. Expand the expression**

**(2x – 3) ^{6}**

Ans. The expansion of the expression (a – b)^{n} is given as

(a – b)^{n} = ^{n}C_{0}a^{n}b^{0} + ^{n}C_{1}a^{n-1}b^{1}+ ^{n}C_{2}a^{n-2}b^{2}+…. ^{n}C_{n}a^{0}b^{n}

(2x – 3)^{6}= ^{6}C_{0}(2x)^{6}3^{0} – ^{6}C_{1}(2x)^{5}3^{1}+ ^{6}C_{2}(2x)^{4}3^{2}– ^{6}C_{3}(2x)^{3}3^{3 }+^{6}C_{4}a(2x)^{2}3^{4}–^{6}C_{5}(2x)3^{5}+^{6}C_{6}3^{6}

=64x^{6 }-576x^{5 }+2160x^{4 }-4320x^{3 }+4860x^{2}-2916x+729

**Solutions of class 11 NCERT maths exercise 8.1 of chapter 8 Binomial Theorem**

**Q4. Expand the expression**

**(x/3 -1/x) ^{5}**

Ans. The given expression is

(x/3 -1/x)^{5}

Applying binomial theorem

(a + b)^{n} = ^{n}C_{0}a^{n}b^{0} +^{n}C_{1}a^{n-1}b^{1}+ ^{n}C_{2}a^{n-2}b^{2}+……..^{n}C_{n}a^{0}b^{n}

(x/3 +1/x)^{5}= ^{5}C_{0}(x/3)^{5} + ^{5}C_{1}(x/3)^{4}(1/x)+ ^{5}C_{2}(x/3)^{3}(1/x)^{2} +^{5}C_{3}(x/3)^{2}(1/x)³+^{5}C_{4}(x/3)(1/x)^{4}+^{5}C_{5}(1/x)^{5}

= x^{5}/243 + 5(x^{4}/81)(1/x) +10(x³/27)(1/x²) +10(x²/9)(1/x³) +5(x/3)(1/x^{4})+1/x^{5}

= x^{5}/243 +5x³/81 + 10x/27 +10/9x + 5/3x³ +1/x^{5}

**Q5. Expand the expression**

**(x+ 1/x) ^{6}**

Ans. Applying the binomial theorem

(a + b)^{n} = ^{n}C_{0}a^{n}b^{0} +^{n}C_{1}a^{n-1}b^{1}+ ^{n}C_{2}a^{n-2}b^{2}+……..^{n}C_{n}a^{0}b^{n}

(x+ 1/x)^{6}= ^{6}C_{0 }x^{6} + ^{6}C_{1}x^{5}(1/x)+ ^{6}C_{2}x^{4}(1/x)^{2} +^{6}C_{3}x^{3}(1/x)³+^{6}C_{4}x²(1/x)^{4}+^{6}C_{5}x(1/x)^{5 }+ (1/x)^{6}

= _{ }x^{6} + 6x^{5}(1/x)+ 15x^{4}(1/x)^{2} +20x^{3}(1/x)³+15x²(1/x)^{4}+6x(1/x)^{5 }+ (1/x)^{6}

= x^{6} + 6x^{4}+ 15x^{2} +20+15/x^{2}+6/x^{4 }+ (1/x)^{6}

**Q6. Using binomial theorem evaluate (96)³**

Ans.Writting 96 in the following form

96 = 100 – 4

Applying the binomial theorem

(a – b)^{n} = ^{n}C_{0}a^{n}b^{0} –^{n}C_{1}a^{n-1}b^{1}+ ^{n}C_{2}a^{n-2}b^{2}-……..^{n}C_{n}a^{0}b^{n}

(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}4³

= 100³ -3(100)²4 +3(100)16 -64

= 1000000 – 120000 +4800 -64

= 884736

**Q7. Using binomial theorem evaluate (102) ^{5}**

Ans.Writting 102 in the following form

102 = 100 + 2

Applying the binomial theorem

(a +b)^{n} = ^{n}C_{0}a^{n}b^{0} +^{n}C_{1}a^{n-1}b^{1}+ ^{n}C_{2}a^{n-2}b^{2}+……..^{n}C_{n}a^{0}b^{n}

(102)^{5}=(100+2)^{5}= ^{5}C_{0 }(100)^{5} +^{5}C_{1}(100)^{4}2+ ^{5}C_{2}(100)³2² +^{5}C_{3}(100)²2³+ ^{5}C_{4}(100)2^{4}+^{5}C_{5}2^{5}

= (100)^{5} +5(100)^{4}2 +10(100)³2² + 10(100)²2³+5(100)2^{4}+2^{5}

= 10000000000+ 5(100000000)2 + 10(1000000)4+10(10000)8 +5(100)16 +32

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

=11040808032

**Solutions of class 11 NCERT maths exercise 8.1 of chapter 8 Binomial Theorem**

**Q8. Using binomial theorem evaluate (101) ^{5}**

Ans.Writting 101 in the following form

101 = 100 + 1

Applying the binomial theorem

(a +b)^{n} = ^{n}C_{0}a^{n}b^{0} +^{n}C_{1}a^{n-1}b^{1}+ ^{n}C_{2}a^{n-2}b^{2}+……..^{n}C_{n}a^{0}b^{n}

(101)^{4}=(100+1)^{4}= ^{4}C_{0 }(100)^{4} +^{4}C_{1}(100)^{3}1+ ^{4}C_{2}(100)²1² +^{4}C_{3}(100)1³+ ^{4}C_{4}1^{4}

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

=100000000 +4000000+60000+400+1

=104060401

**Q9. Using binomial theorem evaluate (99) ^{5}**

Ans.Writting 99 in the following form

99 = 100 – 1

Applying the binomial theorem

(a -b)^{n} = ^{n}C_{0}a^{n}b^{0} –^{n}C_{1}a^{n-1}b^{1}+ ^{n}C_{2}a^{n-2}b^{2}-……..^{n}C_{n}a^{0}b^{n}

(99)^{5}=(100-1)^{5}= ^{5}C_{0 }(100)^{5} –^{5}C_{1}(100)^{4}1+ ^{5}C_{2}(100)³1² –^{5}C_{3}(100)²1³+^{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

=9509900499

**Q10.Using binomial theorem indicate which number is larger (1.1) ^{10000}or 1000**

Ans. Using the binomial theorem, expanding the term (1.1)^{10000}

Writing 1.1 in the form of 1 +0.1

(1.1)^{10000} = (1 + 0.1)^{10000}

= ^{10000}C_{0 }1^{10000} +^{10000}C_{1}1^{999}0.1+……

= 1 + 10000×0.1 +….

= 1+1000+…… = 1001 +……>1000

Therefore (1.1)^{10000}> 1000

**Summary of the class 11 NCERT maths exercise 8.1 of chapter 8 Binomial Theorem**

Class 11 NCERT maths exercise 8.1 of chapter 8 Binomial Theorem is based on the expansion of two terma (a +b)^{n }which is expanded as following

(a +b)^{n} = ^{n}C_{0}a^{n}b^{0} +^{n}C_{1}a^{n-1}b^{1}+ ^{n}C_{2}a^{n-2}b^{2}+……..+^{n}C_{n}a^{0}b^{n}

Expansion of other important expressions

(a -b)^{n} = ^{n}C_{0}a^{n}b^{0} –^{n}C_{1}a^{n-1}b^{1}+ ^{n}C_{2}a^{n-2}b^{2}-……..^{n}C_{n}a^{0}b^{n}

(1+x)^{n }= 1 +^{n}C_{1}x+ ^{n}C_{2}x^{2}+^{n}C_{3}x^{3}……..+x^{n}

(1-x)^{n }= 1 –^{n}C_{1}x+ ^{n}C_{2}x^{2}–^{n}C_{3}x^{3}+……x^{n}

The general term of the expansion (a +b)^{n} is written as follows

T_{r+1 }= ^{n}C_{r}a^{n-r}b^{r}