Careers360 Logo
NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.1 - Binomial Theorem

NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.1 - Binomial Theorem

Edited By Komal Miglani | Updated on May 06, 2025 02:31 PM IST

Squares and cubes of binomials like a+b and ab can be easily evaluated, but what happens when there are higher powers of binomials, like (a+b)n where the value of n is 8, 9, 10, or even more? So, to overcome this difficulty Binomial Theorem introduces a general formula that makes expansion easier for terms involving higher powers of 'n'

This Story also Contains
  1. NCERT Solutions Class 11 Maths Chapter 7 Exercise 7.1
  2. Topics covered in Chapter 7 Binomial Theorem Exercise 7.1
  3. NCERT Solutions of Class 11 Subject Wise
  4. Subject-Wise NCERT Exemplar Solutions

In this exercise, Students are going to learn about the Binomial Theorem, Pascal's triangle, and binomial expansion for the positive integers. Solutions of NCERT are designed to provide detailed and step-by-step solutions to every question. Exercise 7.1 solutions are formulated by subject experts in a very clear and comprehensive manner, which helps students to understand concepts easily. Students can also check NCERT Solutions to get detailed solutions for Science and Maths from Class 6 to Class 12.

Background wave

NCERT Solutions Class 11 Maths Chapter 7 Exercise 7.1

Question 1: Expand the expression. (12x)5

Answer:

Given,

The Expression:

(12x)5

the expansion of this Expression is,

(12x)5=

5C0(1)55C1(1)4(2x)+5C2(1)3(2x)25C3(1)2(2x)3+5C4(1)1(2x)45C5(2x)5

15(2x)+10(4x2)10(8x3)+5(16x4)(32x5)

110x+40x280x3+80x432x5

Question 2: Expand the expression. (2xx2)5

Answer:

Given,

The Expression:

(2xx2)5

the expansion of this Expression is,

(2xx2)5

5C0(2x)55C1(2x)4(x2)+5C2(2x)3(x2)25C3(2x)2(x2)3+5C4(2x)1(x2)45C5(x2)5

32x5(16x4)(x2)+10(8x3)(x24)10(4x2)(x28)+5(2x)(x416)x532

32x540x3+20x5x+5x28x332

Question 3: Expand the expression. (2x3)6

Answer:

Given,

The Expression:

(2x3)6

the expansion of this Expression is,

(2x3)6=

6C0(2x)66C1(2x)5(3)+6C2(2x)4(3)26C3(2x)3(3)3+6C4(2x)2(3)46C5(2x)(3)5+6C6(3)6

64x66(32x5)(3)+15(16x4)(9)20(8x3)(27)+15(4x2)(81)6(2x)(243)+729

64x6576x5+2160x44320x3+4860x22916x+729

Question 4: Expand the expression. (x3+1x)5

Answer:

Given,

The Expression:

(x3+1x)5

the expansion of this Expression is,

(x3+1x)5

5C0(x3)5+5C1(x3)4(1x)+5C2(x3)3(1x)2+5C3(x3)2(1x)3+5C4(x3)1(1x)4+5C5(1x)5

x5243+5(x481)(1x)+10(x327)(1x2)+10(x29)(1x3)+5(x3)(1x4)+1x5

x5243+5x381+10x27+109x+53x2+1x5

Question 5: Expand the expression. (x+1x)6

Answer:

Given,

The Expression:

(x+1x)6

the expansion of this Expression is,

(x+1x)6

6C0(x)6+6C1(x)5(1x)+6C2(x)4(1x)2+6C3(x)3(1x)3+6C4(x)2(1x)4+6C5(x)(1x)5+6C6(1x)6

x6+6(x5)(1x)+15(x4)(1x2)+20(8x3)(1x3)+15(x2)(1x4)+6(x)(1x5)+1x6

x6+6x4+15x2+20+15x2+6x4+1x6

Question 6: Using binomial theorem, evaluate the following:(96)3

Answer:

As 96 can be written as (100-4);

(96)3=(1004)3=3C0(100)33C1(100)2(4)+3C2(100)(4)23C3(4)3

=(100)33(100)2(4)+3(100)(4)2(4)3

=1000000120000+480064

=884736

Question 7: Using binomial theorem, evaluate the following:(102)5

Answer:

As we can write 102 in the form 100+2

(102)5

=(100+2)5

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

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

=11040808032

Question 8: Using binomial theorem, evaluate the following:

(101)4

Answer:

As we can write 101 in the form 100+1

(101)4

=(100+1)4

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

=100000000+4000000+60000+400+1

=104060401

Question 9: Using binomial theorem, evaluate the following:(99)5

Answer:

As we can write 99 in the form 100-1

(99)5

=(1001)5

=5C0(100)55C1(100)4(1)+5C2(100)3(1)25C3(100)2(1)3+5C4(100)1(1)45C5(1)5

=10000000000500000000+10000000100000+5001

=9509900499

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

Answer:

AS we can write 1.1 as 1 + 0.1,

(1.1)10000=(1+0.1)10000

=10000C0+10000C1(1.1)+Otherpositiveterms

=1+10000×1.1+Otherpositiveterm

>1000

Hence,

(1.1)10000>1000

Question 11 :Find (a+b)4(ab)4 . Hence, evaluate(3+2)4(32)4 .

Answer:

Using Binomial Theorem, the expressions (a+b)4 and (ab)4 can be expressed as

(a+b)4=4C0a4+4C1a3b+4C2a2b2+4C3ab3+4C4b4

(ab)4=4C0a44C1a3b+4C2a2b24C3ab3+4C4b4

From Here,

(a+b)4(ab)4=4C0a4+4C1a3b+4C2a2b2+4C3ab3+4C4b44C0a4+4C1a3b4C2a2b2+4C3ab34C4b4

(a+b)4(ab)4=2×(4C1a3b+4C3ab3)

(a+b)4(ab)4=8ab(a2+b2)

Now, Using this, we get

(3+2)4(32)4=8(3)(2)(3+2)=8×6×5=406

Question 12 :Find (x+1)6+(x1)6 . Hence or otherwise evaluate (2+1)6+(21)6.

Answer:

Using Binomial Theorem, the expressions (x+1)4 and (x1)4 can be expressed as ,

(x+1)6=6C0x6+6C1x51+6C2x412+4C3x313+6C4x214+6C5x15+6C616

(x1)6=6C0x66C1x51+6C2x4124C3x313+6C4x2146C5x15+6C616

From Here,

(x+1)6(x1)6=6C0x6+6C1x51+6C2x412+4C3x313+6C4x214+6C5x15+6C616 +6C0x66C1x51+6C2x4124C3x313+6C4x2146C5x15+6C616

(x+1)6+(x1)6=2(6C0x6+6C2x412+6C4x214+6C616)

(x+1)6+(x1)6=2(x6+15x4+15x2+1)

Now, Using this, we get

(2+1)6+(21)6=2((2)6+15(2)4+15(2)2+1)

(2+1)6+(21)6=2(8+60+30+1)=2(99)=198

Question 13: Show that 9n+18n9is divisible by 64, whenever n is a positive integer.

Answer:

If we want to prove that 9n+18n9is divisible by 64, then we have to prove that 9n+18n9=64k

As we know, from binomial theorem,

(1+x)m=mC0+mC1x+mC2x2+mC3x3+....mCmxm

Here putting x = 8 and replacing m by n+1, we get,

9n+1=n+1C0+n+1C18+n+1C282+.......+n+1Cn+18n+1

9n+1=1+8(n+1)+82(n+1C2+n+1C38+n+1C482+.......+n+1Cn+18n1)

9n+1=1+8n+8+64(k)

Now, Using This,

9n+18n9=9+8n+64k98n=64k

Hence

9n+18n9 is divisible by 64.

Question 14: Prove that r=0n3r nCr=4n

Answer:

As we know from Binomial Theorem,

r=0nar nCr=(1+a)n

Here putting a = 3, we get,

r=0n3r nCr=(1+3)n

r=0n3r nCr=4n

Hence Proved.

Also read

Topics covered in Chapter 7 Binomial Theorem Exercise 7.1

The Binomial Theorem for positive integer powers is introduced in this exercise. The formula for the expansion of expressions of the form (x+a)n is discussed in this exercise.

1) Pascal's Triangle

Pascal's triangle is used to represent the Binomial coefficients. It is a triangular array of numbers, which represents each number is the sum of two numbers directly above it in the previous row.

pascal%20triangle%20

The nth  row of pascal triangle contains (x+a)n1 rows coefficients.

3) Binomial Theorem for any Positive Integer n

The expansion of (a+b)n is,

(a+b)n=nC0an+nC1an1b+nC2an2b2++nCn1abn1+nCnbn

Binomial Theorem can also be stated as (a+b)n=k=0nnCkankbk

Some special cases:

a) Taking a=x and b=y, we obtain

(xy)n=[x+(y)]n

Thus (xy)n=nC0xnnC1xn1y+nC2xn2y2++(1)nnCnyn

b) Taking a=1,b=x, we obtain

(1+x)n=nC0+nC1x+nC2x2+nC3x3++nCnxn

c) Taking a=1,b=x, we obtain

(1x)n=nC0nC1x+nC2x2+(1)nnCnxn

Also Read

NEET/JEE Offline Coaching
Get up to 90% Scholarship on your NEET/JEE preparation from India’s Leading Coaching Institutes like Aakash, ALLEN, Sri Chaitanya & Others.
Apply Now

NCERT Solutions of Class 11 Subject Wise

Students can refer subject-wise NCERT solutions. The links to solutions are given below

Subject-Wise NCERT Exemplar Solutions

Students can access the NCERT exemplar solutions to enhance their deep understanding of the topic. These solutions are aligned with the CBSE syllabus and also help in competitive exams.


Frequently Asked Questions (FAQs)

1. What is the weightage of co-ordinate geometry in the CBSE Class 11 Maths ?

Co-ordinate geometry has 10 marks weightage in the CBSE Class 11 Maths exam.

2. Where can I get NCERT exemplar solutions for Class 11 free?
3. Are these NCERT exercise solutions good for revision ?

Yes, NCERT exercise solutions are very helpful for the revision of the important concepts before the exam.

4. Who prepared the NCERT exercise wise solutions ?

NCERT exercise-wise solutions are prepared by subject matter experts who are experienced in this field.

5. What are the important topics covered in the Class 11 Maths exercise 7.1 ?

Binomial expansion for positive integers and pascal triangle are covered in the Class 11 Maths exercise 7.1.

6. Does NCERT solutions for Class 11 Maths contains diagrams ?

Yes, NCERT solutions for Class 11 Maths contain the diagrams and charts where ever it is necessary.

Articles

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top