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Binomial Theorem Class 11th Notes - Free NCERT Class 11 Maths Chapter 8 notes - Download PDF

Binomial Theorem Class 11th Notes - Free NCERT Class 11 Maths Chapter 8 notes - Download PDF

Edited By Ramraj Saini | Updated on Mar 22, 2022 04:54 PM IST

Binomial Theorem belongs to the 8 chapter of NCERT. The NCERT Class 11 Maths chapter 8 notes entirely cover up the main portions of the chapter Binomial Theorem. In the introduction Binomial Theorem Class 11 notes we will learn about how we can find an expanded form of an expression with an index. Binomial Theorem Class 11 notes describe how we get pascal’s triangle from the expansion of 1647326843269where n=1, 2, 3. Class 11 Math chapter 8 notes cover the main topics that are a number of terms of an expansion, how to use combination formula to the expanded form, the middle term of 1647326842959when n is an even or odd.

A Class 11 Math chapter 6 notes help you to find the entire chapter in an easy way. NCERT Notes for Class 11 Maths chapter 6 not only covers the NCERT notes but covers CBSE Class 11 Maths chapter 8 notes also.

After going through Class 11 Binomial Theorem notes

Students can also refer to,

NCERT Class 11 Maths Chapter 8 Notes

We know,

1647326928186

1647326929006

1647326928436

1647326934046

If we observe each expansion,

  • The total number of terms of each expansion is 1 more than the index of1647326928742.

  • In the expansion, the power of the first quantity is gradually decreasing that is by 1 and the power of the second quantity is gradually increased that is by 1.

  • Suppose n is the index of 1647326929313, then the sum of indices of a and b of each term of the expansion is n.

Now arranging of the coefficients of expansion with respect to their index

Index Coefficients

\\ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \\ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ 1 \\ 2 \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ 2 \ \ \ \ \ 1 \\ 3 \ \ \ \ \ \ \ 1 \ \ \ \ \ 3 \ \ \ \ \ \ 3 \ \ \ \ \ \ 1 \\ 4 \ \ \ \ 1 \ \ \ \ \ 4 \ \ \ \ \ 6 \ \ \ \ \ \ 4 \ \ \ \ \ \ 1

If we observe the pattern we can get the coefficient of the next index.

The pattern is given below

1647326929488

Figure: 1

This diagram is known as Pascal’s triangle.

If we apply Pascal’s triangle rule, then the coefficients of 1647326929797 will be

1 5 10 10 5 1

Similarly, the coefficient of 1647326927280 will be

1 6 15 20 15 6 1

Now we can apply the combination formula to find the coefficients.

Binomial coefficient (for a Positive integral index n )1647326932832 where n and r are a positive integer and 1647326933465.

Figure 2 can be rewritten as

Index Coefficients

\\ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^0C_0(=1) \\ \\ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^1C_0(=1) \ \ \ \ \ \ \ \ \ \ \ \ \ ^1C_1(=1) \\ \\ 2 \ \ \ \ \ \ \ \ \ \ \ \ \ ^2C_0(=1) \ \ \ \ \ \ \ \ ^2C_1(=2) \ \ \ \ \ \ \ \ \ ^2C_2(=1) \\ \\ 3 \ \ \ \ \ \ \ \ \ ^3C_0(=1) \ \ \ \ ^3C_1(=3) \ \ \ \ \ \ \ ^3C_3(=3) \ \ \ \ \ \ ^3C_3(=1) \\ \\ 4 \ \ \ \ ^4C_0(=1) \ \ \ \ \ ^4C_1(=4) \ \ \ \ \ \ \ ^4C_2(=6) \ \ \ \ ^4C_3(=4) \ \ \ \ \ ^4C_4(=1)

Binomial Theorem:

Binomial theorem for any positive integer n

1647327126638

Proof:

We will proof the theorem by using principal induction.

Let 1647327125781

Putting 1647327125212

1647327128069

1647327129398

For 1647327126106 it is true.

Assume that it is true for 1647327128628

1647327127505

We will prove that 1647327126927i.e1647327127322is true by using1647327128931.

Now, 1647327127163

1647327129609

1647327129141

Add like terms

1647327132903

Now apply the formula1647327132037,1647327131568, and 1647327130202

1647327129774

Hence proved.

Properties of Binomial coefficient

1. 1647327129960

2. 1647327132323

1647327127804

and

1647327131011

hence prove both are equal.

3. If 1647327126399 only if 1647327130701and 1647327132595

4. 1647327131282

5. If n is a positive integer and x, y are two complex numbers, then 1647327130389

Here are binomial coefficients. 1647327131790

6. Total no. of terms of the given as (n + 1) in the expansion

1647327128279

To find the middle terms of using the binomial theorem

The general equation of the binomial is given as

1647327163900

1647327165162

There are two case

  1. If n is odd

The number of terms of 1647327164516 when 1647327162225 is an odd number is1647327161267. Here 1647327161597 is an even number. So there will be two middle terms. That are 1647327163691 and 1647327163356of the expansion.

  1. If n is even

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The number of terms of 1647327164246 when 1647327161926 is an eve number is1647327160549. Here 1647327160926 is an odd number. So there will be one middle term. That is 1647327166403 of the expansion.

To find the sum of the coefficient of the binomial terms we have to put the value of x numerical is one.

For example

Q if the binomial express is 1647327166887 then find the sum of the binomial coefficient is?

Solution:- Write the given expression 1647327168294 now we have to put the value of 1647327167491and get The coefficient sum =1647327167757

= 1647327167209

= 1

So Sum of the binomial coefficient is 1.

Some Properties of the Binomial coefficients

1647327166081

Put 1647327168026

1647327165413

1647327165885

Put1647327168598

1647327165660

Some Particular expansions

1647327164841

Significance of NCERT Class 11 Math Chapter 8 Notes

Class 11 Binomial Theorem notes will be really helpful to revise the chapter and get a brief overview of the important topics. Also, notes for Class 11 Maths chapter 8 is useful for covering Class 11 CBSE Maths Syllabus and also for competitive exams like BITSAT, and JEE MAINS. Class 11 Maths chapter 8 notes pdf download can be used for preparing in offline mode.

Binomial Theorem Class 11 notes pdf download: link

NCERT Class 11 Notes Chapter Wise.

Subject Wise NCERT Exemplar Solutions

Subject Wise NCERT Solutions

NCERT Books and Syllabus

Frequently Asked Question (FAQs)

1. According to Class 11 Maths chapter 8 notes, define Binomial Theorem?

Binomial Theorem is an expression to find the what is next terms of the higher-order polynomial equation.

2. What is the real-life use of the Binomial Theorem according to NCERT notes for Class 11 Maths chapter 8?

It is the most often used in solving many real-life problems to find the reminder of the polynomial. 

3. What textbook should be followed?

Class 11 Math chapter 8textbook should be followed.

4. From where we can download this note?

The notes can be downloaded from the Binomial Theorem Class 11 notes pdf download.

5. How Binomial Theorem Class 11 notes is helpful?

Class 11 Binomial Theorem Notes help students to revise their important topics and score good marks in exams.

Articles

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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

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0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

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12.89×10−3 kg

 

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

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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\;

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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 .

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Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

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Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

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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.

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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

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