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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 Apr 09, 2025 11:47 PM IST

Students found it relatively easy when they were asked to find the squares or cubes of numbers up to 50. For example, when they have to find (78)6 or (127)5, they face difficulty due to repeated multiplications. This is where the Binomial theorem makes our life easier. In the maths chapter 7 class 11, students will study how to expand (x + y)n or (x - y)n in an easier way, where n is an integer or a rational number. Binomial theorem class 11 notes also discuss how we get Pascal’s triangle from the expansion of (x + y)n.

This Story also Contains
  1. NCERT Class 11 Maths Chapter 7 Notes
  2. Importance of NCERT Class 11 Math Chapter 7 Notes
  3. NCERT Class 11 Notes Chapter Wise
  4. Subject Wise NCERT Exemplar Solutions
  5. Subject Wise NCERT Solutions
  6. NCERT Books and Syllabus

These class 11 maths chapter 7 NCERT notes contain many visual graphics to make learning easier for students. Also, the latest CBSE guidelines have been followed by Careers360 experts when they made these notes. Class 11 Math chapter 7 notes help students to find the entire chapter in one place during the revision after completing the textbook exercises. Students can also check the NCERT Exemplar Class 11 Maths Chapter 7 Binomial Theorem for extra practice purposes.

NCERT Class 11 Maths Chapter 7 Notes

We know,
(a+b)0=1(a+b)1=a+b(a+b)2=a2+2ab+b2(a+b)3=a3+3a2b+3ab2+b3

If we observe each expansion,

  • The total number of terms of each expansion is 1 more than the index of (a+b).

  • 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 increasing by 1.

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

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

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 (a+b)5 will be 1 5 10 10 5 1.

Similarly, the coefficient of (a+b)6 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) nCr=n!(nr)!r!,
where n and r are positive integers and 0r<n.

Figure 2 can be rewritten as:

Index Coefficients

00C0(=1)11C0(=1)1C1(=1)22C0(=1)2C1(=2)2C2(=1)33C0(=1)3C1(=3)3C3(=3)3C3(=1)44C0(=1)4C1(=4)4C2(=6)4C3(=4)4C4(=1)

Binomial Theorem:

Binomial theorem for any positive integer n,
(a+b)n=nC0an+nC1an1b+nC2an2b2++nCn1abn1+nCnbn

Proof:
The proof is obtained by applying the principle of mathematical induction.
Let the given statement be

P(n):(a+b)n=nC0an+nC1an1b+nC2an2b2++nCn1abn1+nCnbn
For n=1, we have

P(1):(a+b)1=1C0a1+1C1b1=a+b
Thus, P(1) is true.
Suppose P(k) is true for some positive integer k, i.e.

(a+b)k=kC0ak+kC1ak1b+kC2ak2b2++kCkbk(1)

We shall prove that P(k+1) is also true, i.e.,

(a+b)k+1=k+1C0ak+1+k+1C1akb+k+1C2ak1b2++k+1Ck+1bk+1

Now, (a+b)k+1=(a+b)(a+b)k

=(a+b)(kC0ak+kC1ak1b+kC2ak2b2++kCk1abk1+kCkbk) from (1)] =kC0ak+1+kC1akb+kC2ak1b2++kCk1a2bk1+kCkabk+kC0akb+kC1ak1b2+kC2ak2b3++kCk1abk+kCkbk+1
[by actual multiplication]

=kC0ak+1+(kC1+kC0)akb+(kC2+kC1)ak1b2++(kCk+kCk1)abk+kCkbk+1[By grouping like terms]=k+1C0aak+1+k+1C1akb+k+1C2ak1 b2++k+1Ckabk+k+1Ck+1bk+1

(by using k+1C0=1,kCr+kCr1=k+1Cr and kCk=1=k+1Ck+1 )
Thus, it has been proved that P(k+1) is true whenever P(k) is true. Therefore, by the principle of mathematical induction, P(n) is true for every positive integer n.

Properties of the Binomial coefficient

1. n!=n(n1)(n2)(n3).4×3×2×1
2. nCr=nCnr
nCr=n!(nr)!r!
and
nCnr=n!(nr)!(nn+r)!
nCnr=n!(nr)!(r)!
Hence, it is proven that both are equal.
3. If nCx=nCy only if x=y and x+y=n
4. nCr+nCr1=n+1Cr
5. If n is a positive integer and x,y are two complex numbers, then (x+y)n=nC0xn+nC1xn1y.nCn1xyn1+nCnyn

Here binomial coefficients are: nC0,nC1,..nCn1,nCn

6. Total no. of terms of the given as (n+1) in the expansion
(x+y)n=nC0xn+nC1xn1y.nCn1xyn1+nCnyn

To find the middle terms of using the binomial theorem

The general equation of the binomial is given as:

(x+y)n=nC0xn+nC1xn1y.nCn1xyn1+nCnyn(x+y)n=T1+T2+T3.Tn+Tn+1

There are two cases.
1. If n is odd.

The number of terms of (a+b)n when n is an odd number is n+1.
Here, n+1 is an even number.
So there will be two middle terms.
That are (n+12)th  and (n+12+1)th terms of the expansion.

2. If n is even

The number of terms of (a+b)n when n is an even number is n+1.
Here, n+1 is an odd number.
So there will be one middle term.
That is (n+1+12)th=(n2+1)th term of the expansion.

To find the sum of the coefficient of the binomial terms, we have to put the value of x numerically as one.

For example
Q. If the binomial expression is (xy+1)4, then find the sum of the binomial coefficients is?

Solution: We have to put the value of x=y=1 and get the coefficient sum.
(xy+1)4
=(11+1)4=1
So the Sum of the binomial coefficients is 1.

Some Properties of the Binomial coefficients

(x+y)n=nC0xn+nC1xn1y..nCn1xyn1+nCnyn
Put x=y=1
(1+1)n=nC011n+nC11n11.nCn11.1n1+nCn1n(2)n=nC0+nC1.nCn1+nCn

(x+y)n=nC0xxn+nC1xxn1y.nCn1xyn1+nCnyn
Put x=y=1
(11)n=nC01n+nC11n1(1)..nCn11.(1)n1+nCn(1)n0=nC0nC1..nCn1(1)n+nCn(1)n

Some Particular expansions:

(1+x)1=1x+x2x3..(1x)1=1+x+x2+x3...(1+x)2=12x+3x24x3.....(1x)2=1+2x+3x2+4x3.....

Importance of NCERT Class 11 Math Chapter 7 Notes

NCERT Class 12 Maths chapter 7 notes are useful in many ways.

  • These notes will give conceptual clarity about the Binomial theorem and build a strong base for the Algebra, Calculus, and Probability chapter, which is not only important for board exams in 11 and 12, but also for other competitive exams like JEE main, NDA, CUET, etc.
  • These notes will teach notes how to use the Binomial theorem efficiently to save time by not doing repeated multiplication.
  • Students can use these notes for a quick revision whenever they need.
  • These notes contain PDF links to other subjects' solutions.
  • The latest CBSE 2025-26 guidelines have been followed in these notes.
  • Experts who have multiple years of experience in these topics have created these notes so that students can learn these concepts easily.
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Subject Wise NCERT Exemplar Solutions

NCERT exemplar is one of the best tools for revision and practice purposes. After completing the exercises, they can compare their solutions with the solved ones.

Subject Wise NCERT Solutions

Students can use the following links to analyze the NCERT textbook solutions for other subjects.

NCERT Books and Syllabus

Students should always have access to the latest CBSE syllabus, and the following link will give them that privilege.

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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

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

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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 :

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2.45×10−3 kg

Option 2)

 6.45×10−3 kg

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 9.89×10−3 kg

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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

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

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

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

Option 4)

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

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K/2\,

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\; K\;

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zero\;

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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)

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

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Option 1)

0.02

Option 2)

3.125 × 10-2

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1.25 × 10-2

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2.5 × 10-2

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decrease twice

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increase two fold

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remain unchanged

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be a function of the molecular mass of the substance.

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twice that in 60 g carbon

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6.023 × 1022

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less than 3

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