# NCERT Solutions for Class 11 Maths Chapter 8 - Binomial Theorem

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 7.

The NCERT Solutions Class 11 Chapter 8 Binomial Theorem can be downloaded at BYJUâ€™S easily. Practising these solutions can help the students clear their doubts as well as solve problems faster. Students can learn new tips and methods to answer a particular question in different ways using NCERT Solutions, which gives them an edge in exam preparation.

The concepts covered in Chapter 8 of the Maths textbook include the study of essential topics such as Positive Integral Indices, Pascalâ€™s Triangle, the Binomial Theorem for any positive integer and some special cases. Students can score high marks in the exams easily by practising the NCERT Class 11 Solutions for all the questions in the textbook. Each solution is solved step-by-step, considering the understanding level of the students. Therefore, it is important to understand the logic set behind each answer and develop a better comprehension of the concepts.

## NCERT Solutions for Class 11 Maths Chapter 8 â€“ Binomial Theorem

### Access Answers to NCERT Class 11 Maths Chapter 8

Exercise 8.1 Page No: 166

Expand each of the expressions in Exercises 1 to 5.

1. (1 â€“ 2x)5

Solution:

From binomial theorem expansion, we can write as

(1 â€“ 2x)5

= 5Co (1)5 â€“ 5C1 (1)4 (2x) + 5C2 (1)3 (2x)2 â€“ 5C3 (1)2 (2x)3 + 5C4 (1)1 (2x)4 â€“ 5C5 (2x)5

= 1 â€“ 5 (2x) + 10 (4x)2 â€“ 10 (8x3) + 5 ( 16 x4) â€“ (32 x5)

= 1 â€“ 10x + 40x2 â€“ 80x3Â + 80x4â€“ 32x5

Solution:

From the binomial theorem, the given equation can be expanded as

3. (2x â€“ 3)6

Solution:

From the binomial theorem, the given equation can be expanded as

Solution:

From the binomial theorem, the given equation can be expanded as

Solution:

From the binomial theorem, the given equation can be expanded as

6. Using the binomial theorem, find (96)3.Â Â

Solution:

Given (96)3

96 can be expressed as the sum or difference of two numbers, and then the binomial theorem can be applied.

The given question can be written as 96 = 100 â€“ 4

(96)3 = (100 â€“ 4)3

= 3C0 (100)3 â€“ 3C1 (100)2 (4) â€“ 3C2 (100) (4)2â€“ 3C3 (4)3

= (100)3 â€“ 3 (100)2 (4) + 3 (100) (4)2 â€“ (4)3

= 1000000 â€“ 120000 + 4800 â€“ 64

= 884736

7. Using the binomial theorem, find (102)5.

Solution:

Given (102)5

102 can be expressed as the sum or difference of two numbers, and then the binomial theorem can be applied.

The given question can be written as 102 = 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) (2)4 + 5C5 (2)5

= (100)5 + 5 (100)4 (2) + 10 (100)3 (2)2 + 5 (100) (2)3 + 5 (100) (2)4 + (2)5

= 1000000000 + 1000000000 + 40000000 + 80000 + 8000 + 32

= 11040808032

8. Using the binomial theorem, find (101)4.

Solution:

Given (101)4

101 can be expressed as the sum or difference of two numbers, and then the binomial theorem can be applied.

The given question can be written as 101 = 100 + 1

(101)4 = (100 + 1)4

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

= (100)4 + 4 (100)3 + 6 (100)2 + 4 (100) + (1)4

= 100000000 + 4000000 + 60000 + 400 + 1

= 104060401

9. Using the binomial theorem, find (99)5m.

Solution:

Given (99)5

99 can be written as the sum or difference of two numbers then the binomial theorem can be applied.

The given question can be written as 99 = 100 -1

(99)5 = (100 â€“ 1)5

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

= (100)5 â€“ 5 (100)4 + 10 (100)3 â€“ 10 (100)2 + 5 (100) â€“ 1

= 1000000000 â€“ 5000000000 + 10000000 â€“ 100000 + 500 â€“ 1

= 9509900499

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

Solution:

By splitting the given 1.1 and then applying the binomial theorem, the first few terms of (1.1)10000 can be obtained as

(1.1)10000 = (1 + 0.1)10000

= (1 + 0.1)10000 C1 (1.1) + other positive terms

= 1 + 10000 Ã— 1.1 + other positive terms

= 1 + 11000 + other positive terms

> 1000

(1.1)10000 > 1000

11. Find (a + b)4Â â€“ (a â€“ b)4. Hence, evaluateÂ

Solution:

Using the binomial theorem, the expression (a + b)4 and (a â€“ b)4Â can be expanded

(a + b)4 = 4C0 a4 + 4C1 a3 b + 4C2 a2 b2 + 4C3 a b3 + 4C4 b4

(a â€“ b)4 = 4C0 a4 â€“ 4C1 a3 b + 4C2 a2 b2 â€“ 4C3 a b3 + 4C4 b4

Now (a + b)4 â€“ (a â€“ b)4 = 4C0 a4 + 4C1 a3 b + 4C2 a2 b2 + 4C3 a b3 + 4C4 b4 â€“ [4C0 a4 â€“ 4C1 a3 b + 4C2 a2 b2 â€“ 4C3 a b3 + 4C4 b4]

= 2 (4C1 a3 b + 4C3 a b3)

= 2 (4a3 b + 4ab3)

= 8ab (a2 + b2)

Now by substituting a = âˆš3 and b = âˆš2, we get

(âˆš3 + âˆš2)4 â€“ (âˆš3 â€“ âˆš2)4 = 8 (âˆš3) (âˆš2) {(âˆš3)2 + (âˆš2)2}

= 8 (âˆš6) (3 + 2)

= 40 âˆš6

12. Find (x + 1)6Â + (x â€“ 1)6. Hence or otherwise evaluateÂ

Â

Solution:

Using binomial theorem, the expressions (x + 1)6 and (x â€“ 1)6 can be expressed as

(x + 1)6 = 6C0 x6 + 6C1 x5 + 6C2 x4 + 6C3 x3 + 6C4 x2 + 6C5 x + 6C6

(x â€“ 1)6 = 6C0 x6 â€“ 6C1 x5 + 6C2 x4 â€“ 6C3 x3 + 6C4 x2 â€“ 6C5 x + 6C6

Now, (x + 1)6 â€“ (x â€“ 1)6 = 6C0 x6 + 6C1 x5 + 6C2 x4 + 6C3 x3 + 6C4 x2 + 6C5 x + 6C6 â€“ [6C0 x6 â€“ 6C1 x5 + 6C2 x4 â€“ 6C3 x3 + 6C4 x2 â€“ 6C5 x + 6C6]

= 2 [6C0 x6 + 6C2 x4 + 6C4 x2 + 6C6]

= 2 [x6 + 15x4 + 15x2 + 1]

Now by substituting x = âˆš2, we get

(âˆš2 + 1)6 â€“ (âˆš2 â€“ 1)6 = 2 [(âˆš2)6 + 15(âˆš2)4 + 15(âˆš2)2 + 1]

= 2 (8 + 15 Ã— 4 + 15 Ã— 2 + 1)

= 2 (8 + 60 + 30 + 1)

= 2 (99)

= 198

13. Show that 9n+1 â€“ 8n â€“ 9 is divisible by 64 whenever n is a positive integer.

Solution:

In order to show that 9n+1 â€“ 8n â€“ 9 is divisible by 64, it has to be shown that 9n+1 â€“ 8n â€“ 9 = 64 k, where k is some natural number.

Using the binomial theorem,

(1 + a)m = mC0 + mC1 a + mC2 a2 + â€¦. + m C m am

For a = 8 and m = n + 1 we get

(1 + 8)n+1 = n+1C0 + n+1C1 (8) + n+1C2 (8)2 + â€¦. + n+1 C n+1 (8)n+1

9n+1 = 1 + (n + 1) 8 + 82 [n+1C2 + n+1C3 (8) + â€¦. + n+1 C n+1 (8)n-1]

9n+1 = 9 + 8n + 64 [n+1C2 + n+1C3 (8) + â€¦. + n+1 C n+1 (8)n-1]

9n+1 â€“ 8n â€“ 9 = 64 k

Where k = [n+1C2 + n+1C3 (8) + â€¦. + n+1 C n+1 (8)n-1] is a natural number

Thus, 9n+1 â€“ 8n â€“ 9 is divisible by 64 whenever n is a positive integer.

Hence proved.

14. Prove thatÂ

Solution:

Exercise 8.2 Page No: 171

Find the coefficient of

1. x5 in (x + 3)8

Solution:

The general term Tr+1Â in the binomial expansion is given by Tr+1Â =Â n C rÂ an-rÂ br

Here x5Â is the Tr+1Â term so a= x, b = 3 and n =8

Tr+1Â =Â 8CrÂ x8-rÂ 3râ€¦â€¦â€¦â€¦â€¦ (i)

To find out x5

We have to equate x5= x8-r

â‡’Â r= 3

Putting the value of r in (I), we get

= 1512 x5

Hence the coefficient of x5= 1512.

2. a5b7Â in (a â€“ 2b)12

Solution:

The general term Tr+1Â in the binomial expansion is given by Tr+1Â =Â n C rÂ an-rÂ br

Here a = a, b = -2b & n =12

Substituting the values, we get

Tr+1Â =Â 12CrÂ a12-rÂ (-2b)râ€¦â€¦â€¦. (i)

To find a5

We equate a12-rÂ =a5

r = 7

Putting r = 7 in (i)

T8Â =Â 12C7Â a5Â (-2b)7

= -101376 a5Â b7

Hence, the coefficient of a5b7= -101376.

Write the general term in the expansion of

3. (x2Â â€“ y)6

Solution:

The general term Tr+1Â in the binomial expansion is given by

Tr+1Â =Â n C rÂ an-rÂ brâ€¦â€¦.. (i)

Here, a = x2 , n = 6 and b = -y

Putting values in (i)

Tr+1Â =Â 6CrÂ xÂ 2(6-r)Â (-1)rÂ yr

= -1r 6cr .x12 â€“ 2r. yr

4. (x2Â â€“ y x)12, x â‰  0

Solution:

The general term Tr+1Â in the binomial expansion is given by Tr+1Â =Â n C rÂ an-rÂ br

Here n = 12, a= x2Â and b = -y x

Substituting the values, we get

Tn+1Â =12CrÂ Ã— x2(12-r)Â (-1)rÂ yrÂ xr

= -1r 12cr .x24 â€“2r. yr

5. Find the 4th term in the expansion of (x â€“ 2y)12.

Solution:

The general term Tr+1Â in the binomial expansion is given by Tr+1Â =Â n C rÂ an-rÂ br

Here, a= x, n =12, r= 3 and b = -2y

By substituting the values, we get

T4Â =Â 12C3Â x9Â (-2y)3

= -1760 x9Â y3

6. Find the 13thÂ term in the expansion ofÂ

Solution:

Find the middle terms in the expansions of

Solution:

Solution:

9. In the expansion of (1 + a)m+n, prove that coefficients of amÂ and anÂ are equal.

Solution:

We know that the general term Tr+1Â in the binomial expansion is given by Tr+1Â =Â nCrÂ an-rÂ br

Here n= m+n, a = 1 and b= a

Substituting the values in the general form

Tr+1Â =Â m+n CrÂ 1m+n-rÂ ar

=Â m+n CrÂ arâ€¦â€¦â€¦â€¦. (i)

Now, we have that the general term for the expression is,

Tr+1Â =Â Â m+n CrÂ ar

Now, for coefficient of am

Tm+1Â =Â Â m+n CmÂ am

Hence, for the coefficient of am, the value of r = m

So, the coefficient isÂ m+n C m

Similarly, the coefficient of anÂ isÂ m+n C n

10. The coefficients of the (r â€“ 1)th, rthÂ and (r + 1)thÂ terms in the expansion of (x + 1)n are in the ratio 1:3:5. Find n and r.

Solution:

The general term Tr+1Â in the binomial expansion is given by Tr+1Â =Â nCrÂ an-rÂ br

Here, the binomial is (1+x)nÂ with a = 1 , b = x and n = n

The (r+1)thÂ term is given by

T(r+1)Â =Â nCrÂ 1n-rÂ xr

T(r+1)Â =Â nCrÂ xr

The coefficient of (r+1)thÂ term isÂ nCr

The rthÂ term is given by (r-1)thÂ term

T(r+1-1)Â =Â nCr-1Â xr-1

TrÂ =Â nCr-1Â xr-1

âˆ´Â the coefficient of rthÂ term isÂ nCr-1

For (r-1)th term, we will take (r-2)thÂ term

Tr-2+1Â =Â nCr-2Â xr-2

Tr-1Â =Â nCr-2Â xr-2

âˆ´Â the coefficient of (r-1)thÂ term isÂ nCr-2

Given that the coefficient of (r-1)th, rthÂ and r+1thÂ term are in ratio 1:3:5

Therefore,

â‡’Â 5r = 3n â€“ 3r + 3

â‡’Â 8rÂ â€“Â 3n â€“ 3 =0â€¦â€¦â€¦â€¦.2

We have 1 and 2 as

n â€“ 4rÂ Â±Â 5 =0â€¦â€¦â€¦â€¦1

8r â€“ 3n â€“ 3 =0â€¦â€¦â€¦â€¦â€¦.2

Multiplying equation 1 by number 2

2n -8r +10 =0â€¦â€¦â€¦â€¦â€¦â€¦.3

2n -8r +10 =0

-3n â€“ 8r â€“ 3 =0

â‡’Â -n = -7

n =7Â and r = 3

11. Prove that the coefficient of xnÂ in the expansion of (1 + x)2nÂ is twice the coefficient of xnÂ in the expansion of (1 + x)2n â€“ 1.

Solution:

The general term Tr+1Â in the binomial expansion is given by Tr+1Â =Â nCrÂ an-rÂ br

The general term for binomial (1+x)2nÂ is

Tr+1Â =Â 2nCrÂ xrÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦..1

To find the coefficient of xn

r = n

Tn+1Â =Â 2nCnÂ xn

The coefficient of xnÂ =Â 2nCn

The general term for binomial (1+x)2n-1Â is

Tr+1Â =Â 2n-1CrÂ xr

To find the coefficient of xn

Putting n = r

Tr+1Â =Â 2n-1CrÂ xn

The coefficient of xnÂ =Â 2n-1Cn

We have to prove

Coefficient of xnÂ in (1+x)2nÂ = 2 coefficient of xnÂ in (1+x)2n-1

Consider LHS = 2nCn

12. Find a positive value of m for which the coefficient of x2Â in the expansion (1 + x)mÂ is 6.

Solution:

The general term Tr+1Â in the binomial expansion is given by Tr+1Â =Â nCrÂ an-rÂ br

Here, a = 1, b = x and n = m

Putting the value

Tr+1Â =Â m CrÂ 1m-rÂ xr

=Â m CrÂ xr

We need the coefficient of x2

âˆ´Â putting r = 2

T2+1Â =Â mC2Â x2

The coefficient of x2Â =Â mC2

Given thatÂ coefficient of x2Â =Â mC2Â = 6

â‡’Â m (m â€“ 1) = 12

â‡’Â m2â€“ m â€“ 12 =0

â‡’Â m2â€“ 4m + 3m â€“ 12 =0

â‡’Â m (m â€“ 4) + 3 (m â€“ 4) = 0

â‡’Â (m+3) (m â€“ 4) = 0

â‡’Â m = â€“ 3, 4

We need the positive value of m, so m = 4

Miscellaneous Exercise Page No: 175

1. Find a, b and n in the expansion of (a + b)nÂ if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Solution:

We know that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by

Tr+1 = nCr an-t br

The first three terms of the expansion are given as 729, 7290 and 30375, respectively. Then we have,

T1 = nC0 an-0 b0 = an = 729â€¦.. 1

T2 = nC1 an-1 b1 = nan-1 b = 7290â€¦. 2

T3 = nC2 an-2 b2 = {n (n -1)/2 }an-2 b2 = 30375â€¦â€¦3

Dividing 2 by 1, we get

$$\begin{array}{l}\frac{na^{n-1}b}{a^{n}} = \frac{7290}{729} \\ \frac{nb}{a} = 10\end{array}$$

Dividing 3 by 2, we get

$$\begin{array}{l}\frac{n(n-1)a^{n-2}b^{2}}{2na^{n-1}b} = \frac{30375}{7290} \\ \frac{(n-1)b}{2a} = \frac{30375}{7290} \\ \frac{(n-1)b}{a} = \frac{30375}{7290} \times 2 =\frac{25}{3} \\ \frac{(nb)}{a} -\frac{b}{a} = \frac{25}{3} \\ 10 -\frac{b}{a} = \frac{25}{3} \\ \frac{b}{a} = 10 â€“ \frac{25}{3} = \frac{5}{3}\end{array}$$

From 4 and 5, we have

n. 5/3 = 10

n = 6

Substituting n = 6 in 1, we get

a6 = 729

a = 3

From 5, we have, b/3 = 5/3

b = 5

Thus a = 3, b = 5 and n = 76

2. Find a if the coefficients of x2Â and x3Â in the expansion of (3 + a x)9Â are equal.

Solution:

3. Find the coefficient of x5Â in the product (1 + 2x)6Â (1 â€“ x)7Â using binomial theorem.

Solution:

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

= 1 + 6 (2x) + 15 (2x)2 + 20 (2x)3 + 15 (2x)4 + 6 (2x)5 + (2x)6

= 1 + 12 x + 60x2 + 160 x3 + 240 x4 + 192 x5 + 64x6

(1 â€“ x)7 = 7C0 â€“ 7C1 (x) + 7C2 (x)2 â€“ 7C3 (x)3 + 7C4 (x)4 â€“ 7C5 (x)5 + 7C6 (x)6 â€“ 7C7 (x)7

= 1 â€“ 7x + 21x2 â€“ 35x3 + 35x4 â€“ 21x5 + 7x6 â€“ x7

(1 + 2x)6 (1 â€“ x)7 = (1 + 12 x + 60x2 + 160 x3 + 240 x4 + 192 x5 + 64x6) (1 â€“ 7x + 21x2 â€“ 35x3 + 35x4 â€“ 21x5 + 7x6 â€“ x7)

192 â€“ 21 = 171

Thus, the coefficient of x5Â in the expression (1+2x)6(1-x)7 is 171.

4. If a and b are distinct integers, prove that a â€“ b is a factor of anÂ â€“ bn, whenever n is a positive integer. [Hint write anÂ = (a â€“ b + b)nÂ and expand]

Solution:

In order to prove that (a â€“ b) is a factor of (an â€“ bn), it has to be proved that

an â€“ bn = k (a â€“ b) where k is some natural number.

a can be written as a = a â€“ b + b

an = (a â€“ b + b)n = [(a â€“ b) + b]n

= nC0 (a â€“ b)n + nC1 (a â€“ b)n-1 b + â€¦â€¦ + n C n bn

an â€“ bn = (a â€“ b) [(a â€“b)n-1 + nC1 (a â€“ b)n-1 b + â€¦â€¦ + n C n bn]

an â€“ bn = (a â€“ b) k

Where k = [(a â€“b)n-1 + nC1 (a â€“ b)n-1 b + â€¦â€¦ + n C n bn] is a natural number

This shows that (a â€“ b) is a factor of (an â€“ bn), where n is a positive integer.

5. EvaluateÂ

Solution:

Using the binomial theorem, the expression (a + b)6 and (a â€“ b)6Â can be expanded

(a + b)6 = 6C0 a6 + 6C1 a5 b + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 a b5 + 6C6 b6

(a â€“ b)6 = 6C0 a6 â€“ 6C1 a5 b + 6C2 a4 b2 â€“ 6C3 a3 b3 + 6C4 a2 b4 â€“ 6C5 a b5 + 6C6 b6

Now (a + b)6 â€“ (a â€“ b)6 =6C0 a6 + 6C1 a5 b + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 a b5 + 6C6 b6 â€“ [6C0 a6 â€“ 6C1 a5 b + 6C2 a4 b2 â€“ 6C3 a3 b3 + 6C4 a2 b4 â€“ 6C5 a b5 + 6C6 b6]

Now by substituting a = âˆš3 and b = âˆš2, we get

(âˆš3 + âˆš2)6 â€“ (âˆš3 â€“ âˆš2)6 = 2 [6 (âˆš3)5 (âˆš2) + 20 (âˆš3)3 (âˆš2)3 + 6 (âˆš3) (âˆš2)5]

= 2 [54(âˆš6) + 120 (âˆš6) + 24 âˆš6]

= 2 (âˆš6) (198)

= 396 âˆš6

6. Find the value ofÂ

Solution:

7. Find an approximation of (0.99)5Â using the first three terms of its expansion.

Solution:

0.99 can be written as

0.99 = 1 â€“ 0.01

Now by applying the binomial theorem, we get

(o. 99)5 = (1 â€“ 0.01)5

= 5C0 (1)5 â€“ 5C1 (1)4 (0.01) + 5C2 (1)3 (0.01)2

= 1 â€“ 5 (0.01) + 10 (0.01)2

= 1 â€“ 0.05 + 0.001

= 0.951

8. Find n, if the ratio of the fifth term from the beginning to the fifth term from the end, in the expansion of , is âˆš6: 1

Solution:

9. Expand using the Binomial TheoremÂ

Solution:

Using the binomial theorem, the given expression can be expanded as

Again by using the binomial theorem to expand the above terms, we get

From equations 1, 2 and 3, we get

10. Find the expansion of (3x2Â â€“ 2ax + 3a2)3Â using binomial theorem.

Solution:

We know that (a + b)3Â = a3Â + 3a2b + 3ab2Â + b3

Putting a = 3x2Â & b = -a (2x-3a), we get

[3x2Â + (-a (2x-3a))]3

= (3x2)3+3(3x2)2(-a (2x-3a)) + 3(3x2) (-a (2x-3a))2Â + (-a (2x-3a))3

= 27x6Â â€“ 27ax4 (2x-3a) + 9a2x2 (2x-3a)2Â â€“ a3(2x-3a)3

=Â 27x6Â â€“ 54ax5Â + 81a2x4Â +Â 9a2x2 (4x2-12ax+9a2) â€“ a3 [(2x)3Â â€“ (3a)3Â â€“ 3(2x)2(3a) + 3(2x)(3a)2]

=Â 27x6Â â€“ 54ax5Â + 81a2x4Â +Â 36a2x4Â â€“ 108a3x3Â + 81a4x2 â€“ 8a3x3Â + 27a6Â + 36a4x2Â â€“ 54a5x

=Â 27x6Â â€“ 54ax5+Â 117a2x4Â â€“ 116a3x3Â + 117a4x2Â â€“ 54a5x + 27a6

Thus,Â (3x2Â â€“ 2ax + 3a2)3

=Â 27x6Â â€“ 54ax5+Â 117a2x4Â â€“ 116a3x3Â + 117a4x2Â â€“ 54a5x + 27a6

## NCERT Solutions for Class 11 Maths Chapter 8 â€“ Binomial Theorem

The Chapter 8 Binomial Theorem of NCERT Solutions for Class 11 covers the topics given below.

8.1 Introduction to Binomial TheoremÂ

8.2 Binomial Theorem for Positive Integral IndicesÂ

Â  Â  Â  Â Pascalâ€™s TriangleÂ

8.2.1 Binomial theorem for any positive integer nÂ

8.2.2 Some special cases

8.3 General and Middle TermsÂ

Exercise 8.1 Solutions 14 Questions

Exercise 8.2 Solutions 12 Questions

Miscellaneous Exercise On Chapter 8 Solutions 10 Questions

## NCERT Solutions for Class 11 Maths Chapter 8 â€“ Binomial Theorem

The unit Algebra houses the chapter Binomial Theorem, adding up to 30 marks of the total 80 marks. A total of 3 exercises, including the miscellaneous exercise, are present in this chapter. Chapter 8 of NCERT Solutions for Class 11 Maths discusses the concepts provided underneath:

1. The expansion of a binomial for any positive integral n is given by the Binomial Theorem, which is (a+b)n = nC0 an + nC1 an â€“ 1b + nC2 an â€“ 2b2 + â€¦+ nCn â€“ 1a.bn â€“ 1 + nCn bn .
2. The coefficients of the expansions are arranged in an array. This array is called Pascalâ€™s triangle.
3. The general term of an expansion (a + b)n is Tr + 1 = nCr an â€“ r. br

Therefore, it is ensured that a student who is thorough with Chapter 8 of Class 11, the Binomial Theorem, will be well-versed in the history of the Binomial Theorem, statement and proof of the binomial theorem for positive integral indices, Pascalâ€™s triangle, general and middle term in binomial expansion as well as simple applications of Binomial theorem.

Disclaimer â€“Â

Dropped Topics â€“Â

8.3 General Middle Terms
Example 17 and Ques. 1â€“3, and 8 (Miscellaneous Exercise)
Last two points in the Summary

## Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 8

Q1

### Explain the concept of the Binomial Theorem covered in Chapter 8 of NCERT Solutions for Class 11 Maths.

The Binomial Theorem is the process of algebraically expanding the power of sums of two or more binomials. The coefficients of binomial terms which are involved in the process of expansion are called binomial coefficients. The introduction of this chapter has definitions of terms which are important for the exams. Students can now study and be updated about the latest syllabus of the CBSE Board using the NCERT Solutions, which are available in PDF format in this article.
Q2

### Will the NCERT Solutions for Class 11 Maths Chapter 8 help students understand the concepts which are important from the exam perspective?

In order to understand the expansion procedure, students can refer to the examples which are present in the NCERT textbook before solving the exercise-wise problems. Each problem in the solutions is solved in a stepwise manner to help students understand the concepts in a better way. By using the solutions PDF, students will be well versed in the method of solving these equations and score well in the exam.
Q3

### In the Binomial Theorem, explain the properties of positive integers covered in the NCERT Solutions for Class 11 Maths Chapter 8.

More than 10 properties are mentioned under the positive integers, which the students can learn using the NCERT Solutions for Class 11 Maths Chapter 8. These properties are important to understand the concept of solving equations efficiently. The question paper in the annual exam would target the chapters which are simple for the students but tricky to solve. For this purpose, students should go through these NCERT Solutions to score good marks in the annual examination.