Binomial Theorem Class 11 Ncert Solutions Pdf Download

*According to the latest update on the term-wise CBSE Syllabus 2021-22, this chapter has been removed.

The NCERT Solutions Class 11 Chapter 8 Binomial Theorem can be downloaded at BYJU'S without any hassle. Practising these solutions can help the students clear their doubts as well as to solve the problems faster. Students can learn new tricks to answer a particular question in different ways giving them an edge with the exam preparation.

The concepts covered in Chapter 8 of the Maths textbook includes the study of essential topics such as Positive Integral Indices, Pascal's Triangle, Binomial theorem for any positive integer and some special cases. Students can score high marks in the exams with ease by practising the NCERT Solutions for all the questions present 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.

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Exercise 8.1 Page No: 166

Expand each of the expressions in Exercises 1 to 5.

1. (1 – 2x)⁵

Solution:

From binomial theorem expansion we can write as

(1 – 2x)⁵

= ⁵C_o (1)⁵ – ⁵C₁ (1)⁴ (2x) + ⁵C₂ (1)³ (2x)² – ⁵C₃ (1)² (2x)³ + ⁵C₄ (1)¹ (2x)⁴ – ⁵C₅ (2x)⁵

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

= 1 – 10x + 40x² – 80x³ + 80x⁴– 32x⁵

Solution:

From binomial theorem, given equation can be expanded as

3. (2x – 3)⁶

Solution:

From binomial theorem, given equation can be expanded as

Solution:

From binomial theorem, given equation can be expanded as

Solution:

From binomial theorem, given equation can be expanded as

6. (96)³

Solution:

Given (96)³

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

The given question can be written as 96 = 100 – 4

(96)³ = (100 – 4)³

= ³C₀ (100)³ – ³C₁ (100)² (4) – ³C₂ (100) (4)²– ³C₃ (4)³

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

= 1000000 – 120000 + 4800 – 64

= 884736

7. (102)⁵

Solution:

Given (102)⁵

102 can be expressed as the sum or difference of two numbers and then 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. (101)⁴

Solution:

Given (101)⁴

101 can be expressed as the sum or difference of two numbers and then 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)² + ⁴C₄ (1)⁴

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

= 100000000 + 400000 + 60000 + 400 + 1

= 1040604001

9. (99)⁵

Solution:

Given (99)⁵

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

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

(99)⁵ = (100 – 1)⁵

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

= (100)⁵ – 5 (100)⁴ + 10 (100)³ – 10 (100)² + 5 (100) – 1

= 1000000000 – 5000000000 + 10000000 – 100000 + 500 – 1

= 9509900499

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

Solution:

By splitting the given 1.1 and then applying 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 × 1.1 + other positive terms

= 1 + 11000 + other positive terms

> 1000

(1.1)¹⁰⁰⁰⁰ > 1000

11. Find (a + b)⁴ – (a – b)⁴. Hence, evaluate

Solution:

Using binomial theorem the expression (a + b)⁴ and (a – b)⁴, can be expanded

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

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

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

= 2 (⁴C₁ a³ b + ⁴C₃ a b³)

= 2 (4a³ b + 4ab³)

= 8ab (a² + b²)

Now by substituting a = √3 and b = √2 we get

(√3 + √2)⁴ – (√3 – √2)⁴ = 8 (√3) (√2) {(√3)² + (√2)²}

= 8 (√6) (3 + 2)

= 40 √6

12. Find (x + 1)⁶ + (x – 1)⁶. Hence or otherwise evaluate

Solution:

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

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

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

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

= 2 [⁶C₀ x⁶ + ⁶C₂ x⁴ + ⁶C₄ x² + ⁶C₆]

= 2 [x⁶ + 15x⁴ + 15x² + 1]

Now by substituting x = √2 we get

(√2 + 1)⁶ – (√2 – 1)⁶ = 2 [(√2)⁶ + 15(√2)⁴ + 15(√2)² + 1]

= 2 (8 + 15 × 4 + 15 × 2 + 1)

= 2 (8 + 60 + 30 + 1)

= 2 (99)

= 198

13. Show that 9ⁿ⁺¹ – 8n – 9 is divisible by 64, whenever n is a positive integer.

Solution:

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

Using binomial theorem,

(1 + a)^m = ^mC₀ + ^mC₁ a + ^mC₂ a² + …. + ^m C _m a^m

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

(1 + 8)ⁿ⁺¹ = ⁿ⁺¹C₀ + ⁿ⁺¹C₁ (8) + ⁿ⁺¹C₂ (8)² + …. + ⁿ⁺¹ C _n+1 (8)ⁿ⁺¹

9ⁿ⁺¹ = 1 + (n + 1) 8 + 8² [ⁿ⁺¹C₂ + ⁿ⁺¹C₃ (8) + …. + ⁿ⁺¹ C _n+1 (8)^n-1]

9ⁿ⁺¹ = 9 + 8n + 64 [ⁿ⁺¹C₂ + ⁿ⁺¹C₃ (8) + …. + ⁿ⁺¹ C _n+1 (8)^n-1]

9ⁿ⁺¹ – 8n – 9 = 64 k

Where k = [ⁿ⁺¹C₂ + ⁿ⁺¹C₃ (8) + …. + ⁿ⁺¹ C _n+1 (8)^n-1] is a natural number

Thus, 9ⁿ⁺¹ – 8n – 9 is divisible by 64, whenever n is positive integer.

Hence the proof

14. Prove that

Solution:

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Exercise 8.2 Page No: 171

Find the coefficient of

1. x⁵ in (x + 3)⁸

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 =ⁿ C _r a^n-r b^r

Here x⁵ is the T_r+1 term so a= x, b = 3 and n =8

T_r+1 =⁸C_r x^8-r 3^r…………… (i)

For finding out x⁵

We have to equate x⁵= x^8-r

⇒ r= 3

Putting value of r in (i) we get

= 1512 x⁵

Hence the coefficient of x⁵= 1512

2. a⁵b⁷ in (a – 2b)¹² .

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 =ⁿ C _r a^n-r b^r

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

Substituting the values, we get

T_r+1 =¹²C_r a^12-r (-2b)^r………. (i)

To find a⁵

We equate a^12-r =a⁵

r = 7

Putting r = 7 in (i)

T₈ =¹²C₇ a⁵ (-2b)⁷

= -101376 a⁵ b⁷

Hence the coefficient of a⁵b⁷= -101376

Write the general term in the expansion of

3. (x² – y)⁶

Solution:

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

T_r+1 =ⁿ C _r a^n-r b^r…….. (i)

Here a = x² , n = 6 and b = -y

Putting values in (i)

T_r+1 =⁶C_r x^2(6-r) (-1)^r y^r

= -1^{r 6}c_r .x^{12 – 2r}. y^r

4. (x² – y x)¹², x ≠ 0.

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 =ⁿ C _r a^n-r b^r

Here n = 12, a= x² and b = -y x

Substituting the values we get

T_n+1 =¹²C_r × x^2(12-r) (-1)^r y^r x^r

= -1^{r 12}c_r .x^{24 –2r}. y^r

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

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 =ⁿ C _r a^n-r b^r

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

By substituting the values we get

T₄ =¹²C₃ x⁹ (-2y)³

= -1760 x⁹ y³

6. Find the 13^th term in the expansion of

Solution:

Find the middle terms in the expansions of

Solution:

Solution:

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

Solution:

We know that the general term T_r+1 in the binomial expansion is given by T_r+1 =ⁿC_r a^n-r b^r

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

Substituting the values in the general form

T_r+1 =^m+n C_r 1^m+n-r a^r

=^m+n C_r a^r…………. (i)

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

T_r+1 =^m+n C_r a^r

Now, For coefficient of a^m

T_m+1 =^m+n C_m a^m

Hence, for coefficient of a^m, value of r = m

So, the coefficient is^m+n C _m

Similarly, Coefficient of aⁿ is^m+n C _n

10. The coefficients of the (r – 1)^th, r^th and (r + 1)^th terms in the expansion of (x + 1)ⁿ are in the ratio 1 : 3 : 5. Find n and r.

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 =ⁿC_r a^n-r b^r

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

The (r+1)^th term is given by

T_(r+1) =ⁿC_r 1^n-r x^r

T_(r+1) =ⁿC_r x^r

The coefficient of (r+1)^th term isⁿC_r

The r^th term is given by (r-1)^th term

T_(r+1-1) =ⁿC_r-1 x^r-1

T_r =ⁿC_r-1 x^r-1

∴ the coefficient of r^th term isⁿC_r-1

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

T_r-2+1 =ⁿC_r-2 x^r-2

T_r-1 =ⁿC_r-2 x^r-2

∴ the coefficient of (r-1)^th term isⁿC_r-2

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

Therefore,

⇒ 5r = 3n – 3r + 3

⇒ 8r – 3n – 3 =0………….2

We have 1 and 2 as

n – 4r ± 5 =0…………1

8r – 3n – 3 =0…………….2

Multiplying equation 1 by number 2

2n -8r +10 =0……………….3

Adding equation 2 and 3

2n -8r +10 =0

-3n – 8r – 3 =0

⇒ -n = -7

n =7 and r = 3

11. Prove that the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1 + x)^{2n – 1}.

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 =ⁿC_r a^n-r b^r

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

T_r+1 =²ⁿC_r x^r …………………..1

To find the coefficient of xⁿ

r = n

T_n+1 =²ⁿC_n xⁿ

The coefficient of xⁿ =²ⁿC_n

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

T_r+1 =^2n-1C_r x^r

To find the coefficient of xⁿ

Putting n = r

T_r+1 =^2n-1C_r xⁿ

The coefficient of xⁿ =^2n-1C_n

We have to prove

Coefficient of xⁿ in (1+x)²ⁿ = 2 coefficient of xⁿ in (1+x)^2n-1

Consider LHS = ²ⁿC_n

12. Find a positive value of m for which the coefficient of x² in the expansion (1 + x)^m is 6.

Solution:

The general term T_r+1 in the binomial expansion is given by T_r+1 =ⁿC_r a^n-r b^r

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

Putting the value

T_r+1 =^m C_r 1^m-r x^r

=^m C_r x^r

We need coefficient of x²

∴ putting r = 2

T₂₊₁ =^mC₂ x²

The coefficient of x² =^mC₂

Given that coefficient of x² =^mC₂ = 6

⇒ m (m – 1) = 12

⇒ m²– m – 12 =0

⇒ m²– 4m + 3m – 12 =0

⇒ m (m – 4) + 3 (m – 4) = 0

⇒ (m+3) (m – 4) = 0

⇒ m = – 3, 4

We need positive value of m so m = 4

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Miscellaneous Exercise Page No: 175

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

Solution:

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

T_r+1 = ⁿC_r a^n-t b^r

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

T₁ = ⁿC₀ a^n-0 b⁰ = aⁿ = 729….. 1

T₂ = ⁿC₁ a^n-1 b¹ = na^n-1 b = 7290…. 2

T₃ = ⁿC₂ a^n-2 b² = {n (n -1)/2 }a^n-2 b² = 30375……3

Dividing 2 by 1 we get

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

Dividing 3 by 2 we get

$\frac{n(n-1)a^{n-2}b^{2}}{2na^{n-1}b} = \frac{30375}{7290} \\ \frac{(n-1)b}{2a} = \frac{30375}{7290} \\ \frac{(n-1)b}{a} = \frac{30375}{7290} \times 2 =\frac{25}{3} \\ \frac{(nb)}{a} -\frac{b}{a} = \frac{25}{3} \\ 10 -\frac{b}{a} = \frac{25}{3} \\ \frac{b}{a} = 10 – \frac{25}{3} = \frac{5}{3}$

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 x² and x³ in the expansion of (3 + a x)⁹ are equal.

Solution:

3. Find the coefficient of x⁵ in the product (1 + 2x)⁶ (1 – x)⁷ 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 – x)⁷ = ⁷C₀ – ⁷C₁ (x) + ⁷C₂ (x)² – ⁷C₃ (x)³ + ⁷C₄ (x)⁴ – ⁷C₅ (x)⁵ + ⁷C₆ (x)⁶ – ⁷C₇ (x)⁷

= 1 – 7x + 21x² – 35x³ + 35x⁴ – 21x⁵ + 7x⁶ – x⁷

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

192 – 21 = 171

Thus, the coefficient of x⁵ in the expression (1+2x)⁶(1-x)7 is 171.

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

Solution:

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

aⁿ – bⁿ = k (a – b) where k is some natural number.

a can be written as a = a – b + b

aⁿ = (a – b + b)ⁿ = [(a – b) + b]ⁿ

= ⁿC₀ (a – b)ⁿ + ⁿC₁ (a – b)^n-1 b + …… + ⁿ C _n bⁿ

aⁿ – bⁿ = (a – b) [(a –b)^n-1 + ⁿC₁ (a – b)^n-1 b + …… + ⁿ C _n bⁿ]

aⁿ – bⁿ = (a – b) k

Where k = [(a –b)^n-1 + ⁿC₁ (a – b)^n-1 b + …… + ⁿ C _n bⁿ] is a natural number

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

5. Evaluate

Solution:

Using binomial theorem the expression (a + b)⁶ and (a – b)⁶, 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 – b)⁶ = ⁶C₀ a⁶ – ⁶C₁ a⁵ b + ⁶C₂ a⁴ b² – ⁶C₃ a³ b³ + ⁶C₄ a² b⁴ – ⁶C₅ a b⁵ + ⁶C₆ b⁶

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

Now by substituting a = √3 and b = √2 we get

(√3 + √2)⁶ – (√3 – √2)⁶ = 2 [6 (√3)⁵ (√2) + 20 (√3)³ (√2)³ + 6 (√3) (√2)⁵]

= 2 [54(√6) + 120 (√6) + 24 √6]

= 2 (√6) (198)

= 396 √6

6. Find the value of

Solution:

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

Solution:

0.99 can be written as

0.99 = 1 – 0.01

Now by applying binomial theorem we get

(o. 99)⁵ = (1 – 0.01)⁵

= ⁵C₀ (1)⁵ – ⁵C₁ (1)⁴ (0.01) + ⁵C₂ (1)³ (0.01)²

= 1 – 5 (0.01) + 10 (0.01)²

= 1 – 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 √6: 1

Solution:

9. Expand using Binomial Theorem

Solution:

Using binomial theorem the given expression can be expanded as

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

From equation 1, 2 and 3 we get

10. Find the expansion of (3x² – 2ax + 3a²)³ using binomial theorem.

Solution:

We know that (a + b)³ = a³ + 3a²b + 3ab² + b³

Putting a = 3x² & b = -a (2x-3a), we get

[3x² + (-a (2x-3a))]³

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

= 27x⁶ – 27ax⁴ (2x-3a) + 9a²x² (2x-3a)² – a³(2x-3a)³

= 27x⁶ – 54ax⁵ + 81a²x⁴ + 9a²x² (4x²-12ax+9a²) – a³ [(2x)³ – (3a)³ – 3(2x)²(3a) + 3(2x)(3a)²]

= 27x⁶ – 54ax⁵ + 81a²x⁴ + 36a²x⁴ – 108a³x³ + 81a⁴x² – 8a³x³ + 27a⁶ + 36a⁴x² – 54a⁵x

= 27x⁶ – 54ax⁵+ 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶

Thus, (3x² – 2ax + 3a²)³

= 27x⁶ – 54ax⁵+ 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶

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NCERT Solutions for Class 11 Maths Chapter 8- Binomial Theorem

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

8.1 Introduction to Binomial Theorem

8.2 Binomial Theorem for Positive Integral Indices

Pascal's Triangle

8.2.1 Binomial theorem for any positive integer n,

8.2.2 Some special cases

8.3 General and Middle Terms

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- 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 is 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 a^{n – 1}b + ⁿC₂ a^{n – 2}b² + …+ ⁿC_{n – 1}a.b^{n – 1} + ⁿC_n bⁿ .
2. The coefficients of the expansions are arranged in an array. This array is called Pascal's triangle.
3. The general term of an expansion (a + b)ⁿ is T_{r + 1} = ⁿC_r a^{n – r} . b^r

Therefore, it is ensured that a student who is thorough with the eighth chapter of Class 11, the Binomial Theorem, will be well versed in the history of Binomial Theorem, statement and proof of the binomial theorem for positive integral indices, Pascal's triangle, General and middle term in binomial expansion as well as simple applications of Binomial theorem.

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

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

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.

Will the NCERT Solutions for Class 11 Maths Chapter 8 help students to 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 are solved in a stepwise manner to help students in understanding the concepts in a better way. By using the solutions PDF, students will be well versed with the method of solving these equations and score well in the exam.

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 the NCERT Solutions if they aspire to score good marks.

Binomial Theorem Class 11 Ncert Solutions Pdf Download

Source: https://byjus.com/ncert-solutions-class-11-maths/chapter-8-binomial-theorem/

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