*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 7.
The NCERT Solutions of the first exercise of Class 11 Chapter 8 are available here. These solutions are present in PDF format to help students with their studies. Exercise 8.1 of NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem is based on the following topics:
- Introduction to Binomial Theorem
- Binomial Theorem for Positive Integral Indices
- Pascal’s Triangle
- Binomial theorem for any positive integer n
- Some special cases
NCERT textbook contains numerous questions which are intended for the students to solve and practise. To score high marks in the Class 11 examination, solving and practising the NCERT Solutions for Class 11 Maths is a must.
NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem Exercise 8.1
Solutions for Class 11 Maths Chapter 8 – Exercise 8.1
Expand each of the expressions in Exercises 1 to 5.
1. (1 – 2x)5
Solution:
From the binomial theorem expansion, we can write as
(1 – 2x)5
= 5Co (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 the binomial theorem, the given equation can be expanded as
3. (2x – 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. (96)3
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 – 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 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. (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)3Â + 4C4 (1)4
= (100)4 + 4 (100)3 + 6 (100)2 + 4 (100) + (1)4
= 100000000 + 400000 + 60000 + 400 + 1
= 104060401
9. (99)5
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 – 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 the 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 × 1.1 + other positive terms
= 1 + 11000 + other positive terms
> 1000
(1.1)10000 > 1000
11. Find (a + b)4 – (a – b)4. Hence, evaluateÂ
Solution:
Using the 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, evaluateÂ
Â
Solution:
Using the 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 shown that 9n+1 – 8n – 9 = 64 k, where k is some natural number
Using the binomial theorem,
(1 + a)m = mC0 + mC1 a + mC2 a2 + …. + 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 + …. + 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 a positive integer.
Hence, the proof.
14. Prove thatÂ
Solution:
Access Other Exercise Solutions of Class 11 Maths Chapter 8 – Binomial Theorem
Exercise 8.2 Solutions 12 Questions
Miscellaneous Exercise on Chapter 8 Solutions 10 Questions
Also explore – NCERT Class 11 Solutions
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