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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
= 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 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 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 positive integer.
Hence the proof
14. Prove that
Solution:
Exercise 8.2 Page No: 171
Find the coefficient of
1. x5 in (x + 3)8
Solution:
The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br
Here x5 is the Tr+1 term so a= x, b = 3 and n =8
Tr+1 = 8Cr x8-r 3r…………… (i)
For finding out x5
We have to equate x5= x8-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 Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br
Here a = a, b = -2b & n =12
Substituting the values, we get
Tr+1 = 12Cr a12-r (-2b)r………. (i)
To find a5
We equate a12-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 Tr+1 in the binomial expansion is given by
Tr+1 = n C r an-r br…….. (i)
Here a = x2 , n = 6 and b = -y
Putting values in (i)
Tr+1 = 6Cr x 2(6-r) (-1)r yr
= -1r 6cr .x12 – 2r. yr
4. (x2 – y x)12, x ≠ 0.
Solution:
The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br
Here n = 12, a= x2 and b = -y x
Substituting the values we get
Tn+1 =12Cr × x2(12-r) (-1)r yr xr
= -1r 12cr .x24 –2r. yr
5. Find the 4th term in the expansion of (x – 2y)12.
Solution:
The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br
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 13th 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 am and an are equal.
Solution:
We know that the general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br
Here n= m+n, a = 1 and b= a
Substituting the values in the general form
Tr+1 = m+n Cr 1m+n-r ar
= m+n Cr ar…………. (i)
Now we have that the general term for the expression is,
Tr+1 = m+n Cr ar
Now, For coefficient of am
Tm+1 = m+n Cm am
Hence, for coefficient of am, 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, rth 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 Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br
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) = nCr 1n-r xr
T(r+1) = nCr xr
The coefficient of (r+1)th term is nCr
The rth term is given by (r-1)th term
T(r+1-1) = nCr-1 xr-1
Tr = nCr-1 xr-1
∴ the coefficient of rth term is nCr-1
For (r-1)th term we will take (r-2)th term
Tr-2+1 = nCr-2 xr-2
Tr-1 = nCr-2 xr-2
∴ the coefficient of (r-1)th term is nCr-2
Given that the coefficient of (r-1)th, rth and r+1th 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 Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br
The general term for binomial (1+x)2n is
Tr+1 = 2nCr xr …………………..1
To find the coefficient of xn
r = n
Tn+1 = 2nCn xn
The coefficient of xn = 2nCn
The general term for binomial (1+x)2n-1 is
Tr+1 = 2n-1Cr xr
To find the coefficient of xn
Putting n = r
Tr+1 = 2n-1Cr xn
The coefficient of xn = 2n-1Cn
We have to prove
Coefficient of xn in (1+x)2n = 2 coefficient of xn in (1+x)2n-1
Consider LHS = 2nCn
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 Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br
Here a = 1, b = x and n = m
Putting the value
Tr+1 = m Cr 1m-r xr
= m Cr xr
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
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, (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 = 729….. 1
T2 = nC1 an-1 b1 = nan-1 b = 7290…. 2
T3 = nC2 an-2 b2 = {n (n -1)/2 }an-2 b2 = 30375……3
Dividing 2 by 1 we get
Dividing 3 by 2 we get
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