RBSE Solutions for Class 11 Maths Chapter 7 Binomial Theorem

RBSE Solutions for Class 11 Maths Chapter 7 Binomial Theorem

Rajasthan Board RBSE Class 11 Maths Chapter 7 Binomial Theorem Ex 7.1

Expand each expression in the following questions (Q. 1 to 5):

Question 1.
(2 – x)³
Solution :
(2 – x)³
= ³C₀ ( 2 )³ + ³C₁ ( 2 )² (-x) + ³C₂ ( 2 ) ( -x )² + ³C₃ ( -x )³
= 1 × 8 + 3 × 4x ( -x ) + 3 × 2 × x² + 1 × ( -x )³
= 8 – 12x + 6x² – x³

Question 2.
(2/x−x/2)5
Solution:

Question 3.
(x/3−1/x)6
Solution:
(x/3−1/x)6,
By using Binomial theorem

Question 4.
(3x + 2y)⁴
Solution:
(3x + 2y)⁴
By using Binomial theorem
= ⁴C₀ ( 3x )⁴ ( 2y )⁰ + ⁴C₁ ( 3x )³ (2 y ) + ⁴C₂ ( 3x )² ( 2y )²
+ ⁴C₃ ( 3x ) ( 2y )³ + ⁴C₄ ( 3x )⁰ ( 2y )⁴
= (1 × 3⁴ × 1 × x⁴ × 1) + (4 × 3³ × 2 × x³ y)+ (6 × 3² × 2² × x² y²)
+ (4 × 3 × 2³ × xy³)+ (1 × 1 × 2⁴ × y⁴)
= 81x⁴ + 216x³y + 216x²y² + 96 xy³ + 16 y⁴

Question 5.
(√x/a−√a/x)6

Solution:
(√x/a−√a/x)6
By using Binomial theorem

Using Binomial theorem find the values of following (Q. 6 to 9) :

Question 6.
(96)³
Solution :
(96)^3
∵ 96 = 100-4
^∴ (96)³ = (100 – 4)³ = [ 100 + (- 4)]³
By using Binomial theorem
96³ = [100 + (- 4)]^3
3C₀ (100)³ ( – 4 )⁰ + ³C₁ (100)² (- 4)¹
+ ³C₂ (100) (- 4)² + ³C₃ (100)⁰ (- 4)³
= 1 x 1000000 x  + 3 x 10000 x (-4)
+ 3 x 100 × 16 + 1 × 1 × (-64)
= 1000000 – 120000 + 4800 – 64
= 1004800 – 120064
= 884736 Hence, (96)³
= 884736

Question 7.
(101)⁴
Solution :
101 = 100 + 1
(101 )⁴ = (100 +1)⁴
By using Binomial theorem
(101)⁴ = (100 + 1)⁴
= ⁴C₀ (100)⁴ (1)⁰ + ⁴C₁ (100)³ (1)¹
+ ⁴C₂ (100)² (1)² + ⁴C₃ (100) (1)³
+ ⁴C₄ (100)⁰ (1)⁴
= 1 x 100000000 x 1 + 4 × 1000000 × 1 + 6
× 10000 x 1 + 4 x 100 × 1 + 1 x 1 x 1
= 100000000 + 4000000 + 60000 + 400+1
= 104060401
Hence, (101)⁴ = 104060401

Question 8.
(99)⁵
Solution :
99 = (100- 1)
99⁵ = (100- 1)⁵ = [100+ (- 1)]⁵
By using Binomial theorem
99⁵ = ⁵C₀ (100)⁵ (-1)⁰ + ⁵C, (100)⁴ (-1)¹
+ ⁵C₂ (100)³ (- 1)² + ⁵C₃ (100)² (- 1)³
+ ⁵C₄ (100) (-1)⁴ + ⁵C₅ (100)⁰ (-1)⁵
= 1 x 10000000000 x 1 + 5 x 100000000
× (- 1) + 10 x 1000000 × 1 + 10 × 10000
× (- 1)+ 5 x 100 × 1 + 1 × 1 × (-1)
= 10000000000 – 500000000 +
10000000 – 100000 + 500 – 1
= 10010000500 – 500100001
= 9509900499
Hence, (99)⁵ = 9509900499

Question 9.
(M)⁶
Solution :
(1.1)⁶ = (1 +0.1)⁶
= ⁶C₀(1)⁶ (0.1)⁰ + ⁶C₁ (1)⁵ (0.1)¹
+ ⁶C₂ (1)⁴ (0.1)² + ⁶C₃ (1)³ (0.3)³
+ ⁶C₄ (1)² (0.1)⁴ + ⁶C₅ (1)¹ (0.1)⁵ + ⁶C₆ (1)⁰ (0.1)⁶
= (1 x 1 x 1) + (6 x 1 x 0-1) + {15 x 1 x (0.1)²}
+ {20 x 1 x (0.1)³} + {15 x 1 x (0.1)⁴}
+ {6 x 1 x (0.1)⁵} + {1 x 1 x (0.1)⁶}
= 1 + 6 x 0.1 + 15 x (0.1)² + 20 x (0.1)³
+ 15 x (0.1)⁴+ 6 x (0.1)⁵ + (0.1)⁶
= 1 + 0.6 + 15 x 0.01 + 20 x 0.001 + 15 x 0.0001
+ 6 x 0.00001 +0.000001
= 1+0-6 + 0-15 + 0020 + 0-0015 + 0-00006 + 0000001
= 1-771561

Question 10.
Using Binomial theorem find which number is greater (M)¹⁰⁰⁰⁰⁰ or 1000.
Solution :
(1.1)¹⁰⁰⁰⁰⁰
= (1+0.1)^1000000
= ¹⁰⁰⁰⁰C₀ (1)¹⁰⁰⁰⁰ (0.1)⁰
+ ⁰⁰⁰⁰c₁ (1)⁹⁹⁹⁹ (0.1)¹ +……….
= 1 × 1 × 1 + 10000 × 1 × 0.1 +………
= 1 +1000 +… other positive numbers
= 1001+… other positive numbers
Hence(1.1)¹⁰⁰⁰⁰= 1001+… other positive numbers
But 1001>1000
Then (1.1)¹⁰⁰⁰⁰>1000.
(1.1)¹⁰⁰⁰⁰ is greater

Question 11.
Expand (a + b)⁴ – (a – b)⁴. Using this find the value of (√3 + √2)⁴ – (√3 – √2)⁴.
Solution :
By using binomial theorem,
(a + b)⁴ = ⁴C₀ a⁴ b⁰ + ⁴C₁ a³ b¹
+ ⁴C₂ a² b²+ ⁴C₃ ab³ + ⁴C₄ a⁰ b⁴
= ⁴C₀ a⁴ + ⁴C₁ a³ b¹ + ⁴C₂ a² b²
+ ⁴C₃ ab³ + ⁴C₄ b⁴ …. (1)
and (a – b)⁴ = ⁴C₀ a⁴ (-b)⁰+ ⁴C₁ a³ (-b)¹
+ ⁴C₂ a² (-b)² + ⁴C₃ a¹ (-b)³ + ⁴C₄ a⁰ (-b)⁴
= ⁴C₀ a⁴ – ⁴C₁ a³ b + ⁴C₂ a² b²
– ⁴C₃ ab⁴ + ⁴C₄ b⁴ …. (2)
From equation (1) and (2) we have,
(a + b)⁴ – (a – b)⁴
= [⁴C₀ a⁴ + ⁴C₁ a³ b +⁴C₂ a² b² + ⁴C₃ ab³ + ⁴C₄ b⁴]
– [⁴C₀ a⁴ – ⁴C₁ a³ b + ⁴C₂ a² b² –⁴C₃ ab³ + ⁴C₄ b⁴]
= ⁴C₀ a⁴ + ⁴C₁ a³ b + ⁴C₂ a² b² + ⁴C₃ ab³ + ⁴C₄ b⁴
– ⁴C₀ a⁴ + ⁴C₁ a³b – ⁴C₂ a²b² + ⁴C₃ ab³ – ⁴C₄ b⁴
= 2. ⁴C₁ a³ b + 2. ⁴C₃ ab³
= 2ab [⁴C₁ a² + ⁴C₃ b²]
= 2ab [4a² + 4b²]  [∴ ⁴C₁ = 4, ⁴C₃ = 4]
= 2ab (a² + b²)
Hence, (a + b)⁴ – (a – b)⁴ = 8ab (a² + b²)
Now, putting a= √3 and b = √2
(√3 + √2)⁴ – (√3 – √2)⁴
= 8 √3 × √2 [(√3)² + (√2)²]
= 8 √6 (3 + 2)
= 8 √6 × 5 – 40 √6
Hence (√3 + √2 )⁴ – (√3 – √2 )^4 = 40 √6

Rajasthan Board RBSE Class 11 Maths Chapter 7 Binomial Theorem Ex 7.2

Question 1.
In the following expansions, find the term as stated :
(i) 5th term of (a + 2x³)¹⁷
(ii) 9th term of (x/y−3y/x2)17
(iii) 6th term of (2/√x−x2/2)9
Solution:
(i) 5th term of  (a + 2x³)¹⁷
We know that (r + 1 )^th term in the expansion of (a + b)ⁿ
T_r+1 = ⁿC_r a^n-r b^r
Here a = ‘a’, b = 2x³, n = 17
r + 1 = 5 ⇒ r = 4
Hence, 5th term = T_5
⇒ T₅ = T_4+1
⇒ T₅ = ¹⁷C₄ a^17-4(23x³)⁴
_⇒ T₅ = 2380 × a¹³ × 2⁴ × x¹²
_⇒ T₅ = 2380 × 16 × a¹³x¹²
_⇒ T₅ = 38080 a¹³ x¹²
Hence,
5th term = 38080 a¹³ x¹² or ¹⁷C₄ a¹³ × 16x¹²

Question 2.
Find the coefficient of:
(i) x^-7 in the expansion of (ax−1/bx2)8
(ii) x⁴ in the expansion of

(x4+1/x3)15
(iii) x⁶ in the expansion (a – bx²)¹⁰
Solution:
(i) Coefficient of x^-7 in the expansion of


In this term for coefficient of x^-7 power of x should be
-7 = 8 – 3r = – 7
Hence, 8 – 3r = – 7
⇒ -3r = – 7 – 8 = – 15
r = 5
Hence, coefficient of x^-7 in the expansion

(ii) Coefficient of x⁴ in the expansion of (x4+1/x3)15

In this term, for coefficient of x⁴, power of x should be 4
So,  60 – 7r = 4
⇒ -7r = 4 – 60 = – 56
⇒ r = 8
Hence, coefficient of x⁴ in the expansion = ¹⁵C₈ = 6435

(iii) Coefficient of x⁶ in the expansion of (a – b²)¹⁰

In this term, for coefficient of x⁶, power of x should be 6.
So,          2r = 6    ⇒ r = 3
Hence, coefficient of x⁶ in the expansion
= (-1)^{3 10}C₃ x a^10-3 b³
= – 1 × 120 × a⁷ b³
= – 120 a⁷ b³

Question 3.
In the following expansions find the term independent of x :

Solution:
(i) Term independent of x in the expansion of

for term independent of x
x^12-3r = x⁰
⇒ 12- 3r = 0
⇒ -3r = -12
⇒ r = 4
Hence, term independent of x = (-1)^{4 12}C₄
= 1 × 495 = 495

(ii) Term independent of x in the expansion of

For the independent of x

Hence, term independent of x
= (-1)^{2 10}C₂ 3²
= 1 × 45 × 9 = 405

(iii) Term independent of x in the expansion of
(√x/3+√3/2x2)10
Let (r + 1)^th term of expansion is constant i.e., independent of x.

(r + 1)^th term is constant i.e., power of x should be zero.
∴ 5- r/2 -2r = 0

(iv) Term independent of x in the expansion of

In this expansion for term independent of x, power of.v should be 0
20 – 4r = 0
⇒ – 4r = -20
⇒ r = 5
∴ Term independent of x in the expansion
= (-1)^{5 10}C₅ =-252

Question 4.
In the following expansion, find the middle term :

Solution:
(i) Middle term in the expansion of (x/2+2y)6 In the expansion of (x/2+2y)6 middle term will be T(n/2+1) th term where x = 6 which is even number.

(ii) Middle term in the expansion of (3a−a3/6)9
Given expression is(3a−a3/6)9
Where n = 9 which is an odd number
∴ Middle term n+1/2+1th term and n+1/2th term
i.e., Middle term is 5^th and 6^th term (T₆ and T₆)

Question 5.

Similarly, if n is odd, then we take (1 + x)²ⁿ⁺¹ in place of (1 + x)ⁿ, then ‘n’ = 2n + 1 is also odd. Now middle term are

Question 6.
If in the expansion of (ax+1/bx)11
coefficient of x⁷ and x⁷ are equal then prove that ab – 1 – 0.
Solution:
(r + 1)^th term in the expansion of (ax+1/bx)11

In this term, for coefficient of x⁷, power of x should be 7.
Hence, 11 – 2r = 7
⇒ 2r = 11-7
⇒ 2r = 4,
⇒ r = 2
Coefficient of x⁷ in the expansion

Question 7.
In the expansion of (1 + y)ⁿ if coefficients of 5^th, 6^th and 7^th terms are in A.P., then find the value of n.
Solution:

Question 8.
In Binomial expansion (x – a)ⁿ second, third and fourth terms are 250, 720 and 1080 respectively. Find x, a and n.
Solution:
Second term in expansion of (x + a)ⁿ
T₂ = ⁿC₁ x^n-1.a
According to question,

Similarly, on dividing (3) by (2), we get

Question 9.
If in the expansion of (1+ a)ⁿ coefficient of three consecutive terms are in ratio 1 : 7 ; 42, then find the value of n.
Solution:
Let in the expansion of (1 + x)ⁿ three consecutive terms are (r – 1 )^th term, r^th term and (r + 1)^th term.
Here, (r – 1)^th term = ⁿC_r – 2a^r-2 and its coefficient ,
=ⁿC_r – 2
Similarly, coefficient of r^th term and (r + 1)^th term are ⁿC_r -1 and ⁿC_r respectively.
Now, given ratio of coefficient is 1 : 7 : 42

or  6r = n – r + 1
or  n – 7r + 1 = 0       …….(2)
On solving (1) and (2), we get
n = 55

Question 10.
Find the positive value of m for which in the expansion of (1 + x)^m coefficient x² is 6.
Solution:
From Binomial theorem
(1 + x)^m = ^mC₀ + ^mC₁x + ^mC₂ x² + ……
Then, coefficient of x² in the expansion of (1 + x )^m = ^mC₂

Rajasthan Board RBSE Class 11 Maths Chapter 7 Binomial Theorem Ex 7.3

Question 1.
If C₀, C₁, C₂,… C_n are coefficients of expansion (1 + x)ⁿ then find the value of :
⁸C₁ + ⁸C₂ + ⁸C₃ + … + ⁸C₈
⁸C₁+ ⁸C₃ + ⁸C₅ + ⁸C₇
Solution:
(i) ⁸C₁ + ⁸C₂ + ⁸C₃ + … + ⁸C₈ We know that
ⁿC₀ + ⁿC₁+ ⁿC₂ + … + ⁿC_n = 2ⁿ
Putting n = 8
⁸C₀ + ⁸C₁ + ⁸C₂ + …. + ⁸C₈ = 2⁸
⇒ 1 +⁸C₁ + ⁸C₂ + ⁸C₃ + … + ⁸C₈ = 2⁸
⇒ ⁸C₁ + ⁸C₂ + …. + ⁸C₈ = 2⁸ – 1 = 255.

(ii) ⁸C₁ + ⁸C₃ + ⁸C₅ + ⁸C₇
We know that
⁸C₁ + ⁸C₃ + ⁸C₅ +… = 2^n-1
Putting n = 8
⁸C₁ + ⁸C₃ + ⁸C₅ + ⁸C₇ = 2^8 -1 = 2⁷ = 128.

Question 2.
If C₀, C₁, C₂,… C_n are coefficients of expansion (1 + x)ⁿ then prove that:
C₀ + 3.C₁ + 5.C₂ + … + (2n + 1). C_n = (n + 1)2ⁿ.
Solution:
L.H.S.
= ⁿC₀ + 3.C₁ + 5.C₂ + … + (2n + 1) ⁿC_n
= ⁿC₀ + (2 + 1) ⁿC₁ + (4 + 1) ⁿC₂ + … (2n + 1) ⁿC_n
= ⁿC₀ + (2. ⁿC₁ + ⁿC₁ + (4. ⁿC₂ + ⁿC₂) + …  (2n.ⁿC_n + ⁿC_n)
= 2. ⁿC₁ + 4 ⁿC₂ + 6 ⁿC₃ + … + 2n ⁿC_n
+ (ⁿC₀ + ⁿC₁ + ⁿC₂ + ⁿC₃ + …ⁿC_n)
= 2[ⁿC₁ + 2 ⁿC₂ + 3.ⁿC₃ +… + ⁿC_n] + (1 + 1)ⁿ

Question 3.
C₀ C₂ + C₁ C₃ + C₂ C₄ + … + C_n _ ₂C_n= (2n)!/(n−2)!(n+2)!
Solution:
From Binomial theorem
(1 + x)ⁿ = C₀ + C₁x + C₂x² + … + C_rx^r  + … + C_nxⁿ … (i)
Again
(x + 1)ⁿ = C₀xⁿ + C₁x^n-1 + C₂x^n-2 + …+ C_rx^n–r+ …+ C_n-1 x + C_n …(ii)
Multiplying equation (i) and (ii) we get
(1 + x)²ⁿ = [C₀ + C₁x + C₂x² + … + C_rx^r  + … + C_nxⁿ]
× [C₀xⁿ + C₁x^n-1 + C₂x^n-2 + …+ C_rx^n–r+ …+ C_n-1 x + C_n]
Comparing cofficient of x^n–r in both sides, we get
C₀C_r + C₁C_r+₁+ C₂C_r +₂ + … + C_n _ _rC_n

Question 4.
C₀ + 2C₁ + 4C₂ + 6C₃ + … + 2nC_n = 1 + n2ⁿ.
Solution:
L.H.S.
= C₀ + 2C₁ + 4C₂ + 6C₃ + … + 2nC_n
= C₀ + 2[C₁ + 2C₂ + 3C₃ + … + nC_n]

Question 5.

Solution:
L.H.S.

Question 6.
If expansion of (1 + x – 2x²)⁶ is denoted by 1 + a₁x+ a₂x² + a₃x³ + … + a₁₂x¹² then prove that a₂ + a₄ + a₆ + … + a₁₂ = 31.
Solution :
Given expansion is :
(1 + x – 2x²)⁶ = 1 + a₁x+ a₂x² + a₃x³ + … + a₁₂x¹² … (i)
Putting x = – 1 in equation (i)
{1 + 1 – 2(1)²}⁶ = 1+ a₁ + a₂ + a₃ + … + a₁₂
⇒ (2 – 2)⁶ = 1 + a₁ + a₂ + a₃ + … + a₁₂
⇒ 1 + a₁ + a₂ + a₃ + … + a₁₂ = 0 … (ii)
Putting x = – 1 in equation (i)
{(1 – 1 – 2(-1)²}⁶ = 1 – a₁ + a₂ + a₃ + … + a₁₂
⇒ (-2)⁶ = 1 – a₁ + a₂ + a₃ + … + a₁₂
⇒ 1- a₁ + a₂ + a₃ + … + a₁₂ = 64       … (iii)
Adding equation (ii) and (iii) we get
⇒ 2 + 2a₂ + 2a₄ + … + a₁₂ = 64
⇒   2(1 + a₂ + a₄ + … + a₁₂ = 64
⇒  1 + a₂ + a₄ + … + a₁₂ = 64/2 =32
⇒ a₂ + a₄ + a₆ … + a₁₂ = 32 – 1 = 31
Hence, a₂ + a₄ + a₆ … + a₁₂ =31.
Hence proved.

Rajasthan Board RBSE Class 11 Maths Chapter 7 Binomial Theorem Ex 7.4

Question 1.
Expand the following Binomials upto fourth term:

Solution:
Expansion upto four terms of (1 + x²)^-2
  

Question 2.
Find the required terms in the following expansions:
(i) Fourth term of (1 – 3x)^-1/3
(ii) Seventh term of (1 + x)^5/2
(iii) Eighth term of (1 + 2x)^-1/2
Solution:
(i) Fourth term of (1 – 3x)^-1/3
= Fourth term of (1 – 3x)^-1/3
= Fourth term of {1 + (- 3x)}^– 1/3

(ii) Seventh term of (1 + x)^5/2

(iii) Eighth term of (1 + 2x)^-1/2

Question 3.
Find the general term of the following expansions:
(i) (a³ – x³)^2/3
(ii) (1 – 2x)^-3/2
(iii) (1 – x)^-p/q
Solution:
(i) General term in the expansion of (a³ – x³)^2/3

(ii) General term in the expansion of (1 – 2x)^-3/2

(iii) General term of in the expansion of (1 – x)^-p/q

Question 4.
If x < 3, find the coefficient of (3 – x)^-8 in the expansion of (3 – x)^{– 5}.
Solution:
(x + 1)th term in the expansion of (3 – x)^-8

Question 5.
Find the coefficient of x⁶ in the expansion of (a + 2bx²)^-3
Solution :
(r + 1)th term in the expansion of (a + 2bx²)^-3

For the coefficient of x⁶ in the this term
2r = 6 ⇒ r = 3
Hence, from equation (i), coefficient of x⁶ in the expansion of (a + 2bx²)^-3

Question 6.
Find the coefficient of x¹⁰ in the expansion of

Solution:

Question 7.
Find the coefficient of x^r in the expansion of (1 – 2x + 3x² – 4x³ + …)ⁿ and if x = 1/2 and n = 1, then find the value of the expression.
Solution:
we know that (1 + x)^-2 = 1 – 2x + 3x² – 4x³ +…+ (-1)^r (r+1)x^r+ ….. (i)
Given progression (1 – 2x + 3x² – 4x³ +…)ⁿ
(1 – 2x + 3x² – 4x³ +…)ⁿ
= {(1 + x)^-2}ⁿ  [From equation (i)]
= (1 + x)^-2n

Question 8.
Prove that (1 + x + x² + x³ + …)² = 1 + 2x + 3x² + …
Solution:
We know that
1 + x + x² + x³ + … = (1 – x)^-1  …. (i)
(1 + x + x² + x³ + …)²
= {(1 – x)^-1}²  [From equation (i)]
= (1 – x)²

Question 9.
Prove that (1 + x + x² + x³ +….)  (1 + 3x + 6x²+ …)
= (1 + 2x + 3x² + …)²
Solution:
We know that
(1 – x)^-1 = 1 + x + x² + x³ +…           … (i)
and (1 – x)^-3 = 1 + 3x + 6x² + 10x³ + … (ii)
L.H.S. = (1 + x + x² + x³ +…)(1 + 3x + 6x² + …)
= {(1 – x)^-1} {(1 – x)^-3}
= {(1 – x)²}^-2
= {(1 – x)^-2}²
= (1 + 2x + 3x² +…)²
= R.H.S.
Hence Proved.

Question 10.
If x = 2y + 3y² + 4y³ + …then express y in series of as ascending powers of x.
Solution:
⇒ x = 2y + 3y² + 4y² + …
⇒ x = (1 + 2y + 3y² + 4y3 + …)- 1
[∵ (1 – x)^-2 = 1 + 2x + 3x² + …]
⇒ (x+ 1) = (1 -y)^-2

Rajasthan Board RBSE Class 11 Maths Chapter 7 Binomial Theorem Ex 7.5

Question 1.
If y is two small as compared to x, then prove

Solution:

Question 2.
If x is too small such that x² and higher powers are negligible then find the value of following expressions :

Solution:

Ignoring x² and terms of higher order

Question 3.
Find the value of:
(i) √3o upto 4 places of decimal
(ii) (1.03)^1/3 upto 4 places of decimal
(iii) 1/(8.16)1/3 4 places of decimal
(iv) Cube root of 126 upto 5 places of decimal
Solution :
(i) Value of √30 upto 4 places of decimal

= 6 (1 – 0 08333 – 0 – 00347 – 0 – 00029 + …]
= 6 (1 – 0 – 08709] = 6(0 – 91291]
= 5.47746 = 5- 4775 (upto four places).

(ii) Value of (1.03)^1/3 upto 4 places of decimal (1.03)^1/3= (1 + 0-03)^1/3



(iv) Value of cube root of 126 upto 5 places of decimal Cube root of 126

Question 4.
If x is approximately equal to 1, then prove that :

Solution:

Question 5.
If p and q are approximately equal, then Prove that

Solution:
∵ p and q are approximately equal
∴      p = q + h
where h is too small whose higher order terms can be neglected.

Rajasthan Board RBSE Class 11 Maths Chapter 7 Binomial Theorem Ex 7.6

Question 1.

Solution:
On comparing the given series with

Question 2.

Solution:
On comparing the given series with the series

Question 3.

Solution:
On comparing the given series with

Question 4.

Solution:
On comparing the given series with

Question 5.

Solution:
On comparing the given series with

Hence, sum of the given series

Question 6.
Prove that :

Solution:
On comparing the given series with

Question 7.
Prove that:

Solution:
On comparing the given series with

Question 8.
Prove that:

Solution:
We Know that

Question 9.
Prove that:

Solution:
From the given series

Question 10.
Prove that

Solution:
R.H.S.

Rajasthan Board RBSE Class 11 Maths Chapter 7 Binomial Theorem Miscellaneous Exercise

Question 1.
The number of terms in the expansion of

(A) 11
(B) 13
(C) 10
(D) 14
Solution :
Terms of R.H.S. of the expansion of (x + a)ⁿ is finite for the positive values of x and number of terms is (n + 1).
So, value of n is 12 in the given expression.
Hence, number of total terms = 12 + 1 = 13.
Hence, option (B) is correct.

Question 2.
The 7th term in the expansion of


Solution :
7th term in the expansion of

Hence, option (C) is correct.

Question 3.
The middle term in the expansion of (a – x)ⁿ is
(A) 56a³ x⁵
(B) -56a³ x⁵
(C) 70a⁴ x⁴
(D) -70a⁴ x⁴
Solution :
If n is even in (a + b)n, then its middle term

Question 4.
The constant term in the expansion

(A) 5th
(B) 4th
(C) 6th
(D) 7th
Solution :
(r + 1)th term in the expansion of

For constant term
x⁹ – 3r = x⁰
⇒ 9 – 3r=0
⇒ r = 3
Hence, r + 1 = 3 + 1 = 4th term is the constant term
Hence, option (B) is correct, Ans.

Question 5.
The general term in the expansion of (x + a)ⁿ
(A) ⁿC_r x^n-r .a^r
(B) ⁿC_r x^r .a^r
(C) ⁿC_n-r x^n-r.a^r
(D) ⁿC_n-r x^r.a^n-r
Solution :
General term in the expansion of (x + a)ⁿ
⇒ T_r+1 = ⁿC_r x^n-r a^r
Hence, option (A) is correct.

Question 6.
The value of term independent of x in the expansion

(A) 264
(B) -264
(C) 7920
(D) -7920
Solution :
(r + 1)th term in the expansion of

Question 7.
The coefficient of x^-17 in the expansion of

(A) 1365
(B) -1365
(C) 3003
(D) -3003
Solution :
(x + 1)th term in the expansion of

Question 8.
If in the expansion of (1 + x)¹⁸, coefficients of (2r + 4)th and (r – 2)th terms are equal then value of r is:
(A) 5
(B) 6
(C) 7
(D) 8
Solution :
In the expansion of (1 + x)¹⁸ coefficients of (2r + 4)th and (r – 2)th terms are ¹⁸C_{2r + 3} and ¹⁸C_{r – 3} respectively
Given :
These are equal
Then, ¹⁸C_{2r + 3}= ¹⁸C_{r – 3}
⇒ 2r + 3 = r – 3
or 2r+ 3= 18 -(r – 3)
For using the statement if
ⁿC_r = ⁿC_p or ⁿC_n-p
⇒ 2r – r = -3 – 3
or 2r + 3 = 18 – r + 3
⇒ r = -6 or r = 6
r is a natural number, so r = – 6 is not possible, then x = 6.
Hence, option (B) is correct. .

Question 9.
If in the expansion of (a + b)ⁿ and (a + b)^{n + 3} ratio of 2nd and 3rd, 3rd and 4th terms are equal, then value of n is :
(A) 5
(B) 6
(C) 3
(D) 4
Solution :
In the expansion of (a + b)ⁿ


Hence, option (A) is correct.

Question 10.
If in the expansion of (1 + x)²ⁿ, coefficient of 3rd and (r + 2)th term are equal, then :
(A) n = 2r
(B) n = 2r – 1
(C) n = 2r + 1
(D) n = r + 1
Solution :
In the expansion of (1 + x)²ⁿ, 3rd term = 2ⁿC_{3r – 1} and r + 2th term = 2ⁿC_{r + 1}
According to question,
2ⁿC_{3r – 1} = 2ⁿC_{r + 1}
Now, 3r – 1 = r + 1 or 3r – 1 = 2n – r – 1
⇒ 3r – r – I + 1 or 3r + r = 2n – 1 + 1
⇒ 2r = 2 or 4r = 2n
⇒ r =1 or 2r = n
Hence, option (A) is correct.

Question 11.
Find the value of term independent of x in the expansion of (2x – 1/x)¹⁰ :
Solution :
(r + 1)th term of the expansion of (2x – 1/x)¹⁰

Hence, value of independent of x in the expansion of

Question 12.
Find the number of term in the expansion of (x + a)²⁰⁰+ (x – a)²⁰⁰ after simplification.
Solution :
∵ (x + a)ⁿ + (x – a)ⁿ
= 2[ⁿC₀ xⁿa⁰ + ⁿC₀ x^n-2 – a² + ⁿC₄x^{n-4 .}a⁴ +…]
where n is an even number then
[Number of terms in the expansion of (x + a)ⁿ+ (x – a)ⁿ]

Here, n = 200
Hence, number of terms in the expansion
Will be

Hence, number of terms is 101 in the expansion.

Question 13.
If the expansion of (1 + x)ⁿ, C₀ + C₁ + C₂ + C₃ + … C_n are coefficients different terms then find the value C₀ + C₂ + C₄… .
Solution :
(1 + x)ⁿ= ⁿC₀ 1ⁿ + ⁿC₁ 1ⁿ + ⁿC₁ 1^n-1 x^1
nC₂ 1^n-2 x² + ⁿC₃ 1^n-3x³ +….
Putting x = 1
(1 + 1)ⁿ = ⁿC₀ + ⁿC₁ +ⁿC₂ + ⁿC₃ + ….
Putting x = – 1
(1 – 1)ⁿ = ⁿC₀ – ⁿC₁ +ⁿC₂ – ⁿC₃ + ….
Here ⁿC₀ + ⁿC₁ +ⁿC₂ + ⁿC₃ + …. = 2ⁿ … (i)
“C0 – “Ci + ”C2 – “C3 + … = 0 … (ii)
Adding equation (i) and (ii)
2[ⁿC₀ + ⁿC₂ + ⁿC₄ + …] = 2ⁿ
⇒ ⁿC₀ + ⁿC₂ + ⁿC₄ + … = 2^{n – 1}
or C₀ + C₂ + C₄ + … = 2^{n – 1}

Question 14.
Find the value of
³⁰C₁ + ³⁰C₂ + ³⁰C₃ +… + ³⁰C₃₀ .
Solution :
(1 + x)ⁿ = ⁿC₀ + ⁿC₁x+ⁿC₂x²+ ⁿC₃x³+ ….
Putting x = 1
(1 + 1)ⁿ = ⁿC₀ + ⁿC₁ +ⁿC₂ + ⁿC₃ + …. +  ⁿC_n
or    2ⁿ = ⁿC₀ + ⁿC₁ +ⁿC₂ + ⁿC₃ + …. + ⁿC_n
Here, putting n = 30.
2³⁰ = 1+ ³⁰C₁ +³⁰C₂ + ³⁰C₃ + …. + ³⁰C₃o
⇒ 1+ ³⁰C₁ +³⁰C₂ + ³⁰C₃ + …. + ³⁰C₃o  = 2³⁰
⇒ ³⁰C₁ + ³⁰C₂ + ³⁰C₃ + … + ³⁰c₃₀ = 2³⁰ – 1

Question 15.
Find the middle term in the expansion of

Solution :

Question 16
In the product of expansion of (1 + 2x)⁶ (1 – x)⁷, find the coefficient of x⁵.
Solution :
Given expression
= (1 + 2x)⁶ (1 – x)⁷
∵ (1 + 2x)⁶ = ⁶C₀ + ⁶C₁ (2x)¹ + ⁶C₂ (2x)²
+ ⁶C₃ (2x)³ + ⁶C₄ (2x)⁴+ ⁶C₅ (2x)⁵ + ⁶C₆ (2x)⁶
= 1 + 6 × 2x + 15 × 4x² + 20 × 8x³
+ 15 × 16x⁴+ 6 × 32 x 5+ 1 × 64x⁶
or (1 + 2x)⁶ = 1 + 12x + 60x² + 160x³
+ 240x⁴+ 192x⁵ + 64x⁶ … (i)
[∵ ⁶C₀ = 1, ⁶C₁ = 6, ⁶C₂ = 15, ⁶C₃ = 20, ⁶C₄ = 15, ⁶C₅ = 6]
and (1 – x)⁷ = ⁷C₀ – ⁷C₁x + ⁷C₂x²– ⁷C₃x³ + ⁷C₄x⁴ – ⁷C₅x⁶ +,..
= 1 – 7x + 21x² – 35x³+ 35x⁴-21x⁵ +….. (ii)
[⁷C₀ = 1, ⁷C₁= 7, ⁷C₂ = 21, ⁷C₃ = 35, ⁷C₄= 35, ⁷C₅ = 21]
Now (1 + 2x)⁶ (1 – x)⁷
= [1 + 12x + 60x² + 160x³+ 240x⁴ + 192x⁴+ …]
= [1 – 7x + 21x² – 35x³ + 35x⁴ – 21x⁵ + …]
Coefficient of x⁵ in the above expansion
[- 21 + 12 x 35 – 60 × 35 + 160 × 21+ 240 (-7) + 192 × 1]
= [- 21 + 420 – 2100 + 3360 – 1680 + 192]
= [-3801 + 3972]= 171.

Question 17.
If in the expansion of (1 + x)²ⁿ coefficient of 2nd, 3nd and 4th terms are in A.P. then prove that 2n² – 9n + 7 = 0.
Solution :
In the expansion of (1 + x)²ⁿ, coefficient of 2nd, 3rd and 4th terms are
T₂= T₁ + 1= ²ⁿC₁
T₃ = T₂ + 1 = ²ⁿC₂
T_{4  =} T₃ +1 =  ²ⁿC₃
According to questions,

⇒ 6 (2n – 1) = 6 + 2(2n – 1) (n – 1)
⇒ 12n – 6 = 6 + 2(2n² – 3n + 1)
⇒ 4 n² – 6n + 2 – 12n + 6 + 6 = 0
⇒ 4n² – 18n +14 = 0
⇒ 2n² – 9n + 7 = 0.
Hence Proved.

Question 18.

Solution :

 

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