RBSE Solutions for Class 11 Maths Chapter 7 Binomial Theorem Miscellaneous Exercise

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 :

RBSE Solutions for Class 11 Maths