Binomial Theorem
Class 11th Mathematics RS Aggarwal Solution
- (1 – 2x)5 Using binomial theorem, expand each of the following:
- (2x – 3)6 Using binomial theorem, expand each of the following:
- (3x + 2y)5 Using binomial theorem, expand each of the following:
- (2x – 3y)4 Using binomial theorem, expand each of the following:
- ( {2x}/{3} - frac {3}/{2x} ) ^{6} Using binomial theorem, expand each of the…
- ( x^{2} - {3}/{x} ) ^{7} Using binomial theorem, expand each of the…
- ( x - {1}/{y} ) ^{5} Using binomial theorem, expand each of the following:…
- ( root {x} + sqrt{y} ) ^{8} Using binomial theorem, expand each of the…
- ( cube root {x} - root [3]{y} ) ^{6} Using binomial theorem, expand each of the…
- (1 + 2x – 3x2)4 Using binomial theorem, expand each of the following:…
- ( 1 + {x}/{2} - frac {2}/{x} ) ^{4} , x not equal 0 Using binomial theorem,…
- (3x2 – 2ax + 3a2)3 Using binomial theorem, expand each of the following:…
- ( root {2}+1 ) ^{6} + ( sqrt{2}-1 ) ^{6} Evaluate :
- ( root {3}+1 ) ^{5} - ( sqrt{3}-1 ) ^{5} Evaluate :
- ( 2 + root {3} ) ^{7} + ( 2 - sqrt{3} ) ^{7} Evaluate :
- ( root {3} + sqrt{2} ) ^{6} - ( sqrt{3} - sqrt{2} ) ^{6} Evaluate :…
- Prove that sum _ { r = 0 } ^{n}^{n}c_{r} c. 3^{t} = 4^{n}
- Using binominal theorem, evaluate each of the following :(i) (101)4 (ii)…
- Using binomial theorem, prove that (23n - 7n -1) is divisible by 49, where n…
- Prove that ( 2 + root {x} ) ^{4} + ( 2 - sqrt{x} ) ^{4} = 2 ( 16+24x+x^{2} )…
- Find the 7th term in the expansion of ( {4x}/{5} + frac {5}/{2x} ) ^{8} .…
- Find the 9th term in the expansion of ( {a}/{b} - frac {b}/{ 2a^{2} } )…
- Find the 16th term in the expansion of ( root {x} - sqrt{y} ) ^{17} .…
- Find the 13th term in the expansion of ( 9x - {1}/{ 3 root {x} } ) ^{18} ,…
- Find the coefficients of x7 and x8 in the expansion of ( 2 + {x}/{3} )…
- Find the ratio of the coefficient of x15 to the term independent of x in the…
- Show that the ratio of the coefficient of x10 in the expansion of (1 – x2)10…
- Find the term independent of x in the expansion of (91 + x + 2x3) (…
- Find the coefficient of x in the expansion of (1 – 3x + 7x2) (1 – x)16.…
- Find the coefficient of(i)x5 in the expansion of (x + 3)8(ii) x6 in the…
- Show that the term containing x3 does not exist in the expansion of ( 3x -…
- Show that the expansion of ( 2x^{2} - {1}/{x} ) ^{20} does not contain…
- Show that the expansion of ( x^{2} + {1}/{x} ) ^{12} does not contain…
- Write the general term in the expansion of(x2 – y)6
- Find the 5th term from the end in the expansion of ( x - {1}/{x} ) ^{12}…
- Find the 4th term from the end in the expansion of ( {4x}/{5} - frac…
- Find the 4th term from the beginning and end in the expansion of ( cube root…
- Find the middle term in the expansion of :(i) (3 + x)6(ii) ( {x}/{3} + 3y…
- (x2 + a2)5 Find the two middle terms in the expansion of :
- ( x^{4} - {1}/{ x^{3} } ) ^{11} Find the two middle terms in the expansion…
- ( {p}/{x} + frac {x}/{p} ) ^{9} Find the two middle terms in the expansion…
- ( 3x - { x^{3} }/{6} ) ^{9} Find the two middle terms in the expansion of :…
- ( 2x + {1}/{ 3x^{2} } ) ^{9} Find the term independent of x in the…
- ( { 3x^{2} }/{2} - frac {1}/{3x} ) ^{6} Find the term independent of x in…
- ( x - {1}/{ x^{2} } ) ^{3n} Find the term independent of x in the expansion…
- ( 3x - {2}/{ x^{2} } ) ^{15} Find the term independent of x in the…
- Find the coefficient of x5 in the expansion of (1 + x)3 (1 – x)6.…
- Find numerically the greatest term in the expansion of (2 + 3x)9, where x =…
- If the coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)2n…
- Find the 6th term of the expansion (y1/2 + x1/3)n, if the binomial coefficient…
- If the 17th and 18th terms in the expansion of (2 + a)50 are equal, find the…
- Find the coefficient of x4 in the expansion of (1 + x)n (1 – x)n. Deduce that…
- Prove that the coefficient of xn in the binomial expansion of (1 + x)2n is…
- Find the middle term in the expansion of ( {p}/{2} + 2 ) ^{8}…
- Show that the term independent of x in the expansion of ( x - {1}/{x} )…
- If the coefficients of x2 and x3 in the expansion of (3 + px)9 are the same…
- Show that the coefficient of x-3 in the expansion of ( x c. 2 ) ^{21} is…
- Show that the middle term in the expansion of is 252.
- Show that the coefficient of x4 in the expansion of is {405}/{256} .…
- Prove that there is no term involving x6 in the expansion of.
- Show that the coefficient of x4 in the expansion of (1 + 2x + x2)5 is 212.…
- Write the number of terms in the expansion of
- Which term is independent of x in the expansion of ( x - {1}/{ 3x^{2} } )…
- Write the coefficient of the middle term in the expansion of (1 + x)2n.…
- Write the coefficient of x7y2 in the expansion of (x + 2y)9
- If the coefficients of (r – 5)th and (2r – 1)th terms in the expansion of (1 +…
- Write the 4th term from the end in the expansion of ( {3}/{ x^{2} } - frac…
- Find the coefficient of xn in the expansion of (1 + x) (1 – x)n.
- In the binomial expansion of (a + b)n, the coefficients of the 4th and 13th…
- Find the positive value of m for which the coefficient of x2 in the expansion…
Exercise 10a
Question 1.Using binomial theorem, expand each of the following:
(1 – 2x)5
Answer:
To find: Expansion of (1 – 2x)5
Formula used: (i) 
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
We have, (1 – 2x)5
⇒ [ 5C0(1)5] + [5C1(1)5-1(-2x)1] + [5C2(1)5-2(-2x)2] + [5C3(1)5-3(-2x)3]+ [5C4(1)5-4(-2x)4] + [5C5(-2x)5]
⇒ 1 – 5(2x) + 10(4x2) – 10(8x3) + 5(16x4) – 1(32x5)
⇒ 1 – 10x + 40x2 – 80x3 + 80x4 – 32x5
On rearranging
Ans) –32x5 + 80x4 – 80x3 + 40x2 – 10x + 1
Question 2.
Using binomial theorem, expand each of the following:
(2x – 3)6
Answer:
To find: Expansion of (2x – 3)6
Formula used: (i)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
We have, (2x – 3)6
⇒ [6C0(2x)6]+[6C1(2x)6-1(-3)1]+[6C2(2x)6-2(-3)2]+[6C3(2x)6-3(-3)3]+ [6C4(2x)6-4(-3)4] + [6C5(2x)6-5(-3)5] + [6C6(-3)6]
⇒ [(1) (64x6)] – [(6)(32x5)(3)] + [15(16x4)(9)] – [20(8x3)(27)] + [15(4x2)(81)] – [(6)(2x)(243)] + [(1)(729)]
⇒ 64x6 – 576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729
Ans) 64x6 – 576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729
Question 3.
Using binomial theorem, expand each of the following:
(3x + 2y)5
Answer:
To find: Expansion of (3x + 2y)5
Formula used: (i)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
We have, (3x + 2y)5
⇒ [5C0(3x)5-0] + [5C1(3x)5-1(2y)1] + [5C2(3x)5-2(2y)2] + [5C3(3x)5-3(2y)3]+ [5C4(3x)5-4(2y)4] + [5C5(2y)5]
⇒ [1(243x5)] + [5(81x4)(2y)] + [10(27x3)(4y2)] + [10(9x2)(8y3)] + [5(3x)(16y4)] + [1(32y5)]
⇒ 243x5 + 810x4y + 1080x3y2 + 720x2y3 + 240xy4 + 32y5
Ans) 243x5 + 810x4y + 1080x3y2 + 720x2y3 + 240xy4 + 32y5
Question 4.
Using binomial theorem, expand each of the following:
(2x – 3y)4
Answer:
To find: Expansion of (2x – 3y)4
Formula used: (i)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
We have, (2x – 3y)4
⇒ [4C0(2x)4-0] + [4C1(2x)4-1(-3y)1] + [4C2(2x)4-2(-3y)2] + [4C3(2x)4-3(-3y)3]+ [4C4(-3y)4]
⇒ [1(16x4)] – [4(8x3)(3y)] + [6(4x2)(9y2)] – [4(2x)(27y3)] + [1(81y4)]
⇒ 16x4 – 96x3y + 216x2y2 – 216xy3 + 81y4
Ans) 16x4 – 96x3y + 216x2y2 – 216xy3 + 81y4
Question 5.
Using binomial theorem, expand each of the following:
Answer:
To find: Expansion of 
Formula used: (i)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
We have, 
Ans)
Question 6.
Using binomial theorem, expand each of the following:
Answer:
To find: Expansion of 
Formula used: (i) 
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
We have,
Ans)
Question 7.
Using binomial theorem, expand each of the following:
Answer:
To find: Expansion of 
Formula used: (i)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
We have,
⇒ 5C0(x)5-0 + 5C1(x)5-1
+ 5C2(x)5-2
+ 5C3(x)5-3
+ 5C4(x)5-4
+ 5C5
⇒
Ans) 
Question 8.
Using binomial theorem, expand each of the following:
Answer:
To find: Expansion of 
Formula used: (i)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
We have,
We can write 
as 
and 
as 
Now, we have to solve for 
Ans) 
+ 8
+ 28
+ 56
+ 70
+ 56
+ 28(x)1(y)3 + 8
+ (y)4
Question 9.
Using binomial theorem, expand each of the following:
Answer:
To find: Expansion of 
Formula used: (i)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
We have, 
We can write 
as 
and 
as 
Now, we have to solve for 
Ans)
Question 10.
Using binomial theorem, expand each of the following:
(1 + 2x – 3x2)4
Answer:
To find: Expansion of (1 + 2x – 3x2)4
Formula used: (i)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
We have, (1 + 2x – 3x2)4
Let (1+2x) = a and (-3x2) = b … (i)
Now the equation becomes (a + b)4
⇒ [ 4C0(a)4-0] + [4C1(a)4-1(b)1] + [4C2(a)4-2(b)2] + [4C3(a)4-3(b)3]+ [4C4(b)4]
⇒ [4C0(a)4] + [4C1(a)3(b)1] + [4C2(a)2(b)2] + [4C3(a)(b)3]+ [4C4(b)4]
(Substituting value of b from eqn. i )
(Substituting value of b from eqn. i )
… (ii)
We need the value of a4,a3 and a2, where a = (1+2x)
For (1+2x)4, Applying Binomial theorem
(1+2x)4⇒ 
⇒ 1 + 8x + 24x2 + 32x3 + 16x4
We have (1+2x)4 = 1 + 8x + 24x2 + 32x3 + 16x4 … (iii)
For (a+b)3 , we have formula a3+b3+3a2b+3ab2
For, (1+2x)3 , substituting a = 1 and b = 2x in the above formula
⇒ 13+ (2x) 3+3(1)2(2x) +3(1) (2x) 2
⇒ 1 + 8x3 + 6x + 12x2
⇒ 8x3 + 12x2 + 6x + 1 … (iv)
For (a+b)2 , we have formula a2+2ab+b2
For, (1+2x)2 , substituting a = 1 and b = 2x in the above formula
⇒ (1)2 + 2(1)(2x) + (2x)2
⇒ 1 + 4x + 4x2
⇒ 4x2 + 4x + 1 … (v)
Putting the value obtained from eqn. (iii),(iv) and (v) in eqn. (ii)
⇒ 1 + 8x + 24x2 + 32x3 + 16x4 - 96x5 - 144x4 - 72x3 - 12x2 + 216x6 + 216x5 + 54x4 -108x6 - 216x7 + 81x8
On rearranging
Ans) 81x8 - 216x7 + 108x6 + 120x5 - 74x4 - 40x3 + 12x2 +8x+ 1
Question 11.
Using binomial theorem, expand each of the following:
Answer:
To find: Expansion of 
Formula used: (i)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
We have,
Let 
= a and 
= b … (i)
Now the equation becomes (a + b)4
(Substituting value of b from eqn. i )
(Substituting value of a from eqn. i )
…(ii)
We need the value of a4,a3 and a2, where a =
For 
, Applying Binomial theorem
= 
On rearranging the above eqn.
We have, = 
+ 
x3 + 
x2 + 2x + 1
For, (a+b)3 , we have formula a3+b3+3a2b+3ab2
For, 
, substituting a = 1 and b = 
in the above formula
For, (a+b)2 , we have formula a2+2ab+b2
For, 
, substituting a = 1 and b = in the above formula
… (v)
Putting the value obtained from eqn. (iii),(iv) and (v) in eqn. (ii)
On rearranging
Ans) 
+ 
+ 
- 
+ 
Question 12.
Using binomial theorem, expand each of the following:
(3x2 – 2ax + 3a2)3
Answer:
To find: Expansion of (3x2 – 2ax + 3a2)3
Formula used: (i)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
We have, (3x2 – 2ax + 3a2)3
Let, (3x2 – 2ax) = p … (i)
The equation becomes (p + 3a2)3
Substituting the value of p from eqn. (i)
… (ii)
We need the value of p3 and p2, where p = 3x2 – 2ax
For, (a+b)3 , we have formula a3+b3+3a2b+3ab2
For, (3x2 – 2ax)3 , substituting a = 3x2 and b = –2ax in the above formula
⇒ 27x6 – 8a3x3 – 54ax5 + 36a2x4 … (iii)
For, (a+b)2 , we have formula a2+2ab+b2
For, (3x2 – 2ax)3 , substituting a = 3x2 and b = –2ax in the above formula
⇒ 9x4 – 12x3a + 4a2x2 … (iv)
Putting the value obtained from eqn. (iii) and (iv) in eqn. (ii)
⇒ 27x6 – 8a3x3 – 54ax5 + 36a2x4 + 81a2x4 – 108x3a3 + 36a4x2 + 81a4x2 – 54a5x + 27a6
On rearranging
Ans) 27x6 – 54ax5 + 117a2x4 – 116x3a3 + 117a4x2 – 54a5x + 27a6
Question 13.
Evaluate :
Answer:
To find: Value of 
Formula used: (i)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
(a+1)6 = 
⇒ 6C0a6 + 6C1a5 + 6C2a4 + 6C3a3 + 6C4a2 + 6C5a + 6C6 … (i)
(a-1)6 = 
⇒ 6C0a6 - 6C1a5 + 6C2a4 - 6C3a3 + 6C4a2 - 6C5a + 6C6 … (ii)
Adding eqn. (i) and (ii)
(a+1)6 + (a-1)6 = [6C0a6 + 6C1a5 + 6C2a4 + 6C3a3 + 6C4a2 + 6C5a + 6C6] + [6C0a6 - 6C1a5 + 6C2a4 - 6C3a3 + 6C4a2 - 6C5a + 6C6]
⇒ 2[6C0a6 + 6C2a4 + 6C4a2 + 6C6]
⇒ 2
⇒ 2[(1)a6 + (15)a4 + (15)a2 + (1)]
⇒ 2[a6 + 15a4 + 15a2 + 1] = (a+1)6 + (a-1)6
Putting the value of a = 
in the above equation
= 2[
6 + 154 + 15
2 + 1]
⇒ 2[8 + 15(4) + 15(2) + 1]
⇒ 2[8 + 60 + 30 + 1]
⇒ 2[99]
⇒ 198
Ans) 198
Question 14.
Evaluate :
Answer:
To find: Value of 
Formula used: (I)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
(a+1)5 = 5C0a5 + 5C1a5-11 + 5C2a5-212 + 5C3a5-313 + 5C4a5-414 + 5C515
⇒ 5C0a5 + 5C1a4 + 5C2a3 + 5C3a2 + 5C4a + 5C5… (i)
(a-1)5
⇒ 5C0a5 - 5C1a4 + 5C2a3 - 5C3a2 + 5C4a - 5C5 … (ii)
Substracting (ii) from (i)
(a+1)5 - (a-1)5 = [5C0a5 + 5C1a4 + 5C2a3 + 5C3a2 + 5C4a + 5C5] - [5C0a5 - 5C1a4 + 5C2a3 - 5C3a2 + 5C4a - 5C5]
⇒ 2[5C1a4 + 5C3a2 + 5C5]
⇒ 2
⇒ 2[(5)a4 + (10)a2 + (1)]
⇒ 2[5a4 + 10a2 + 1] = (a+1)5 - (a-1)5
Putting the value of a = 
in the above equation
= 2[5
4 + 10
2 + 1]
⇒ 2[(5)(9) + (10)(3) + 1]
⇒ 2[45+30+1]
⇒ 152
Ans) 152
Question 15.
Evaluate :
Answer:
To find: Value of 
Formula used: (i)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
(a+b)7 = 
⇒ 7C0a7 + 7C1a6b + 7C2a5b2 + 7C3a4b3 + 7C4a3b4 + 7C5a2b5 + 7C6a1b6 + 7C7b7 … (i)
(a-b)7 = 
⇒ 7C0a7 - 7C1a6b + 7C2a5b2 - 7C3a4b3 + 7C4a3b4 - 7C5a2b5 + 7C6a1b6 - 7C7b7 … (ii)
Adding eqn. (i) and (ii)
(a+b)7 + (a-b)7 = [7C0a7 + 7C1a6b + 7C2a5b2 + 7C3a4b3 + 7C4a3b4 + 7C5a2b5 + 7C6a1b6 + 7C7b7] + [7C0a7 - 7C1a6b + 7C2a5b2 - 7C3a4b3 + 7C4a3b4 - 7C5a2b5 + 7C6a1b6 - 7C7b7]
⇒ 2[7C0a7 + 7C2a5b2 + 7C4a3b4 + 7C6a1b6]
⇒ 2
⇒ 2[(1)a7 + (21)a5b2 + (35)a3b4 + (7)ab6]
⇒ 2[a7 + 21a5b2 + 35a3b4 + 7ab6] = (a+b)7 + (a-b)7
Putting the value of a = 2 and b = in the above equation
= 2
= 2[128 + 21(32)(3)+ 35(8)(9) + 7(2)(27)]
= 2[128 + 2016 + 2520 + 378]
= 10084
Ans) 10084
Question 16.
Evaluate :
Answer:
To find: Value of 
Formula used: (i) 
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
(a+b)6 = 6C0a6 + 6C1a6-1b + 6C2a6-2b2 + 6C3a6-3b3 + 6C4a6-4b4 + 6C5a6-5b5 + 6C6b6
⇒ 6C0a6 + 6C1a5b + 6C2a4b2 + 6C3a3b3 + 6C4a2b4 + 6C5ab5 + 6C6b6 … (i)
(a-b)6
⇒ 6C0a6 - 6C1a5b + 6C2a4b2 - 6C3a3b3 + 6C4a2b4 - 6C5ab5 + 6C6b6 … (ii)
Substracting (ii) from (i)
(a+b)6 - (a-b)6 = [6C0a6 + 6C1a5b + 6C2a4b2 + 6C3a3b3 + 6C4a2b4 + 6C5ab5 + 6C6b6] – [6C0a6 - 6C1a5b + 6C2a4b2 - 6C3a3b3 + 6C4a2b4 - 6C5ab5 + 6C6b6]
= 2[6C1a5b + 6C3a3b3 + 6C5ab5]
= 2
= 2[(6)a5b + (20)a3b3 + (6)ab5]
⇒ (a+b)6 - (a-b)6 = 2[(6)a5b + (20)a3b3 + (6)ab5]
Putting the value of a = and b = in the above equation
⇒ 2
⇒ 2
⇒ 
Ans) 
Question 17.
Prove that
Answer:
To prove: 
Formula used: 
Proof: In the above formula if we put a = 1 and b = 3, then we will get
Therefore,
Hence Proved.
Question 18.
Using binominal theorem, evaluate each of the following :
(i) (101)4 (ii) (98)4
(iii)(1.2)4
Answer:
(i) (101)4
To find: Value of (101)4
Formula used: (i)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
101 = (100+1)
Now (101)4 = (100+1)4
(100+1)4 = 
= 104060401
Ans) 104060401
(ii) (98)4
To find: Value of (98)4
Formula used: (I)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
98 = (100-2)
Now (98)4 = (100-2)4
(100-2)4
= 92236816
Ans) 92236816
(iii) (1.2)4
To find: Value of (1.2)4
Formula used: (i)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
1.2 = (1 + 0.2)
Now (1.2)4 = (1 + 0.2)4
(1+0.2)4
= 2.0736
Ans) 2.0736
Question 19.
Using binomial theorem, prove that (23n - 7n -1) is divisible by 49, where n N.
Answer:
To prove: (23n - 7n -1) is divisible by 49, where n N
Formula used: (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
(23n - 7n -1) = (23)n – 7n – 1
⇒ 8n - 7n – 1
⇒ (1+7)n – 7n – 1
⇒ nC01n + nC11n-17 + nC21n-272 + …… +nCn-17n-1 + nCn7n – 7n – 1
⇒ nC0 + nC17 + nC272 + …… +nCn-17n-1 + nCn7n – 7n – 1
⇒ 1 + 7n + 72[nC2 + nC37 + … + nCn-1 7n-3 + nCn 7n-2] -7n -1
⇒ 72[nC2 + nC37 + … + nCn-1 7n-3 + nCn 7n-2]
⇒ 49[nC2 + nC37 + … + nCn-1 7n-3 + nCn 7n-2]
⇒ 49K, where K = (nC2 + nC37 + … + nCn-1 7n-3 + nCn7n-2)
Now, (23n - 7n -1) = 49K
Therefore (23n - 7n -1) is divisible by 49
Question 20.
Prove that 
Answer:
To prove:
Formula used: (i)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn
(a+b)4 = 4C0a4 + 4C1a4-1b + 4C2a4-2b2 + 4C3a4-3b3 + 4C4b4
⇒ 4C0a4 + 4C1a3b + 4C2a2b2 + 4C3a1b3 + 4C4b4 … (i)
(a-b)4 = 4C0a4 + 4C1a4-1(-b) + 4C2a4-2(-b)2 +4C3a4-3(-b)3+4C4(-b)4
⇒ 4C0a4 - 4C1a3b + 4C2a2b2 - 4C3ab3 + 4C4b4 … (ii)
Adding (i) and (ii)
(a+b)4 + (a-b)7 = [4C0a4 + 4C1a3b + 4C2a2b2 + 4C3a1b3 + 4C4b4] + [4C0a4 - 4C1a3b + 4C2a2b2 - 4C3ab3 + 4C4b4]
⇒ 2[4C0a4 + 4C2a2b2 + 4C4b4]
⇒ 2
⇒ 2[(1)a4 + (6)a2b2 + (1)b4]
⇒ 2[a4 + 6a2b2 + b4]
Therefore, (a+b)4 + (a-b)7 = 2[a4 + 6a2b2 + b4]
Now, putting a = 2 and b = 
in the above equation.
= 2[(2)4 + 6(2)2 ()2 + ()4]
= 2(16+24x+x2)
Hence proved.
Question 21.
Find the 7th term in the expansion of
.
Answer:
To find: 7th term in the expansion of 
Formula used: (i)
(ii) Tr+1 = nCr an-r br
For 7th term, r+1=7
⇒ r = 6
In,
7th term = T6+1
⇒ (28)
⇒ 
Ans) 
Question 22.
Find the 9th term in the expansion of 
.
Answer:
To find: 9th term in the expansion of 
Formula used: (i)
(ii) Tr+1 = nCr an-r br
For 9th term, r+1=9
⇒ r = 8
In,
9th term = T8+1
⇒ 12C8
⇒ 
⇒ 495
⇒ 
Ans) 
Question 23.
Find the 16th term in the expansion of
.
Answer:
To find: 16th term in the expansion of 
Formula used: (i)
(ii) Tr+1 = nCr an-r br
For 16th term, r+1=16
⇒ r = 15
In, 
16th term = T15+1
⇒ -136x
Ans) -136x
Question 24.
Find the 13th term in the expansion of 
.
Answer:
To find: 13th term in the expansion of 
Formula used: (i)
(ii) Tr+1 = nCr an-r br
For 13th term, r+1=13
⇒ r = 12
In, 
13th term = T12+1
⇒ 18564
Question 25.
Find the coefficients of x7 and x8 in the expansion of 
.
Answer:
To find : coefficients of x7 and x8
Formula :
Here, a=2, 
We have,
To get a coefficient of x7, we must have,
x7 = xr
• r = 7
Therefore, the coefficient of x7
And to get the coefficient of x8 we must have,
x8 = xr
• r = 8
Therefore, the coefficient of x8
Conclusion :
• coefficient of x7
• coefficient of x8
Question 26.
Find the ratio of the coefficient of x15 to the term independent of x in the expansion of 
.
Answer:
To Find: the ratio of the coefficient of x15 to the term independent of x
Formula :
Here, a=x2, 
and n=15
We have a formula,
To get coefficient of x15 we must have,
(x)30-3r = x15
• 30 - 3r = 15
• 3r = 15
• r = 5
Therefore, coefficient of x15
Now, to get coefficient of term independent of xthat is coefficient of x0 we must have,
(x)30-3r = x0
• 30 - 3r = 0
• 3r = 30
• r = 10
Therefore, coefficient of x0
But 
……….
Therefore, the coefficient of x0
Therefore,
Hence, coefficient of x15 : coefficient of x0 = 1:32
Conclusion : The ratio of coefficient of x15 to coefficient of x0 = 1:32
Question 27.
Show that the ratio of the coefficient of x10 in the expansion of (1 – x2)10 and the term independent of x in the expansion of 
is 1 : 32.
Answer:
To Prove : coefficient of x10 in (1-x2)10: coefficient of x0 in 
= 1:32
For (1-x2)10 ,
Here, a=1, b=-x2 and n=15
We have formula,
To get coefficient of x10 we must have,
(x)2r = x10
• 2r = 10
• r = 5
Therefore, coefficient of x10
For ,
Here, a=x, 
and n=10
We have a formula,
Now, to get coefficient of term independent of xthat is coefficient of x0 we must have,
(x)10-2r = x0
• 10 - 2r = 0
• 2r = 10
• r = 5
Therefore, coefficient of x0
Therefore,
Hence,
coefficient of x10 in (1-x2)10: coefficient of x0 in = 1:32
Question 28.
Find the term independent of x in the expansion of (91 + x + 2x3) 
.
Answer:
To Find : term independent of x, i.e. coefficient of x0
Formula : 
We have a formula,
Therefore, the expansion of is given by,
Now,
Multiplying the second bracket by 91 , x and 2x3
In the first bracket, there will be a 6th term of x0 having coefficient 
While in the second and third bracket, the constant term is absent.
Therefore, the coefficient of term independent of x, i.e. constant term in the above expansion
=-91(2)5 (252)
Conclusion : coefficient of term independent of x =-91(2)5 (252)
Question 29.
Find the coefficient of x in the expansion of (1 – 3x + 7x2) (1 – x)16.
Answer:
To Find : coefficient of x
Formula :
We have a formula,
Therefore, expansion of (1-x)16 is given by,
Now,
Multiplying the second bracket by 1 , (-3x) and 7x2
In the above equation terms containing x are
and -3x
Therefore, the coefficient of x in the above expansion
=-16-3
=-19
Conclusion : coefficient of x =-19
Question 30.
Find the coefficient of
(i)x5 in the expansion of (x + 3)8
(ii) x6 in the expansion of 
.
(iii) x-15 in the expansion of 
.
(iv) a7b5 in the expansion of (a – 2b)12.
Answer:
(i) Here, a=x, b=3 and n=8
We have a formula,
To get coefficient of x5 we must have,
(x)8-r = x5
• 8 - r = 5
• r = 3
Therefore, coefficient of x5
= 1512
(ii) Here, a=3x2, 
and n=9
We have a formula,
To get coefficient of x6 we must have,
(x)18-3r = x6
• 18 - 3r = 6
• 3r = 12
• r = 4
Therefore, coefficient of x6
= 126×3
= 378
(iii) Here, a=3x2, 
and n=10
We have a formula,
To get coefficient of x-15 we must have,
(x)20-5r = x-15
• 20 - 5r = -15
• 5r = 35
• r = 7
Therefore, coefficient of x-15
But 
……….
Therefore, qthe coefficient of x-15
(iv) Here, a=a, b=-2b and n=12
We have formula,
To get coefficient of a7b5 we must have,
(a)12-r (b)r = a7b5
• r = 5
Therefore, coefficient of a7b5
= 792. (-32)
= -25344
Question 31.
Show that the term containing x3 does not exist in the expansion of 
.
Answer:
For 
,
a=3x, 
and n=8
We have a formula,
To get coefficient of x3 we must have,
(x)8-2r = (x)3
• 8 - 2r = 3
• 2r = 5
• r = 2.5
As 
is not possible
Therefore, the term containing x3 does not exist in the expansion of
Question 32.
Show that the expansion of 
does not contain any term involving x9.
Answer:
For 
,
a=2x2, 
and n=20
We have a formula,
To get coefficient of x9 we must have,
(x)40-3r = (x)9
• 40 - 3r = 9
• 3r = 31
• r = 10.3333
As 
is not possible
Therefore, the term containing x9 does not exist in the expansion of
Question 33.
Show that the expansion of 
does not contain any term involving x-1.
Answer:
For 
,
a=x2, 
and n=12
We have a formula,
To get coefficient of x-1 we must have,
(x)24-3r = (x)-1
• 24 - 3r = -1
• 3r = 25
• r = 8.3333
As 
is not possible
Therefore, the term containing x-1 does not exist in the expansion of
Question 34.
Write the general term in the expansion of
(x2 – y)6
Answer:
To Find : General term, i.e. tr+1
For (x2 - y)6
a=x2, b=-y and n=6
General term tr+1 is given by,
Conclusion : General term 
Question 35.
Find the 5th term from the end in the expansion of 
.
Answer:
To Find : 5th term from the end
Formulae :
•
• 
For 
,
a=x, and n=12
As n=12 ,therefore there will be total (12+1)=13 terms in the expansion
Therefore,
5th term from the end = (13-5+1)th i.e. 9th term from the starting.
We have a formula,
For t9 , r=8
……….
Therefore, a 5th term from the end
Conclusion : 5th term from the end
Question 36.
Find the 4th term from the end in the expansion of 
.
Answer:
To Find : 4th term from the end
Formulae :
•
•
For 
,
, 
and n=9
As n=9 ,therefore there will be total (9+1)=10 terms in the expansion
Therefore,
4th term from the end = (10-4+1)th, i.e. 7th term from the starting.
We have a formula,
For t7 , r=6
……….
Therefore, a 4th term from the end
Conclusion : 4th term from the end
Question 37.
Find the 4th term from the beginning and end in the expansion of 
.
Answer:
To Find :
I. 4th term from the beginning
II. 4th term from the end
Formulae :
•
•
For 
,
, 
and n=9
As n=n ,therefore there will be total (n+1) terms in the expansion
Therefore,
I. For the 4th term from the starting.
We have a formula,
For t4 , r=3
Therefore, a 4th term from the starting 
Now,
II. For the 4th term from the end
We have a formula,
For t(n-2) , r = (n-2)-1 = (n-3)
……….
Therefore, a 4th term from the end 
Conclusion :
I. 4th term from the beginning
II. 4th term from the end
Question 38.
Find the middle term in the expansion of :
(i) (3 + x)6
(ii) 
(iii) 
(iv) 
Answer:
(i) For (3 + x)6,
a=3, b=x and n=6
As n is even, 
is the middle term
Therefore, the middle term 
General term tr+1 is given by,
Therefore, for 4th , r=3
Therefore, the middle term is
= (20). (27) x3
= 540 x3
(ii) For 
,
, b=3y and n=8
As n is even, is the middle term
Therefore, the middle term 
General term tr+1 is given by,
Therefore, for 5th , r=4
Therefore, the middle term is
= (70). x4 y4
(iii) For 
,
, 
and n=10
As n is even, is the middle term
Therefore, the middle term 
General term tr+1 is given by,
Therefore, for 6th , r=5
Therefore, the middle term is
= -252
(iv) For 
,
a=x2, and n=10
As n is even, is the middle term
Therefore, the middle term
General term tr+1 is given by,
Therefore, for the 6th middle term, r=5
Therefore, the middle term is
= -252 (32) x5
= -8064 x5
Question 39.
Find the two middle terms in the expansion of :
(x2 + a2)5
Answer:
For (x2 + a2)5,
a= x2, b= a2 and n=5
As n is odd, there are two middle terms i.e.
I. 
and II. 
General term tr+1 is given by,
I. The first, middle term is 
Therefore, for the 3rd middle term, r=2
Therefore, the first middle term is
= 10. a4. x6
II. The second middle term is 
Therefore, for the 4th middle term, r=3
Therefore, the second middle term is
………
= 10. a6. x4
Question 40.
Find the two middle terms in the expansion of :
Answer:
For 
,
a= x4, 
and n=11
As n is odd, there are two middle terms i.e.
II. and II.
General term tr+1 is given by,
I. The first middle term is 
Therefore, for the 6th middle term, r=5
Therefore, the first middle term is
= - 462. x9
II. The second middle term is 
Therefore, for the 7th middle term, r=6
Therefore, the second middle term is
………
= 462. x2
Question 41.
Find the two middle terms in the expansion of :
Answer:
For 
,
, 
and n=9
As n is odd, there are two middle terms i.e.
I. and II.
General term tr+1 is given by,
I. The first middle term is 
Therefore, for 5th middle term, r=4
Therefore, the first middle term is
= 126p. x-1
II. The second middle term is 
Therefore, for the 6th middle term, r=5
Therefore, the second middle term is
………
Question 42.
Find the two middle terms in the expansion of :
Answer:
For 
,
a=3x, 
and n=9
As n is odd, there are two middle terms i.e.
I. and II.
General term tr+1 is given by,
I. The first middle term is
Therefore, for 5th middle term, r=4
Therefore, the first middle term is
II. The second middle term is
Therefore, for the 6th middle term, r=5
Therefore, the second middle term is
………
Question 43.
Find the term independent of x in the expansion of :
Answer:
To Find : term independent of x, i.e. x0
For 
a=2x, 
and n=9
We have a formula,
Now, to get coefficient of term independent of xthat is coefficient of x0 we must have,
(x)9-3r = x0
• 9 - 3r = 0
• 3r = 9
• r = 3
Therefore, coefficient of x0
Conclusion : coefficient of x0
Question 44.
Find the term independent of x in the expansion of :
Answer:
To Find : term independent of x, i.e. x0
For 
, 
and n=6
We have a formula,
Now, to get coefficient of term independent of xthat is coefficient of x0 we must have,
(x)12-3r = x0
• 12 - 3r = 0
• 3r = 12
• r = 4
Therefore, coefficient of x0
………
Conclusion : coefficient of x0
Question 45.
Find the term independent of x in the expansion of :
Answer:
To Find : term independent of x, i.e. x0
For 
a=x, 
and N=3n
We have a formula,
Now, to get coefficient of term independent of xthat is coefficient of x0 we must have,
(x)3n-3r = x0
• 3n - 3r = 0
• 3r = 3n
• r = n
Therefore, coefficient of x0
Conclusion : coefficient of x0
Question 46.
Find the term independent of x in the expansion of :
Answer:
To Find : term independent of x, i.e. x0
For 
a=3x, 
and n=15
We have a formula,
Now, to get coefficient of term independent of xthat is coefficient of x0 we must have,
(x)15-3r = x0
• 15 - 3r = 0
• 3r = 15
• r = 5
Therefore, coefficient of x0
Conclusion : coefficient of x0
Question 47.
Find the coefficient of x5 in the expansion of (1 + x)3 (1 – x)6.
Answer:
To Find : coefficient of x5
For (1+x)3
a=1, b=x and n=3
We have a formula,
For (1-x)6
a=1, b=-x and n=6
We have formula,
We have a formula ,
By using this formula, we get,×
Coefficients of x5 are
x0.x5 = 1× (-6)=-6
x1.x4 = 3×15=45
x2.x3 = 3×(-20)=-60
x3.x2 = 1×15=15
Therefore, Coefficients of x5 = -6+45-60+15 = -6
Conclusion : Coefficients of x5 = -6
Question 48.
Find numerically the greatest term in the expansion of (2 + 3x)9, where 
.
Answer:
To Find : numerically greatest term
For (2+3x)9,
a=2, b=3x and n=9
We have relation,
tr+1 ≥ tr or 
we have a formula,
At x = 3/2
• 90 ≥ 13r
• r ≤ 6.923
therefore, r=6 and hence the 7th term is numerically greater.
By using formula,
Conclusion : the 7th term is numerically greater with value 
Question 49.
If the coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)2n are in AP, show that 2n2 – 9n + 7 = 0.
Answer:
For (1 + x)2n
a=1, b=x and N=2n
We have, 
For the 2nd term, r=1
………
Therefore, the coefficient of 2nd term = (2n)
For the 3rd term, r=2
……….(n! = n. (n-1)!)
Therefore, the coefficient of 3rd term = (n)(2n-1)
For the 4th term, r=3
……….(n! = n. (n-1)!)
Therefore, the coefficient of 3rd term 
As the coefficients of 2nd, 3rd and 4th terms are in A.P.
Therefore,
2×coefficient of 3rd term = coefficient of 2nd term + coefficient of the 4th term
Dividing throughout by (2n),
• 3 (2n-1) = 3 + (2n-1)(n-1)
• 6n – 3 = 3 + (2n2 - 2n – n + 1)
• 6n – 3 = 3 + 2n2 - 3n + 1
• 3 + 2n2 - 3n + 1 - 6n + 3 = 0
• 2n2 - 9n + 7 = 0
Conclusion : If the coefficients of 2nd, 3rd and 4th terms of (1 + x)2n are in A.P. then 2n2 - 9n + 7 = 0
Question 50.
Find the 6th term of the expansion (y1/2 + x1/3)n, if the binomial coefficient of the 3rd term from the end is 45.
Answer:
Given : 3rd term from the end =45
To Find : 6th term
For (y1/2 + x1/3)n,
a= y1/2, b= x1/3
We have,
As n=n, therefore there will be total (n+1) terms in the expansion.
3rd term from the end = (n+1-3+1)th i.e. (n-1)th term from the starting
For (n-1)th term, r = (n-1-1) = (n-2)
………..
Therefore 3rd term from the end 
Therefore coefficient 3rd term from the end 
• 90 = n (n-1)
• 10 (9) = n (n-1)
Comparing both sides, n=10
For 6th term, r=5
Conclusion : 6th term 
Question 51.
If the 17th and 18th terms in the expansion of (2 + a)50 are equal, find the value of a.
Answer:
Given : 
To Find : value of a
For (2 + a)50
A=2, b=a and n=50
We have, 
For the 17th term, r=16
For the 18th term, r=17
As 17th and 18th terms are equal
……..
……..
• a = 1122
Conclusion : value of a = 1122
Question 52.
Find the coefficient of x4 in the expansion of (1 + x)n (1 – x)n. Deduce that C2 = C0C4 – C1C3 + C2C2 – C3C1 + C4C0, where Cr stands for nCr.
Answer:
To Find : Coefficients of x4
For (1+x)n
a=1, b=x
We have a formula,
For (1-x)n
a=1, b=-x and n=n
We have formula,
Coefficients of x4 are
x0.x4
= C0C4
x1.x3
= - C1C3
x2.x2
= C2C2
x3.x1
= - C3C1
x4.x0
= C4C0
Therefore, Coefficient of x4
= C4C0 - C1C3 + C2C2 - C3C1 + C4C0
Let us assume, n=4, it becomes
4C44C0 - 4C14C3 + 4C24C2 - 4C34C1 + 4C44C0
We know that ,
By using above formula, we get,
4C44C0 - 4C14C3 + 4C24C2 - 4C34C1 + 4C44C0
= (1)(1) – (4)(4) + (6)(6) – (4)(4) + (1)(1)
= 1 – 16 + 36 – 16 + 1
= 6
= 4C2
Therefore, in general,
C4C0 - C1C3 + C2C2 - C3C1 + C4C0 = C2
Therefore, Coefficient of x4 = C2
Conclusion :
• Coefficient of x4 = C2
• C4C0 - C1C3 + C2C2 - C3C1 + C4C0 = C2
Question 53.
Prove that the coefficient of xn in the binomial expansion of (1 + x)2n is twice the coefficient of xn in the binomial expansion of (1 + x)2n-1.
Answer:
To Prove : coefficient of xn in (1+x)2n = 2 × coefficient of xn in (1+x)2n-1
For (1+x)2n,
a=1, b=x and m=2n
We have a formula,
To get the coefficient of xn, we must have,
xn = xr
• r = n
Therefore, the coefficient of xn
………
………..
………cancelling n
Therefore, the coefficient of xn in (1+x)2n
………eq(1)
Now for (1+x)2n-1,
a=1, b=x and m=2n-1
We have formula,
To get the coefficient of xn, we must have,
xn = xr
• r = n
Therefore, the coefficient of xn in (1+x)2n-1
…..multiplying and dividing by 2
Therefore,
coefficient of xn in (1+x)2n-1 = � × coefficient of xn in (1+x)2n
or coefficient of xn in (1+x)2n = 2 × coefficient of xn in (1+x)2n-1
Hence proved.
Question 54.
Find the middle term in the expansion of 
Answer:
Given : 
, b=2 and n=8
To find : middle term
Formula :
• The middle term 
• 
Here, n is even.
Hence,
Therefore, 
tthe erm is the middle term.
For 
, r=4
We have,
Conclusion : The middle term is 
.
Exercise 10b
Question 1.Show that the term independent of x in the expansion of 
is -252.
Answer:
To show: the term independent of x in the expansion of 
is -252.
Formula Used:
General term, Tr+1 of binomial expansion
is given by,
Tr+1
nCr xn-r yr where
nCr
Now, finding the general term of the expression, , we get
Tr+110Cr
For finding the term which is independent of x,
10-2r=5
r=5
Thus, the term which would be independent of x is T6
T610C5
T610C5
T6
10C5
T6
T6
T6
T6
T6=252
Thus, the term independent of x in the expansion of is -252.
Question 2.
If the coefficients of x2 and x3 in the expansion of (3 + px)9 are the same then prove that 
.
Answer:
To prove: that. If the coefficients of x2 and x3 in the expansion of (3 + px)9 are the same then .
Formula Used:
General term, Tr+1 of binomial expansionis given by,
Tr+1nCr xn-r yr where
nCr
Now, finding the general term of the expression, (3 + px)9 , we get
Tr+19Cr
For finding the term which has 
in it, is given by
r=2
Thus, the coefficients of x2 are given by,
T39C2
T39C2
For finding the term which has in it, is given by
r=3
Thus, the coefficients of x3 are given by,
T39C3
T39C3
As the coefficients of x2 and x3 in the expansion of (3 + px)9 are the same.
9C3
9C2
9C3
9C2
Thus, the value of p for which coefficients of x2 and x3 in the expansion of (3 + px)9 are the same is 
Question 3.
Show that the coefficient of x-3 in the expansion of 
is -330.
Answer:
To show: that the coefficient of x-3 in the expansion of is -330.
Formula Used:
General term, Tr+1 of binomial expansionis given by,
Tr+1nCr xn-r yr where
nCr
Now, finding the general term of the expression, , we get
Tr+111Cr
For finding the term which has 
in it , is given by
11-2r=3
2r=14
r=7
Thus, the term which the term which has in it isT8
T811C7
T8
11C7
T8
T6
T6=-330
Thus, the coefficient of x-3 in the expansion of is -330.
Question 4.
Show that the middle term in the expansion of 
is 252.
Answer:
To show: that the middle term in the expansion of is 252.
Formula Used:
General term, Tr+1 of binomial expansionis given by,
Tr+1nCr xn-r yr where
nCr
Total number of terms in the expansion is 11
Thus, the middle term of the expansion is T6 and is given by,
T6
10C5
T610C5
T6
T6=252
Thus, the middle term in the expansion of 
is 252.
Question 5.
Show that the coefficient of x4 in the expansion of 
is 
.
Answer:
To show: that the coefficient of x4 in the expansion of 
is -330.
Formula Used:
General term, Tr+1 of binomial expansionis given by,
Tr+1nCr xn-r yr where
nCr
Now, finding the general term of the expression, , we get
Tr+110Cr
For finding the term which has 
in it, is given by
10-3r=4
3r=6
r=2
Thus, the term which has in it isT3
T310C2
T3
T3
T3
Thus, the coefficient of x4 in the expansion of is 
Question 6.
Prove that there is no term involving x6 in the expansion of
.
Answer:
To prove: that there is no term involving x6 in the expansion of
Formula Used:
General term, Tr+1 of binomial expansionis given by,
Tr+1nCr xn-r yr where
nCr
Now, finding the general term of the expression, , we get
Tr+111Cr
For finding the term which has 
in it, is given by
22-2r-r=6
3r=16
Since, 
is not possible as r needs to be a whole number
Thus, there is no term involving x6 in the expansion of
.
Question 7.
Show that the coefficient of x4 in the expansion of (1 + 2x + x2)5 is 212.
Answer:
To show: that the coefficient of x4 in the expansion of (1 + 2x + x2)5 is 212.
Formula Used:
We have,
(1 + 2x + x2)5=(1 +x+ x+ x2)5
=(1 +x+ x(1+x))5
=(1 +x)5(1 +x)5
=(1 +x)10
General term, Tr+1 of binomial expansionis given by,
Tr+1nCr xn-r yr where s
nCr
Now, finding the general term,
Tr+110Cr
10-r=4
r=6
Thus, the coefficient of x4 in the expansion of (1 + 2x + x2)5 is given by,
10C4
10C4
10C4=210
Thus, the coefficient of x4 in the expansion of (1 + 2x + x2)5 is 210
Question 8.
Write the number of terms in the expansion of 
Answer:
To find: the number of terms in the expansion of
Formula Used:
Binomial expansion of is given by,
Thus,
So, the no. of terms left would be 6
Thus, the number of terms in the expansion of is 6
Question 9.
Which term is independent of x in the expansion of
?
Answer:
To find: the term independent of x in the expansion of
?
Formula Used:
A general term, Tr+1 of binomial expansionis given by,
Tr+1nCr xn-r yr where
nCr
Now, finding the general term of the expression, , we get
Tr+19Cr
Tr+19Cr
Tr+19Cr
For finding the term which is independent of x,
9-3r=0
r=3
Thus, the term which would be independent of x is T4
Thus, the term independent of x in the expansion of is T4 i.e 4th term
Question 10.
Write the coefficient of the middle term in the expansion of (1 + x)2n.
Answer:
To find: that the middle term in the expansion of is 252.
Formula Used:
A general term, Tr+1 of binomial expansionis given by,
Tr+1nCr xn-r yr where
nCr
Total number of terms in the expansion is 11
Thus, the middle term of the expansion is T6 and is given by,
T610C5
T610C5
T6
T6=252
Thus, the middle term in the expansion of is 252.
Question 11.
Write the coefficient of x7y2 in the expansion of (x + 2y)9
Answer:
To find: the coefficient of x7y2 in the expansion of (x + 2y)9
Formula Used:
A general term, Tr+1 of binomial expansionis given by,
Tr+1nCr xn-r yr where
nCr
Now, finding the general term of the expression, (x + 2y)9, we get
Tr+19Cr
The value of r for which coefficient of x7y2 is defined
r=2
Hence, the coefficient of x7y2 in the expansion of (x + 2y)9 is given by:
T39C3
T39C3
T3
T3
T3=336
Thus, the coefficient of x7y2 in the expansion of (x + 2y)9 is 336
Question 12.
If the coefficients of (r – 5)th and (2r – 1)th terms in the expansion of (1 + x)34 are equal, find the value of r.
Answer:
To find: the value of r with respect to the binomial expansion of (1 + x)34 where the coefficients of the (r – 5)th and (2r – 1)th terms are equal to each other
Formula Used:
The general term, Tr+1 of binomial expansionis given by,
Tr+1nCr xn-r yr where
nCr
Now, finding the (r – 5)th term, we get
Tr-534Cr-6
Thus, the coefficient of (r – 5)th term is 34Cr-6
Now, finding the (2r – 1)th term, we get
T2r-134C2r-2
Thus, coefficient of (2r – 1)th term is 34C2r-2
As the coefficients are equal, we get
34C2r-234Cr-6
2r-2=r-6
r=-4
Value of r=-4 is not possible
2r-2+r-6=34
3r=42
r=14
Thus, value of r is 14
Question 13.
Write the 4th term from the end in the expansion of 
Answer:
To find: 4th term from the end in the expansion of 
Formula Used:
A general term, Tr+1 of binomial expansionis given by,
Tr+1nCr xn-r yr where
nCr
Total number of terms in the expansion is 8
Thus, the 4th term of the expansion is T5 and is given by,
T57C5
T5
T5
T5
Thus, a 4th term from the end in the expansion of is T5
Question 14.
Find the coefficient of xn in the expansion of (1 + x) (1 – x)n.
Answer:
To find: the coefficient of xn in the expansion of (1 + x) (1 – x)n.
Formula Used:
Binomial expansion of is given by,
Thus,
Thus, the coefficient of 
is,
nCn-nCn-1 (If n is even)
-nCn+nCn-1 (If n is odd)
Thus, the coefficient of is,nCn-nCn-1 (If n is even)and -nCn+nCn-1 (If n is odd)
Question 15.
In the binomial expansion of (a + b)n, the coefficients of the 4th and 13th terms are equal to each other. Find the value of n.
Answer:
To find: the value of n with respect to the binomial expansion of (a + b)n where the coefficients of the 4th and 13th terms are equal to each other
Formula Used:
A general term, Tr+1 of binomial expansionis given by,
Tr+1nCr xn-r yr where
nCr
Now, finding the 4th term, we get
T4nC3
Thus, the coefficient of a 4th term is nC3
Now, finding the 13th term, we get
T13nC12
Thus, coefficient of 4th term is nC12
As the coefficients are equal, we get
nC12= nC3
Also, nCr= nCn-r
nCn-12=nC3
n-12=3
n=15
Thus, value of n is 15
Question 16.
Find the positive value of m for which the coefficient of x2 in the expansion of (1 + x)m is 6.
Answer:
To find: the positive value of m for which the coefficient of x2 in the expansion of (1 + x)m is 6.
Formula Used:
General term, Tr+1 of binomial expansionis given by,
Tr+1nCr xn-r yr where
nCr
Now, finding the general term of the expression, (1 + x)m , we get
Tr+1mCr
Tr+1mCr
The coefficient of 
is mC2
mC2=6
=6
m=3,-2
Since m cannot be negative. Therefore,
m=3
Thus, positive value of m is 3 for which the coefficient of x2 in the expansion of (1 + x)m is 6