# RD Sharma Solutions for Class 11 Chapter 18 - Binomial Theorem

An algebraic expression containing two terms is called a binomial expression. The RD Sharma Class 11 Maths solutions are developed by our experienced faculty team at BYJUâ€™S, which is created for the purpose of clarifying studentâ€™s doubts, as per their convenience. These solutions also provide guidance to students for solving problems confidently, which in turn, helps in improving their problem-solving skills, that is very important from the examination point of view. For students wishing to learn the right steps of solving such problems, the RD Sharma Class 11 Maths Solutions is made available in the pdf format, students can easily download the pdf from the links given below.

Chapter 18 – Binomial Theorem contains two exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

• Binomial theorem for positive integral index.
• Some important conclusions from the binomial theorem.
• General term and middle terms in a binomial expansion.

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EXERCISE 18.1 PAGE NO: 18.11

1. Using binomial theorem, write down the expressions of the following:

Solution:

(i) (2x + 3y) 5

Let us solve the given expression:

(2x + 3y) 5 = 5C0 (2x)5 (3y)0 + 5C1 (2x)4 (3y)1 + 5C2 (2x)3 (3y)2 + 5C3 (2x)2 (3y)3 + 5C4 (2x)1 (3y)4 + 5C5 (2x)0 (3y)5

= 32x5 + 5 (16x4) (3y) + 10 (8x3) (9y)2 + 10 (4x)2 (27y)3 + 5 (2x) (81y4) + 243y5

= 32x5 + 240x4y + 720x3y2 + 1080x2y3 + 810xy4 + 243y5

(ii) (2x â€“ 3y) 4

Let us solve the given expression:

(2x â€“ 3y) 4 = 4C0 (2x)4 (3y)04C1 (2x)3 (3y)1 + 4C2 (2x)2 (3y)24C3 (2x)1 (3y)3 + 4C4 (2x)0 (3y)4

= 16x4 â€“ 4 (8x3) (3y) + 6 (4x2) (9y2) â€“ 4 (2x) (27y3) + 81y4

= 16x4 â€“ 96x3y + 216x2y2 â€“ 216xy3 + 81y4

(iv) (1 â€“ 3x) 7

Let us solve the given expression:

(1 â€“ 3x) 7 = 7C0 (3x)07C1 (3x)1 + 7C2 (3x)27C3 (3x)3 + 7C4 (3x)4 –Â  7C5 (3x)5 +Â 7C6 (3x)67C7 (3x)7

= 1 â€“ 7 (3x) + 21 (9x)2 â€“ 35 (27x3) + 35 (81x4) â€“ 21 (243x5) + 7 (729x6) â€“ 2187(x7)

= 1 â€“ 21x + 189x2 â€“ 945x3 + 2835x4 â€“ 5103x5 + 5103x6 â€“ 2187x7

(viii) (1 + 2x â€“ 3x2)5

Let us solve the given expression:

Let us consider (1 + 2x) and 3x2 as two different entities and apply the binomial theorem.

(1 + 2x â€“ 3x2)5 = 5C0 (1 + 2x)5 (3x2)05C1 (1 + 2x)4 (3x2)1 + 5C2 (1 + 2x)3 (3x2)25C3 (1 + 2x)2 (3x2)3 + 5C4 (1 + 2x)1 (3x2)45C5 (1 + 2x)0 (3x2)5

= (1 + 2x)5 â€“ 5(1 + 2x)4 (3x2) + 10 (1 + 2x)3 (9x4) â€“ 10 (1 + 2x)2 (27x6) + 5 (1 + 2x) (81x8) â€“ 243x10

= 5C0 (2x)0 + 5C1 (2x)1 + 5C2 (2x)2 + 5C3 (2x)3 + 5C4 (2x)4 + 5C5 (2x)5 â€“ 15x2 [4C0 (2x)0 + 4C1 (2x)1 + 4C2 (2x)2 + 4C3 (2x)3 + 4C4 (2x)4] + 90x4 [1 + 8x3 + 6x + 12x2] â€“ 270x6(1 + 4x2 + 4x) + 405x8 + 810x9 â€“ 243x10

= 1 + 10x + 40x2 + 80x3 + 80x4 + 32x5 â€“ 15x2 â€“ 120x3 â€“ 3604 â€“ 480x5 â€“ 240x6 + 90x4 + 720x7 + 540x5 + 1080x6 â€“ 270x6 â€“ 1080x8 â€“ 1080x7 + 405x8 + 810x9 â€“ 243x10

= 1 + 10x + 25x2 â€“ 40x3 â€“ 190x4 + 92x5 + 570x6 â€“ 360x7 â€“ 675x8 + 810x9 â€“ 243x10

(x) (1 â€“ 2x + 3x2)3

Let us solve the given expression:

2. Evaluate the following:

Solution:

Let us solve the given expression:

= 2 [5C0 (2âˆšx)0 + 5C2 (2âˆšx)2 + 5C4 (2âˆšx)4]

= 2 [1 + 10 (4x) + 5 (16x2)]

= 2 [1 + 40x + 80x2]

Let us solve the given expression:

= 2 [6C0 (âˆš2)6 + 6C2 (âˆš2)4 + 6C4 (âˆš2)2 + 6C6 (âˆš2)0]

= 2 [8 + 15 (4) + 15 (2) + 1]

= 2 [99]

= 198

Let us solve the given expression:

= 2 [5C1 (34) (âˆš2)1 + 5C3 (32) (âˆš2)3 + 5C5 (30) (âˆš2)5]

= 2 [5 (81) (âˆš2) + 10 (9) (2âˆš2) + 4âˆš2]

= 2âˆš2 (405 + 180 + 4)

= 1178âˆš2

Let us solve the given expression:

= 2 [7C0 (27) (âˆš3)0 + 7C2 (25) (âˆš3)2 + 7C4 (23) (âˆš3)4 + 7C6 (21) (âˆš3)6]

= 2 [128 + 21 (32)(3) + 35(8)(9) + 7(2)(27)]

= 2 [128 + 2016 + 2520 + 378]

= 2 [5042]

= 10084

Let us solve the given expression:

= 2 [5C1 (âˆš3)4 + 5C3 (âˆš3)2 + 5C5 (âˆš3)0]

= 2 [5 (9) + 10 (3) + 1]

= 2 [76]

= 152

Let us solve the given expression:

= (1 â€“ 0.01)5 + (1 + 0.01)5

= 2 [5C0 (0.01)0 + 5C2 (0.01)2 + 5C4 (0.01)4]

= 2 [1 + 10 (0.0001) + 5 (0.00000001)]

= 2 [1.00100005]

= 2.0020001

Let us solve the given expression:

= 2 [6C1 (âˆš3)5 (âˆš2)1 + 6C3 (âˆš3)3 (âˆš2)3 + 6C5 (âˆš3)1 (âˆš2)5]

= 2 [6 (9âˆš3) (âˆš2) + 20 (3âˆš3) (2âˆš2) + 6 (âˆš3) (4âˆš2)]

= 2 [âˆš6 (54 + 120 + 24)]

= 396 âˆš6

= 2 [a8 + 6a6 â€“ 6a4 + a4 + 1 â€“ 2a2]

= 2a8 + 12a6 â€“ 10a4 â€“ 4a2 + 2

3. Find (a + b) 4 â€“ (a â€“ b) 4. Hence, evaluate (âˆš3 + âˆš2)4 â€“ (âˆš3 – âˆš2)4.

Solution:

Firstly, let us solve the given expression:

(a + b) 4 â€“ (a â€“ b) 4

The above expression can be expressed as,

(a + b) 4 â€“ (a â€“ b) 4 = 2 [4C1 a3b1 + 4C3 a1b3]

= 2 [4a3b + 4ab3]

= 8 (a3b + ab3)

Now,

Let us evaluate the expression:

(âˆš3 + âˆš2)4 â€“ (âˆš3 -âˆš2)4

So consider, a = âˆš3 and b = âˆš2 we get,

(âˆš3 + âˆš2)4 â€“ (âˆš3 -âˆš2)4 = 8 (a3b + ab3)

= 8 [(âˆš3)3 (âˆš2) + (âˆš3) (âˆš2)3]

= 8 [(3âˆš6) + (2âˆš6)]

= 8 (5âˆš6)

= 40âˆš6

4. Find (x + 1) 6 + (x â€“ 1) 6. Hence, or otherwise evaluate (âˆš2 + 1)6 + (âˆš2 â€“ 1)6.

Solution:

Firstly, let us solve the given expression:

(x + 1) 6 + (x â€“ 1) 6

The above expression can be expressed as,

(x + 1) 6 + (x â€“ 1) 6 = 2 [6C0 x6 + 6C2 x4 + 6C4 x2 + 6C6 x0]

= 2 [x6 + 15x4 + 15x2 + 1]

Now,

Let us evaluate the expression:

(âˆš2 + 1)6 + (âˆš2 â€“ 1)6

So consider, x = âˆš2 then we get,

(âˆš2 + 1)6 + (âˆš2 â€“ 1)6 = 2 [x6 + 15x4 + 15x2 + 1]

= 2 [(âˆš2)6 + 15 (âˆš2)4 + 15 (âˆš2)2 + 1]

= 2 [8 + 15 (4) + 15 (2) + 1]

= 2 [8 + 60 + 30 + 1]

= 198

5. Using binomial theorem evaluate each of the following:

(i) (96)3

(ii) (102)5

(iii) (101)4

(iv) (98)5

Solution:

(i) (96)3

We have,

(96)3

Let us express the given expression as two different entities and apply the binomial theorem.

(96)3 = (100 – 4)3

= 3C0 (100)3 (4)03C1 (100)2 (4)1 + 3C2 (100)1 (4)23C3 (100)0 (4)3

= 1000000 â€“ 120000 + 4800 â€“ 64

= 884736

(ii) (102)5

We have,

(102)5

Let us express the given expression as two different entities and apply the binomial theorem.

(102)5 = (100 + 2)5

= 5C0 (100)5 (2)0 + 5C1 (100)4 (2)1 + 5C2 (100)3 (2)2 + 5C3 (100)2 (2)3 + 5C4 (100)1 (2)4 + 5C5 (100)0 (2)5

= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32

= 11040808032

(iii) (101)4

We have,

(101)4

Let us express the given expression as two different entities and apply the binomial theorem.

(101)4 = (100 + 1)4

= 4C0 (100)4 + 4C1 (100)3 + 4C2 (100)2 + 4C3 (100)1 + 4C4 (100)0

= 100000000 + 4000000 + 60000 + 400 + 1

= 104060401

(iv) (98)5

We have,

(98)5

Let us express the given expression as two different entities and apply the binomial theorem.

(98)5 = (100 – 2)5

= 5C0 (100)5 (2)05C1 (100)4 (2)1 + 5C2 (100)3 (2)25C3 (100)2 (2)3 + 5C4 (100)1 (2)45C5 (100)0 (2)5

= 10000000000 – 1000000000 + 40000000 – 800000 + 8000 â€“ 32

= 9039207968

6. Using binomial theorem, prove that 23n â€“ 7n â€“ 1 is divisible by 49, where n âˆˆ N.

Solution:

Given:

23n â€“ 7n â€“ 1

So, 23n â€“ 7n â€“ 1 = 8n â€“ 7n â€“ 1

Now,

8n â€“ 7n â€“ 1

8n = 7n + 1

= (1 + 7) n

= nC0 + nC1 (7)1 + nC2 (7)2 + nC3 (7)3 + nC4 (7)2 + nC5 (7)1 + â€¦ + nCn (7) n

8n = 1 + 7n + 49 [nC2 + nC3 (71) + nC4 (72) + â€¦ + nCn (7) n-2]

8n â€“ 1 â€“ 7n = 49 (integer)

So now,

8n â€“ 1 â€“ 7n is divisible by 49

Or

23n â€“ 1 â€“ 7n is divisible by 49.

Hence proved.

EXERCISE 18.2 PAGE NO: 18.37

1. Find the 11th term from the beginning and the 11th term from the end in the expansion of (2x â€“ 1/x2)25.

Solution:

Given:

(2x â€“ 1/x2)25

The given expression contains 26 terms.

So, the 11thÂ term from the end is the (26 âˆ’ 11 + 1) thÂ term from the beginning.

In other words, the 11thÂ term from the end is the 16thÂ term from the beginning.

Then,

T16 = T15+1 = 25C15 (2x)25-15 (-1/x2)15

= 25C15 (210) (x)10 (-1/x30)

= – 25C15 (210 / x20)

Now we shall find the 11th term from the beginning.

T11 = T10+1 = 25C10 (2x)25-10 (-1/x2)10

= 25C10 (215) (x)15 (1/x20)

= 25C10 (215 / x5)

2. Find the 7th term in the expansion of (3x2 â€“ 1/x3)10.

Solution:

Given:

(3x2 â€“ 1/x3)10

Let us consider the 7th term as T7

So,

T7 = T6+1

= 10C6 (3x2)10-6 (-1/x3)6

= 10C6 (3)4 (x)8 (1/x18)

= [10Ã—9Ã—8Ã—7Ã—81] / [4Ã—3Ã—2Ã—x10]

= 17010 / x10

âˆ´ The 7th term of the expression (3x2 â€“ 1/x3)10 is 17010 / x10.

3. Find the 5th term in the expansion of (3x â€“ 1/x2)10.

Solution:

Given:

(3x â€“ 1/x2)10

The 5th term from the end is the (11 â€“ 5 + 1)th, is., 7th term from the beginning.

So,

T7 = T6+1

= 10C6 (3x)10-6 (-1/x2)6

= 10C6 (3)4 (x)4 (1/x12)

= [10Ã—9Ã—8Ã—7Ã—81] / [4Ã—3Ã—2Ã—x8]

= 17010 / x8

âˆ´ The 5th term of the expression (3x â€“ 1/x2)10 is 17010 / x8.

4. Find the 8th term in the expansion of (x3/2 y1/2 â€“ x1/2 y3/2)10.

Solution:

Given:

(x3/2 y1/2 â€“ x1/2 y3/2)10

Let us consider the 8th term as T8

So,

T8 = T7+1

= 10C7 (x3/2 y1/2)10-7 (-x1/2 y3/2)7

= -[10Ã—9Ã—8]/[3Ã—2] x9/2 y3/2 (x7/2 y21/2)

= -120 x8y12

âˆ´ The 8th term of the expression (x3/2 y1/2 â€“ x1/2 y3/2)10 is -120 x8y12.

5. Find the 7th term in the expansion of (4x/5 + 5/2x) 8.

Solution:

Given:

(4x/5 + 5/2x) 8

Let us consider the 7th term as T7

So,

T7 = T6+1

âˆ´ The 7th term of the expression (4x/5 + 5/2x) 8 is 4375/x4.

6. Find the 4th term from the beginning and 4th term from the end in the expansion of (x + 2/x) 9.

Solution:

Given:

(x + 2/x) 9

LetÂ Tr+1Â be the 4th term from the end.

Then,Â Tr+1Â is (10 âˆ’ 4 + 1)th, i.e., 7th, term from the beginning.

7. Find the 4th term from the end in the expansion of (4x/5 â€“ 5/2x)Â 9.

Solution:

Given:

(4x/5 â€“ 5/2x)Â 9

Let Tr+1Â be theÂ 4th term from the end of the given expression.

Then, Tr+1Â is (10 âˆ’ 4 + 1)th term, i.e., 7th term, from the beginning.

T7 = T6+1

âˆ´ The 4th term from the end is 10500/x3.

8. Find the 7th term from the end in the expansion of (2x2 â€“ 3/2x) 8.

Solution:

Given:

(2x2 â€“ 3/2x) 8

Let Tr+1Â be theÂ 4th term from the end of the given expression.

Then, Tr+1Â is (9 âˆ’ 7 + 1)th term, i.e., 3rd term, from the beginning.

T3 = T2+1

âˆ´ The 7th term from the end is 4032 x10.

9. Find the coefficient of:

(i) Â x10Â in the expansion of (2x2 â€“ 1/x)20

(ii) x7 in the expansion of (x â€“ 1/x2)40

(iii) x-15 in the expansion of (3x2 â€“ a/3x3)10

(iv) x9 in the expansion of (x2 â€“ 1/3x)9

(v) xm in the expansion of (x + 1/x)n

(vi) x in the expansion of (1 â€“ 2x3 + 3x5) (1 + 1/x)8

(vii) a5b7 in the expansion of (a â€“ 2b)12

(viii) x in the expansion of (1 â€“ 3x + 7x2) (1 – x)16

Solution:

(i) Â x10Â in the expansion of (2x2 â€“ 1/x)20

Given:

(2x2 â€“ 1/x)20

If Â x10Â occurs in the (rÂ + 1)th term in the given expression.

Then, we have:

Tr+1Â = nCr xn-r ar

(ii) x7 in the expansion of (x â€“ 1/x2)40

Given:

(x â€“ 1/x2)40

If x7Â occurs at the (rÂ + 1) th term in the given expression.

Then, we have:

Tr+1Â = nCr xn-r ar

40 âˆ’ 3rÂ =7

3r = 40 â€“ 7

3r = 33

r = 33/3

= 11

(iii) x-15 in the expansion of (3x2 â€“ a/3x3)10

Given:

(3x2 â€“ a/3x3)10

IfÂ xâˆ’15Â occurs at the (rÂ + 1)th term in the given expression.

Then, we have:

Tr+1Â = nCr xn-r ar

(iv) x9 in the expansion of (x2 â€“ 1/3x)9

Given:

(x2 â€“ 1/3x)9

IfÂ x9Â occurs at the (rÂ + 1)th term in the above expression.

Then, we have:

Tr+1Â = nCr xn-r ar

ForÂ thisÂ termÂ toÂ containÂ x9,Â weÂ mustÂ have:

18 âˆ’ 3rÂ = 9

3r = 18 – 9

3r = 9

r = 9/3

= 3

(v) xm in the expansion of (x + 1/x)n

Given:

(x + 1/x)n

IfÂ xmÂ occurs at the (rÂ + 1)th term in the given expression.

Then, we have:

Tr+1Â = nCr xn-r ar

(vi) x in the expansion of (1 â€“ 2x3 + 3x5) (1 + 1/x)8

Given:

(1 â€“ 2x3 + 3x5) (1 + 1/x)8

IfÂ xÂ occurs at the (rÂ + 1)th term in the given expression.

Then, we have:

(1 â€“ 2x3 + 3x5) (1 + 1/x)8 = (1 â€“ 2x3 + 3x5) (8C0 + 8C1 (1/x) + 8C2 (1/x)2 + 8C3 (1/x)3 + 8C4 (1/x)4 + 8C5 (1/x)5 + 8C6 (1/x)6 + 8C7 (1/x)7 + 8C8 (1/x)8)

So, â€˜xâ€™ occurs in the above expression at -2x3.8C2 (1/x2) + 3x5.8C4 (1/x4)

âˆ´ Coefficient of x = -2 (8!/(2!6!)) + 3 (8!/(4! 4!))

= -56 + 210

= 154

(vii) a5b7 in the expansion of (a â€“ 2b)12

Given:

(a â€“ 2b)12

IfÂ a5b7Â occurs at the (rÂ + 1)th term in the given expression.

Then, we have:

Tr+1Â = nCr xn-r ar

(viii) x in the expansion of (1 â€“ 3x + 7x2) (1 – x)16

Given:

(1 â€“ 3x + 7x2) (1 – x)16

IfÂ xÂ occurs at the (rÂ + 1)th term in the given expression.

Then, we have:

(1 â€“ 3x + 7x2) (1 – x)16 = (1 â€“ 3x + 7x2) (16C0 + 16C1 (-x) + 16C2 (-x)2 + 16C3 (-x)3 + 16C4 (-x)4 + 16C5 (-x)5 + 16C6 (-x)6 + 16C7 (-x)7 + 16C8 (-x)8 + 16C9 (-x)9 + 16C10 (-x)10 + 16C11 (-x)11 + 16C12 (-x)12 + 16C13 (-x)13 + 16C14 (-x)14 + 16C15 (-x)15 + 16C16 (-x)16)

So, â€˜xâ€™ occurs in the above expression at 16C1 (-x) â€“ 3x16C0

âˆ´ Coefficient of x = -(16!/(1! 15!)) â€“ 3(16!/(0! 16!))

= -16 â€“ 3

= -19

10. Which term in the expansion of contains x and y to one and the same power.

Solution:

Let us consider Tr+1Â th term in the given expansion contains x and y to one and the same power.

Then we have,

Tr+1 = nCr xn-r ar

11. Does the expansion of (2x2 â€“ 1/x) contain any term involving x9?

Solution:

Given:

(2x2 â€“ 1/x)

IfÂ x9Â occurs at the (rÂ + 1)th term in the given expression.

Then, we have:

Tr+1 = nCr xn-r ar

For this term to contain x9, we must have

40 â€“ 3r = 9

3r = 40 – 9

3r = 31

r = 31/3

It is not possible, since r is not an integer.

Hence, there is no term with x9 in the given expansion.

12. Show that the expansion of (x2 + 1/x)12 does not contain any term involving x-1.

Solution:

Given:

(x2 + 1/x)12

IfÂ x-1Â occurs at the (rÂ + 1)th term in the given expression.

Then, we have:

Tr+1 = nCr xn-r ar

For this term to contain x-1, we must have

24 â€“ 3r = -1

3r = 24 + 1

3r = 25

r = 25/3

It is not possible, since r is not an integer.

Hence, there is no term with x-1 in the given expansion.

13. Find the middle term in the expansion of:

(i) (2/3x â€“ 3/2x)20

(ii) (a/x + bx)12

(iii) (x2 â€“ 2/x)10

(iv) (x/a â€“ a/x)10

Solution:

(i) (2/3x â€“ 3/2x)20

We have,

(2/3x â€“ 3/2x)20 where, n = 20 (even number)

So the middle term is (n/2 + 1) = (20/2 + 1) = (10 + 1) = 11. ie., 11th term

Now,

T11 = T10+1

= 20C10 (2/3x)20-10 (3/2x)10

= 20C10 210/310 Ã— 310/210 x10-10

= 20C10

Hence, the middle term is 20C10.

(ii) (a/x + bx)12

We have,

(a/x + bx)12 where, n = 12 (even number)

So the middle term is (n/2 + 1) = (12/2 + 1) = (6 + 1) = 7. ie., 7th term

Now,

T7 = T6+1

= 924 a6b6

Hence, the middle term is 924 a6b6.

(iii) (x2 â€“ 2/x)10

We have,

(x2 â€“ 2/x)10 where, n = 10 (even number)

So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term

Now,

T6 = T5+1

Hence, the middle term is -8064x5.

(iv) (x/a â€“ a/x)10

We have,

(x/a â€“ a/x) 10 where, n = 10 (even number)

So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term

Now,

T6 = T5+1

Hence, the middle term is -252.

14. Find the middle terms in the expansion of:

(i) (3x â€“ x3/6)9

(ii) (2x2 â€“ 1/x)7

(iii) (3x â€“ 2/x2)15

(iv) (x4 â€“ 1/x3)11

Solution:

(i) (3x â€“ x3/6)9

We have,

(3x â€“ x3/6)9 where, n = 9 (odd number)

So the middle terms are ((n+1)/2) = ((9+1)/2) = 10/2 = 5 and

((n+1)/2 + 1) = ((9+1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6

The terms are 5th and 6th.

Now,

T5 = T4+1

Hence, the middle term are 189/8 x17 and -21/16 x19.

(ii) (2x2 â€“ 1/x)7

We have,

(2x2 â€“ 1/x)7 where, n = 7 (odd number)

So the middle terms are ((n+1)/2) = ((7+1)/2) = 8/2 = 4 and

((n+1)/2 + 1) = ((7+1)/2 + 1) = (8/2 + 1) = (4 + 1) = 5

The terms are 4th and 5th.

Now,

Hence, the middle term are -560x5 and 280x2.

(iii) (3x â€“ 2/x2)15

We have,

(3x â€“ 2/x2)15 where, n = 15 (odd number)

So the middle terms are ((n+1)/2) = ((15+1)/2) = 16/2 = 8 and

((n+1)/2 + 1) = ((15+1)/2 + 1) = (16/2 + 1) = (8 + 1) = 9

The terms are 8th and 9th.

Now,

Hence, the middle term are (-6435Ã—38Ã—27)/x6 and (6435Ã—37Ã—28)/x9.

(iv) (x4 â€“ 1/x3)11

We have,

(x4 â€“ 1/x3)11

where, n = 11 (odd number)

So the middle terms are ((n+1)/2) = ((11+1)/2) = 12/2 = 6 and

((n+1)/2 + 1) = ((11+1)/2 + 1) = (12/2 + 1) = (6 + 1) = 7

The terms are 6th and 7th.

Now,

T7 = T6+1

Hence, the middle term are -462x9 and 462x2.

15. Find the middle terms in the expansion of:

(i) (x â€“ 1/x)10

(ii) (1 â€“ 2x + x2)n

(iii) (1 + 3x + 3x2 + x3)2n

(iv) (2x â€“ x2/4)9

(v) (x â€“ 1/x)2n+1

(vi) (x/3 + 9y)10

(vii) (3 â€“ x3/6)7

(viii) (2ax â€“ b/x2)12

(ix) (p/x + x/p)9

(x) (x/a â€“ a/x)10

Solution:

(i) (x â€“ 1/x)10

We have,

(x â€“ 1/x)10 where, n = 10 (even number)

So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term

Now,

T6 = T5+1

Hence, the middle term is -252.

(ii) (1 â€“ 2x + x2)n

We have,

(1 â€“ 2x + x2)n = (1 – x)2n where, n is an even number.

So the middle term is (2n/2 + 1) = (n + 1)th term.

Now,

Tn = Tn+1

= 2nCn (-1)n (x)n

= (2n)!/(n!)2 (-1)n xn

Hence, the middle term is (2n)!/(n!)2 (-1)n xn.

(iii) (1 + 3x + 3x2 + x3)2n

We have,

(1 + 3x + 3x2 + x3)2n = (1 + x)6n where, n is an even number.

So the middle term is (n/2 + 1) = (6n/2 + 1) = (3n + 1)th term.

Now,

T2n = T3n+1

= 6nC3n x3n

= (6n)!/(3n!)2 x3n

Hence, the middle term is (6n)!/(3n!)2 x3n.

(iv) (2x â€“ x2/4)9

We have,

(2x â€“ x2/4)9 where, n = 9 (odd number)

So the middle terms are ((n+1)/2) = ((9+1)/2) = 10/2 = 5 and

((n+1)/2 + 1) = ((9+1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6

The terms are 5th and 6th.

Now,

T5 = T4+1

And,

T6 = T5+1

Hence, the middle term is 63/4 x13 and -63/32 x14.

(v) (x â€“ 1/x)2n+1

We have,

(x â€“ 1/x)2n+1 where, n = (2n + 1) is an (odd number)

So the middle terms are ((n+1)/2) = ((2n+1+1)/2) = (2n+2)/2 = (n + 1) and

((n+1)/2 + 1) = ((2n+1+1)/2 + 1) = ((2n+2)/2 + 1) = (n + 1 + 1) = (n + 2)

The terms are (n + 1)th and (n + 2)th.

Now,

Tn = Tn+1

And,

Tn+2 = Tn+1+1

Hence, the middle term is (-1)n.2n+1Cn x and (-1)n+1.2n+1Cn (1/x).

(vi) (x/3 + 9y)10

We have,

(x/3 + 9y)10 where, n = 10 is an even number.

So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. i.e., 6th term.

Now,

T6 = T5+1

Hence, the middle term is 61236x5y5.

(vii) (3 â€“ x3/6)7

We have,

(3 â€“ x3/6)7 where, n = 7 (odd number).

So the middle terms are ((n+1)/2) = ((7+1)/2) = 8/2 = 4 and

((n+1)/2 + 1) = ((7+1)/2 + 1) = (8/2 + 1) = (4 + 1) = 5

The terms are 4th and 5th.

Now,

T4 = T3+1

= 7C3 (3)7-3 (-x3/6)3

= -105/8 x9

And,

T5 = T4+1

= 9C4 (3)9-4 (-x3/6)4

Hence, the middle terms are -105/8 x9 and 35/48 x12.

(viii) (2ax â€“ b/x2)12

We have,

(2ax â€“ b/x2)12 where, n = 12 is an even number.

So the middle term is (n/2 + 1) = (12/2 + 1) = (6 + 1) = 7. i.e., 7th term.

Now,

T7 = T6+1

Hence, the middle term is (59136a6b6)/x6.

(ix) (p/x + x/p)9

We have,

(p/x + x/p)9 where, n = 9 (odd number).

So the middle terms are ((n+1)/2) = ((9+1)/2) = 10/2 = 5 and

((n+1)/2 + 1) = ((9+1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6

The terms are 5th and 6th.

Now,

T5 = T4+1

And,

T6 = T5+1

= 9C5 (p/x)9-5 (x/p)5

Hence, the middle terms are 126p/x and 126x/p.

(x) (x/a â€“ a/x)10

We have,

(x/a â€“ a/x) 10 where, n = 10 (even number)

So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term

Now,

T6 = T5+1

Hence, the middle term is -252.

16. Find the term independent of x in the expansion of the following expressions:

(i) (3/2 x2 â€“ 1/3x)9

(ii) (2x + 1/3x2)9

(iii) (2x2 â€“ 3/x3)25

(iv) (3x â€“ 2/x2)15

(v) ((âˆšx/3) + âˆš3/2x2)10

(vi) (x â€“ 1/x2)3n

(vii) (1/2 x1/3 + x-1/5)8

(viii) (1 + x + 2x3) (3/2x2 â€“ 3/3x)9

(ix) (âˆ›x + 1/2âˆ›x)18, x > 0

(x) (3/2x2 â€“ 1/3x)6

Solution:

(i) (3/2 x2 â€“ 1/3x)9

Given:

(3/2 x2 â€“ 1/3x)9

If (rÂ + 1)th term in the given expression is independent of x.

Then, we have:

Tr+1 = nCr xn-r ar

For this term to be independent of x, we must have

18 â€“ 3r = 0

3r = 18

r = 18/3

= 6

So, the required term is 7th term.

We have,

T7 = T6+1

= 9C6 Ã— (39-12)/(29-6)

= (9Ã—8Ã—7)/(3Ã—2) Ã— 3-3 Ã— 2-3

= 7/18

Hence, the term independent of x is 7/18.

(ii) (2x + 1/3x2)9

Given:

(2x + 1/3x2)9

If (rÂ + 1)th term in the given expression is independent of x.

Then, we have:

Tr+1 = nCr xn-r ar

For this term to be independent of x, we must have

9 â€“ 3r = 0

3r = 9

r = 9/3

= 3

So, the required term is 4th term.

We have,

T4 = T3+1

= 9C3 Ã— (26)/(33)

= 9C3 Ã— 64/27

Hence, the term independent of x is 9C3 Ã— 64/27.

(iii) (2x2 â€“ 3/x3)25

Given:

(2x2 â€“ 3/x3)25

If (rÂ + 1)th term in the given expression is independent of x.

Then, we have:

Tr+1 = nCr xn-r ar

= 25Cr (2x2)25-r (-3/x3)r

= (-1)r 25Cr Ã— 225-r Ã— 3r x50-2r-3r

For this term to be independent of x, we must have

50 â€“ 5r = 0

5r = 50

r = 50/5

= 10

So, the required term is 11th term.

We have,

T11 = T10+1

= (-1)10 25C10 Ã— 225-10 Ã— 310

= 25C10 (215 Ã— 310)

Hence, the term independent of x is 25C10 (215 Ã— 310).

(iv) (3x â€“ 2/x2)15

Given:

(3x â€“ 2/x2)15

If (rÂ + 1)th term in the given expression is independent of x.

Then, we have:

Tr+1 = nCr xn-r ar

= 15Cr (3x)15-r (-2/x2)r

= (-1)r 15Cr Ã— 315-r Ã— 2r x15-r-2r

For this term to be independent of x, we must have

15 â€“ 3r = 0

3r = 15

r = 15/3

= 5

So, the required term is 6th term.

We have,

T6 = T5+1

= (-1)5 15C5 Ã— 315-5 Ã— 25

= -3003 Ã— 310 Ã— 25

Hence, the term independent of x is -3003 Ã— 310 Ã— 25.

(v) ((âˆšx/3) + âˆš3/2x2)10

Given:

((âˆšx/3) + âˆš3/2x2)10

If (rÂ + 1)th term in the given expression is independent of x.

Then, we have:

Tr+1 = nCr xn-r ar

For this term to be independent of x, we must have

(10-r)/2 â€“ 2r = 0

10 â€“ 5r = 0

5r = 10

r = 10/5

= 2

So, the required term is 3rd term.

We have,

T3 = T2+1

Hence, the term independent of x is 5/4.

(vi) (x â€“ 1/x2)3n

Given:

(x â€“ 1/x2)3n

If (rÂ + 1)th term in the given expression is independent of x.

Then, we have:

Tr+1 = nCr xn-r ar

= 3nCr x3n-r (-1/x2)r

= (-1)r 3nCr x3n-r-2r

For this term to be independent of x, we must have

3n â€“ 3r = 0

r = n

So, the required term is (n+1)th term.

We have,

(-1)n 3nCn

Hence, the term independent of x is (-1)n 3nCn

(vii) (1/2 x1/3 + x-1/5)8

Given:

(1/2 x1/3 + x-1/5)8

If (rÂ + 1)th term in the given expression is independent of x.

Then, we have:

Tr+1 = nCr xn-r ar

For this term to be independent of x, we must have

(8-r)/3 â€“ r/5 = 0

(40 â€“ 5r â€“ 3r)/15 = 0

40 â€“ 5r â€“ 3r = 0

40 â€“ 8r = 0

8r = 40

r = 40/8

= 5

So, the required term is 6th term.

We have,

T6 = T5+1

= 8C5 Ã— 1/(28-5)

= (8Ã—7Ã—6)/(3Ã—2Ã—8)

= 7

Hence, the term independent of x is 7.

(viii) (1 + x + 2x3) (3/2x2 â€“ 3/3x)9

Given:

(1 + x + 2x3) (3/2x2 â€“ 3/3x)9

If (rÂ + 1)th term in the given expression is independent of x.

Then, we have:

(1 + x + 2x3) (3/2x2 â€“ 3/3x)9 =

= 7/18 â€“ 2/27

= (189 – 36)/486

= 153/486 (divide by 9)

= 17/54

Hence, the term independent of x is 17/54.

(ix) (âˆ›x + 1/2âˆ›x)18, x > 0

Given:

(âˆ›x + 1/2âˆ›x)18, x > 0

If (rÂ + 1)th term in the given expression is independent of x.

Then, we have:

Tr+1 = nCr xn-r ar

For this term to be independent of r, we must have

(18-r)/3 â€“ r/3 = 0

(18 â€“ r â€“ r)/3 = 0

18 â€“ 2r = 0

2r = 18

r = 18/2

= 9

So, the required term is 10th term.

We have,

T10 = T9+1

= 18C9 Ã— 1/29

Hence, the term independent of x is 18C9 Ã— 1/29.

(x) (3/2x2 â€“ 1/3x)6

Given:

(3/2x2 â€“ 1/3x)6

If (rÂ + 1)th term in the given expression is independent of x.

Then, we have:

Tr+1 = nCr xn-r ar

For this term to be independent of r, we must have

12 â€“ 3r = 0

3r = 12

r = 12/3

= 4

So, the required term is 5th term.

We have,

T5 = T4+1

Hence, the term independent of x is 5/12.

17. If the coefficients of (2r + 4)th and (r – 2)th terms in the expansion of (1 + x)18 are equal, find r.

Solution:

Given:

(1 + x)18

We know, the coefficient of the r term in the expansion of (1 + x)n is nCr-1

So, the coefficients of the (2r + 4) and (r – 2) terms in the given expansion are 18C2r+4-1 and 18Cr-2-1

For these coefficients to be equal, we must have

18C2r+4-1 = 18Cr-2-1

18C2r+3 = 18Cr-3

2r + 3 = r â€“ 3 (or) 2r + 3 + r â€“ 3 = 18 [Since, nCr = nCs => r = s (or) r + s = n]

2r â€“ r = -3 â€“ 3 (or) 3r = 18 â€“ 3 + 3

r = -6 (or) 3r = 18

r = -6 (or) r = 18/3

r = -6 (or) r = 6

âˆ´ r = 6 [since, r should be a positive integer.]

18. If the coefficients of (2rÂ + 1)th term and (rÂ + 2)th term in the expansion of (1 +Â x)43Â are equal, findÂ r.

Solution:

Given:

(1 +Â x)43

We know, the coefficient of the r term in the expansion of (1 + x)n is nCr-1

So, the coefficients of the (2r + 1) and (r + 2) terms in the given expansion are 43C2r+1-1 and 43Cr+2-1

For these coefficients to be equal, we must have

43C2r+1-1 = 43Cr+2-1

43C2r = 43Cr+1

2r = r + 1 (or) 2r + r + 1 = 43 [Since, nCr = nCs => r = s (or) r + s = n]

2r – r = 1 (or) 3r + 1 = 43

r = 1 (or) 3r = 43 â€“ 1

r = 1 (or) 3r = 42

r = 1 (or) r = 42/3

r = 1 (or) r = 14

âˆ´ r = 14 [since, value â€˜1â€™ gives the same term]

19. Prove that the coefficient of (rÂ + 1)th term in the expansion of (1 +Â x)nÂ + 1Â is equal to the sum of the coefficients ofÂ rth and (rÂ + 1)th terms in the expansion of (1 +Â x)n.

Solution:

We know, the coefficients of (r + 1)th term in (1 + x)n+1 is n+1Cr

So, sum of the coefficients of the rth and (r + 1)th terms in (1 + x)n is

(1 + x)n = nCr-1 + nCr

= n+1Cr [since, nCr+1 + nCr = n+1Cr+1]

Hence proved.

### Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 18 – Binomial Theorem

Exercise 18.1 Solutions

Exercise 18.2 Solutions