# Class 11 RD Sharma Solution – Chapter 18 Binomial Theorem- Exercise 18.2 | Set 1

• Last Updated : 14 Jul, 2021

### Question 1. Find the 11th term from the beginning and the 11th term from the end in the expansion of (2x – 1/x2)25.

Solution:

We are given, (2x – 1/x2)25.

The given expression contains 25 + 1 = 26 terms.

So, the 11th term from the end is the (26 − 11 + 1) th term = 16th term from the beginning.

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

= 25C15 (210) (x)10 (−1/x30)

= – 25C15 (210/ x20)

Now, the 11th term from the beginning is,

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

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

= 25C10 (215/ x5)

### Question 2. Find the 7th term in the expansion of (3x2 – 1/x3)10.

Solution:

We are given, (3x2 – 1/x3)10

The 7th term of the expression is given by,

T7 = T6+1

= 10C6 (3x2)10−6 (−1/x3)6

= 10C6 (3)4 (x)8 (1/x18) = 17010/ x10

### Question 3. Find the 5th term from the end in the expansion of (3x – 1/x2)10.

Solution:

We are given, (3x – 1/x2)10

The 5th term from the end is the (11 – 5 + 1)th = 7th term from the beginning.

So, T7 = T6+1

= 10C6 (3x)10-6 (–1/x2)6

= 10C6 (3)4 (x)4 (1/x12) = 17010/ x8

### Question 4. Find the 8th term in the expansion of (x3/2 y1/2 – x1/2 y3/2)10.

Solution:

We are given, (x3/2 y1/2 – x1/2 y3/2)10

The 8th term of the expression is given by,

T8 = T7+1

= 10C7 (x3/2 y1/2)10–7 (–x1/2 y3/2)7 = –120 x8y12

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

Solution:

We are given, (4x/5 + 5/2x)8.

The 8th term of the expression is given by,

T7 = T6+1  = 4375/ x4

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

Solution:

We are given, (x + 2/x)9.

The given expression contains 9 + 1 = 10 terms.

So, the 4th term from the end is the (10 − 4 + 1) th term = 7th term from the beginning.

Hence, T7 = T6+1 = 9C6 (x)9-6 (2/x)6 = 5376/ x3

Now, the 4th term from the beginning is,

T4 = T3+1 = 9C3 (2x)9-3 (2/x)3 = 672 x3

### Question 7. Find the 4th term from the end in the expansion of (4x/5 – 5/2x)9.

Solution:

We are given, (4x/5 – 5/2x)9

The 4th term from the end is (10 − 4 + 1)th term = 7th term, from the beginning.

T7 = T6+1  = 10500/ x3

### Question 8. Find the 7th term from the end in the expansion of (2x2 – 3/2x)8.

Solution:

We are given, (2x2 – 3/2x)8

The 7th term from the end is (9 − 7 + 1)th term = 3rd term, from the beginning.

T3 = T2+1  = 4032 x10

### (i) x10 in the expansion of (2x2 – 1/x)20

Solution:

We are given, (2x2 – 1/x)20

We know, the (r+1)th term of the expression is given by,

Tr+1 = nCr xn-r ar

= 20Cr (2x2)20-r (-1/x)r

= (-1)r 20Cr (2)20-r x40-2r-r

If x10 exists in the expansion, we must have,

=> 40 − 3r = 10

=> 3r = 30

=> r = 10

Coefficient of x10 = (-1)10 20C10 (2)20-10

= 20C10 (2)10

### (ii) x7 in the expansion of (x – 1/x2)40

Solution:

We are given, (x – 1/x2)40

We know, the (r+1)th term of the expression is given by,

Tr+1 = nCr xn-r ar

= 40Cr (x)40-r (-1/x2)r

= (-1)r 40Cr x40-r-2r

If x7 exists in the expansion, we must have,

=> 40 − 3r = 7

=> 3r = 33

=> r = 11

Coefficient of x7 = (-1)11 40C11

= − 40C11

### (iii) x-15 in the expansion of (3x2 – a/3x3)10

Solution:

We are given, (3x2 – a/3x3)10

We know, the (r+1)th term of the expression is given by,

Tr+1 = nCr xn-r ar

= 10Cr (3x2)10-r (a/3x3)r

= (-1)r 10Cr 310-r-r x20-2r-3r ar

If x-15 exists in the expansion, we must have,

=> 20 − 5r = −15

=> 5r = 35

=> r = 7

Coefficient of x-15 = (-1)7 10C7 310-14 a7 = −40a7/27

### (iv) x9 in the expansion of (x2 – 1/3x)9

Solution:

We are given, (x2 – 1/3x)9

We know, the (r+1)th term of the expression is given by,

Tr+1 = nCr xn-r ar

= 9Cr (x2)9-r (-1/3x)r

= (-1)r 9Cr x18-2r-r (1/3)r

If x9 exists in the expansion, we must have,

=> 18 − 3r = 9

=> 3r = 9

=> r = 3

Coefficient of x9 = (-1)3 9C3 (1/3)3 = −28/9

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

Solution:

We are given, (x + 1/x)n

We know, the (r + 1)th term of the expression is given by,

Tr+1 = nCr xn- r (1/xr)

= nCr xn- 2r

If xm exists in the expansion, we must have,

=> n – 2r = m

=> r = (n – m)/2

Coefficient of xm = nC(n-m)/2 ### (vi) x in the expansion of  (1 – 2x3 + 3x5) (1 + 1/x)8

Solution:

We are given, (1 – 2x3 + 3x5) (1 + 1/x)8

We know, the (r + 1)th term of the expression is given by,

= (1 – 2x3 + 3x5) (8C0 + 8C1 (1/x) + 8C2 (1/x2) + 8C3 (1/x3) + 8C4 (1/x4) + 8C5 (1/x5) + 8C6 (1/x6) + 8C7 (1/x7) + 8C8 (1/x8))

Here, x appear in the above expression at -2 x3 8C2 (1/x2) + 3x5.8C4 (1/x4)

So, coefficient of x = = – 56 + 210

= 154

### (vii) a5b7 in the expansion of (a – 2b)12

Solution:

We are given, (a – 2b)12

We know, the (r+1)th term of the expression is given by,

Tr+1 = nCr xn-r ar

= (-1)r 12Cr (a)12-r (2b)r

If a5b7 exists in the expansion, we must have,

=> 12 − r = 5

=> r = 7

Coefficient of a5b7 = (-1)7 12C7 (2)7 = − 101376

### (viii) x in the expansion of (1 – 3x + 7 x2) (1 – x)16

Solution:

We are given, (1 – 3x + 7 x2) (1 – x)16

We know, the (r + 1)th term of the expression is given by,

= (1 – 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)

Here, x appear in the above expression at 16C1 (-x) – 3x 16C0

So, the coefficient of x = = – 16 – 3

= – 19

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

Solution:

We are given, We know, the (r+1)th term of the expression is given by,

Tr+1 = nCr xn-r ar = 21Cr x7-r/2 y2r/3-7/2

If x and y have same power, we must have,

=> 7 − r/2 = 2r/3 − 7/2

=> 7r/6 = 21/2

=> r = 9

Hence the required term is 9 + 1 = 10th term.

### Question 11. Does the expansion of (2x2 – 1/x)20 contain any term involving x9?

Solution:

We are given, (2x2 – 1/x)20

We know, the (r+1)th term of the expression is given by,

Tr+1 = nCr xn-r ar

= 20Cr (2x2)20-r (1/x)r

= 20Cr (2)20-r x40-2r-r

If x9 exists in the expansion, we must have,

=> 40 − 3r = 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.

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

Solution:

We have, (x2 + 1/x)12

We know, the (r+1)th term of the expression is given by,

Tr+1 = nCr xn-r ar

= 12Cr (x2)12-r (1/x)r

= 12Cr x24-2r-r

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

=> 24 – 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.

### (i) (2/3x – 3/2x)20

Solution:

We have,

(2/3x – 3/2x)20 where, n = 20 (which is an even number)

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

Now,

T11 = T10+1

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

= 20C10 (210/310) × (310/210) x10-10

= 20C10

### (ii) (x2 – 2/x)10

Solution:

We have,

(x2 – 2/x)10 where, n = 10 (which is an even number)

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

Now,

T6 = T5+1

= 10C5 (x2)10-5 (-2/x)5 = − 8064 x5

### (iii) (x/a – a/x)10

Solution:

We have,

(x/a – a/x)10 where, n = 10 (even number).

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

Now,

T6 = T5+1

= 10C5 (x/a)10-5 (-a/x)5 = −252

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