## RD Sharma Solutions Class 11 Maths Chapter 18 – Free PDF Download

**RD Sharma Solutions for Class 11 Maths Chapter 18 Binomial Theorem **are provided here to boost in-depth knowledge of concepts among students. 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 BYJU’S for the purpose of clarifying students’ doubts immediately. These solutions also provide guidance to students to solve problems confidently, which in turn, helps in improving their problem-solving skills, which is very important from the examination point of view.

**Chapter 18 – Binomial Theorem** contains two exercises, and RD Sharma Solutions provide accurate solutions to the questions present in each exercise. Students who wish to learn the right steps for solving problems with ease can make use of RD Sharma Solutions in PDF format updated for the 2023-24 syllabus from the links given below. 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 terms and middle terms in a binomial expansion.

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EXERCISE 18.1 PAGE NO: 18.11

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

**Solution:**

**(i) (2x + 3y) ^{ 5}**

Let us solve the given expression

(2x + 3y)^{ 5} = ^{5}C_{0} (2x)^{5} (3y)^{0} + ^{5}C_{1} (2x)^{4} (3y)^{1} + ^{5}C_{2} (2x)^{3} (3y)^{2} + ^{5}C_{3} (2x)^{2} (3y)^{3} + ^{5}C_{4} (2x)^{1} (3y)^{4} + ^{5}C_{5} (2x)^{0} (3y)^{5}

= 32x^{5} + 5 (16x^{4}) (3y) + 10 (8x^{3}) (9y)^{2} + 10 (4x)^{2} (27y)^{3} + 5 (2x) (81y^{4}) + 243y^{5}

*= 32x ^{5} + 240x^{4}y + 720x^{3}y^{2} + 1080x^{2}y^{3} + 810xy^{4} + 243y^{5}*

**(ii) (2x – 3y) ^{ 4}**

Let us solve the given expression

(2x – 3y)^{ 4} = ^{4}C_{0} (2x)^{4} (3y)^{0} – ^{4}C_{1} (2x)^{3} (3y)^{1} + ^{4}C_{2} (2x)^{2} (3y)^{2} – ^{4}C_{3} (2x)^{1} (3y)^{3} + ^{4}C_{4} (2x)^{0} (3y)^{4}

= 16x^{4} – 4 (8x^{3}) (3y) + 6 (4x^{2}) (9y^{2}) – 4 (2x) (27y^{3}) + 81y^{4}

*= 16x ^{4} – 96x^{3}y + 216x^{2}y^{2} – 216xy^{3} + 81y^{4}*

**(iv) (1 – 3x) ^{ 7}**

Let us solve the given expression

(1 – 3x)^{ 7} = ^{7}C_{0} (3x)^{0} – ^{7}C_{1} (3x)^{1} + ^{7}C_{2} (3x)^{2} – ^{7}C_{3} (3x)^{3} + ^{7}C_{4} (3x)^{4} – ^{7}C_{5} (3x)^{5} + ^{7}C_{6} (3x)^{6} – ^{7}C_{7} (3x)^{7}

= 1 – 7 (3x) + 21 (9x)^{2} – 35 (27x^{3}) + 35 (81x^{4}) – 21 (243x^{5}) + 7 (729x^{6}) – 2187(x^{7})

*= 1 – 21x + 189x ^{2} – 945x^{3} + 2835x^{4} – 5103x^{5} + 5103x^{6} – 2187x^{7}*

**(viii) (1 + 2x – 3x ^{2})^{5}**

Let us solve the given expression

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

(1 + 2x – 3x^{2})^{5} = ^{5}C_{0} (1 + 2x)^{5} (3x^{2})^{0} – ^{5}C_{1} (1 + 2x)^{4} (3x^{2})^{1} + ^{5}C_{2} (1 + 2x)^{3} (3x^{2})^{2} – ^{5}C_{3} (1 + 2x)^{2} (3x^{2})^{3} + ^{5}C_{4} (1 + 2x)^{1} (3x^{2})^{4} – ^{5}C_{5} (1 + 2x)^{0} (3x^{2})^{5}

= (1 + 2x)^{5} – 5(1 + 2x)^{4} (3x^{2}) + 10 (1 + 2x)^{3} (9x^{4}) – 10 (1 + 2x)^{2} (27x^{6}) + 5 (1 + 2x) (81x^{8}) – 243x^{10}

= ^{5}C_{0} (2x)^{0} + ^{5}C_{1} (2x)^{1} + ^{5}C_{2} (2x)^{2} + ^{5}C_{3} (2x)^{3} + ^{5}C_{4} (2x)^{4} + ^{5}C_{5} (2x)^{5 }– 15x^{2} [^{4}C_{0} (2x)^{0} + ^{4}C_{1} (2x)^{1} + ^{4}C_{2} (2x)^{2} + ^{4}C_{3} (2x)^{3} + ^{4}C_{4} (2x)^{4}] + 90x^{4} [1 + 8x^{3} + 6x + 12x^{2}] – 270x^{6}(1 + 4x^{2} + 4x) + 405x^{8} + 810x^{9} – 243x^{10}

= 1 + 10x + 40x^{2} + 80x^{3} + 80x^{4} + 32x^{5} – 15x^{2} – 120x^{3} – 360^{4} – 480x^{5} – 240x^{6} + 90x^{4} + 720x^{7} + 540x^{5} + 1080x^{6} – 270x^{6} – 1080x^{8} – 1080x^{7} + 405x^{8} + 810x^{9} – 243x^{10}

*= 1 + 10x + 25x ^{2} – 40x^{3} – 190x^{4} + 92x^{5} + 570x^{6} – 360x^{7} – 675x^{8} + 810x^{9} – 243x^{10}*

**(x) **(1 – 2x + 3x^{2})^{3}

Let us solve the given expression

**2. Evaluate the following:**

**Solution:**

Let us solve the given expression

= 2 [^{5}C_{0} (2**√**x)^{0} + ^{5}C_{2} (2**√**x)^{2} + ^{5}C_{4} (2**√**x)^{4}]

= 2 [1 + 10 (4x) + 5 (16x^{2})]

*= 2 [1 + 40x + 80x ^{2}]*

Let us solve the given expression

= 2 [^{6}C_{0} (**√**2)^{6} + ^{6}C_{2} (**√**2)^{4} + ^{6}C_{4} (**√**2)^{2} + ^{6}C_{6} (**√**2)^{0}]

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

= 2 [99]

*= 198*

Let us solve the given expression

= 2 [^{5}C_{1} (3^{4}) (**√**2)^{1} + ^{5}C_{3} (3^{2}) (**√**2)^{3} + ^{5}C_{5} (3^{0}) (**√**2)^{5}]

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

= 2**√**2 (405 + 180 + 4)

*= 1178 √2*

Let us solve the given expression

= 2 [^{7}C_{0} (2^{7}) (**√**3)^{0} + ^{7}C_{2} (2^{5}) (**√**3)^{2} + ^{7}C_{4} (2^{3}) (**√**3)^{4} + ^{7}C_{6} (2^{1}) (**√**3)^{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 [^{5}C_{1} (**√**3)^{4} + ^{5}C_{3} (**√**3)^{2} + ^{5}C_{5} (**√**3)^{0}]

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

= 2 [76]

*= 152*

Let us solve the given expression

= (1 – 0.01)^{5} + (1 + 0.01)^{5}

= 2 [^{5}C_{0} (0.01)^{0} + ^{5}C_{2} (0.01)^{2} + ^{5}C_{4} (0.01)^{4}]

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

= 2 [1.00100005]

*= 2.0020001*

Let us solve the given expression

= 2 [^{6}C_{1} (**√**3)^{5} (**√**2)^{1} + ^{6}C_{3} (**√**3)^{3} (**√**2)^{3} + ^{6}C_{5} (**√**3)^{1} (**√**2)^{5}]

= 2 [6 (9**√**3) (**√**2) + 20 (3**√**3) (2**√**2) + 6 (**√**3) (4**√**2)]

= 2 [**√**6 (54 + 120 + 24)]

*= 396 √6*

= 2 [a^{8} + 6a^{6} – 6a^{4} + a^{4} + 1 – 2a^{2}]

*= 2a ^{8} + 12a^{6} – 10a^{4} – 4a^{2} + 2*

**3. Find (a + b) ^{ 4} – (a – b)^{ 4}. Hence, evaluate (√3 + √2)^{4} – (√3 – √2)^{4}. **

**Solution:**

Firstly, let us solve the given expression

(a + b)^{ 4} – (a – b)^{ 4}

The above expression can be expressed as,

(a + b)^{ 4} – (a – b)^{ 4} = 2 [^{4}C_{1} a^{3}b^{1} + ^{4}C_{3} a^{1}b^{3}]

= 2 [4a^{3}b + 4ab^{3}]

= 8 (a^{3}b + ab^{3})

Now,

Let us evaluate the expression

(√3 + √2)^{4} – (√3 -√2)^{4}

So consider, a = √3 and b = √2 we get,

(√3 + √2)^{4} – (√3 -√2)^{4} = 8 (a^{3}b + ab^{3})

= 8 [(√3)^{3} (√2) + (√3) (√2)^{3}]

= 8 [(3√6) + (2√6)]

= 8 (5√6)

*= 40√6*

**4. Find (x + 1) ^{ 6} + (x – 1)^{ 6}. Hence, or otherwise, evaluate (√2 + 1)^{6} + (√2 – 1)^{6}. **

**Solution:**

Firstly, let us solve the given expression

(x + 1)^{ 6} + (x – 1)^{ 6}

The above expression can be expressed as,

(x + 1)^{ 6} + (x – 1)^{ 6} = 2 [^{6}C_{0} x^{6} + ^{6}C_{2} x^{4} + ^{6}C_{4} x^{2} + ^{6}C_{6} x^{0}]

= 2 [x^{6} + 15x^{4} + 15x^{2} + 1]

Now,

Let us evaluate the expression

(√2 + 1)^{6} + (√2 – 1)^{6}

So consider x = √2 then, we get,

(√2 + 1)^{6} + (√2 – 1)^{6} = 2 [x^{6} + 15x^{4} + 15x^{2} + 1]

= 2 [(√2)^{6} + 15 (√2)^{4} + 15 (√2)^{2} + 1]

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

= 2 [8 + 60 + 30 + 1]

*= 198*

**5. Using the 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}

= ^{3}C_{0} (100)^{3} (4)^{0} – ^{3}C_{1} (100)^{2} (4)^{1} + ^{3}C_{2} (100)^{1} (4)^{2} – ^{3}C_{3} (100)^{0} (4)^{3}

= 1000000 – 120000 + 4800 – 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}

= ^{5}C_{0} (100)^{5} (2)^{0} + ^{5}C_{1} (100)^{4} (2)^{1} + ^{5}C_{2} (100)^{3} (2)^{2} + ^{5}C_{3} (100)^{2} (2)^{3} + ^{5}C_{4} (100)^{1} (2)^{4} + ^{5}C_{5} (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}

= ^{4}C_{0} (100)^{4} + ^{4}C_{1} (100)^{3} + ^{4}C_{2} (100)^{2} + ^{4}C_{3} (100)^{1} + ^{4}C_{4} (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}

= ^{5}C_{0} (100)^{5} (2)^{0} – ^{5}C_{1} (100)^{4} (2)^{1} + ^{5}C_{2} (100)^{3} (2)^{2} – ^{5}C_{3} (100)^{2} (2)^{3} + ^{5}C_{4} (100)^{1} (2)^{4} – ^{5}C_{5} (100)^{0} (2)^{5}

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

*= 9039207968*

**6. Using binomial theorem, prove that 2 ^{3n} – 7n – 1 is divisible by 49, where n ∈ N.**

**Solution:**

Given:

2^{3n} – 7n – 1

So, 2^{3n} – 7n – 1 = 8^{n} – 7n – 1

Now,

8^{n} – 7n – 1

8^{n} = 7n + 1

= (1 + 7)^{ n}

= ^{n}C_{0} + ^{n}C_{1} (7)^{1} + ^{n}C_{2} (7)^{2} + ^{n}C_{3} (7)^{3} + ^{n}C_{4} (7)^{2} + ^{n}C_{5} (7)^{1} + … + ^{n}C_{n} (7)^{ n}

8^{n} = 1 + 7n + 49 [^{n}C_{2} + ^{n}C_{3} (7^{1}) + ^{n}C_{4 }(7^{2}) + … + ^{n}C_{n} (7)^{ n-2}]

8^{n} – 1 – 7n = 49 (integer)

So now,

8^{n} – 1 – 7n is divisible by 49

Or

*2 ^{3n} – 1 – 7n is divisible by 49.*

Hence proved.

EXERCISE 18.2 PAGE NO: 18.37

**1. Find the 11 ^{th} term from the beginning and the 11^{th} term from the end in the expansion of (2x – 1/x^{2})^{25}.**

**Solution:**

Given:

(2x – 1/x^{2})^{25}

The given expression contains 26 terms.

So, the 11^{th} term from the end is the (26 − 11 + 1) ^{th} term from the beginning.

In other words, the 11^{th} term from the end is the 16^{th} term from the beginning.

Then,

T_{16} = T_{15+1} = ^{25}C_{15} (2x)^{25-15} (-1/x^{2})^{15}

= ^{25}C_{15} (2^{10}) (x)^{10} (-1/x^{30})

= – ^{25}C_{15} (2^{10} / x^{20})

Now, we shall find the 11^{th} term from the beginning.

T_{11} = T_{10+1} = ^{25}C_{10} (2x)^{25-10} (-1/x^{2})^{10}

= ^{25}C_{10} (2^{15}) (x)^{15} (1/x^{20})

*= ^{25}C_{10} (2^{15} / x^{5})*

**2. Find the 7 ^{th} term in the expansion of (3x^{2} – 1/x^{3})^{10}.**

**Solution:**

Given:

(3x^{2} – 1/x^{3})^{10}

Let us consider the 7^{th} term as T_{7}

So,

T_{7} = T_{6+1}

= ^{10}C_{6} (3x^{2})^{10-6} (-1/x^{3})^{6}

= ^{10}C_{6} (3)^{4} (x)^{8} (1/x^{18})

= [10×9×8×7×81] / [4×3×2×x^{10}]

= 17010 / x^{10}

*∴ The 7 ^{th} term of the expression (3x^{2} – 1/x^{3})^{10} is 17010 / x^{10}.*

**3.** **Find the 5 ^{th} term in the expansion of (3x – 1/x^{2})^{10}.**

**Solution:**

Given:

(3x – 1/x^{2})^{10}

The 5^{th} term from the end is the (11 – 5 + 1)th, is., 7^{th} term from the beginning.

So,

T_{7} = T_{6+1}

= ^{10}C_{6} (3x)^{10-6} (-1/x^{2})^{6}

= ^{10}C_{6} (3)^{4} (x)^{4} (1/x^{12})

= [10×9×8×7×81] / [4×3×2×x^{8}]

= 17010 / x^{8}

*∴ The 5 ^{th} term of the expression (3x – 1/x^{2})^{10} is 17010 / x^{8}.*

**4. Find the 8 ^{th} term in the expansion of (x^{3/2} y^{1/2} – x^{1/2} y^{3/2})^{10}.**

**Solution:**

Given:

(x^{3/2} y^{1/2} – x^{1/2} y^{3/2})^{10}

Let us consider the 8^{th} term as T_{8}

So,

T_{8} = T_{7+1}

= ^{10}C_{7} (x^{3/2} y^{1/2})^{10-7} (-x^{1/2} y^{3/2})^{7}

= -[10×9×8]/[3×2] x^{9/2} y^{3/2} (x^{7/2} y^{21/2})

= -120 x^{8}y^{12}

*∴ The 8 ^{th} term of the expression (x^{3/2} y^{1/2} – x^{1/2} y^{3/2})^{10} is -120 x^{8}y^{12}.*

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

**Solution:**

Given:

(4x/5 + 5/2x)^{ 8}

Let us consider the 7^{th} term as T_{7}

So,

T_{7} = T_{6+1}

*∴ The 7 ^{th} term of the expression (4x/5 + 5/2x)^{ 8} is 4375/x^{4}.*

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

**Solution:**

Given:

(x + 2/x)^{ 9}

Let T_{r+1} be the 4th term from the end.

Then, T_{r+1} is (10 − 4 + 1)th, i.e., 7th, the term from the beginning.

**7. Find the 4 ^{th} term from the end in the expansion of (4x/5 – 5/2x)^{ 9}.**

**Solution:**

Given:

(4x/5 – 5/2x)^{ 9}

Let T_{r+1} be the_{ }4th term from the end of the given expression.

Then, T_{r+1 }is (10 − 4 + 1)th term, i.e., the 7th term, from the beginning.

T_{7} = T_{6+1}

*∴ The 4 ^{th} term from the end is 10500/x^{3}.*

**8. Find the 7th term from the end in the expansion of (2x ^{2} – 3/2x)^{ 8}.**

**Solution:**

Given:

(2x^{2} – 3/2x)^{ 8}

Let T_{r+1} be the_{ }4th term from the end of the given expression.

Then, T_{r+1 }is (9 − 7 + 1)th term, i.e., the 3rd term, from the beginning.

T_{3} = T_{2+1}

*∴ The 7 ^{th} term from the end is 4032 x^{10}.*

**9. Find the coefficient of:**

**(i) x ^{10} in the expansion of (2x^{2} – 1/x)^{20}**

**(ii) x ^{7} in the expansion of (x – 1/x^{2})^{40} **

**(iii) x ^{-15} in the expansion of (3x^{2} – a/3x^{3})^{10}**

**(iv) x ^{9} in the expansion of (x^{2} – 1/3x)^{9}**

**(v) x ^{m} in the expansion of (x + 1/x)^{n}**

**(vi) x in the expansion of (1 – 2x ^{3} + 3x^{5}) (1 + 1/x)^{8} **

**(vii) a ^{5}b^{7} in the expansion of (a – 2b)^{12} **

**(viii) x in the expansion of (1 – 3x + 7x ^{2}) (1 – x)^{16}**

**Solution:**

**(i) **x^{10} in the expansion of (2x^{2} – 1/x)^{20}

Given:

(2x^{2} – 1/x)^{20}

If x^{10 }occurs in the (r + 1)th term in the given expression.

Then, we have:

T_{r+1 }= ^{n}C_{r} x^{n-r} a^{r}

**(ii) **x^{7} in the expansion of (x – 1/x^{2})^{40}

Given:

(x – 1/x^{2})^{40}

If x^{7 }occurs at the (r + 1) th term in the given expression.

Then, we have:

T_{r+1 }= ^{n}C_{r} x^{n-r} a^{r}

40 − 3r =7

3r = 40 – 7

3r = 33

r = 33/3

= 11

**(iii) **x^{-15} in the expansion of (3x^{2} – a/3x^{3})^{10}

Given:

(3x^{2} – a/3x^{3})^{10 }

If x^{−15} occurs at the (r + 1)th term in the given expression.

Then, we have:

T_{r+1 }= ^{n}C_{r} x^{n-r} a^{r}

**(iv) **x^{9} in the expansion of (x^{2} – 1/3x)^{9}

Given:

(x^{2} – 1/3x)^{9}

If x^{9} occurs at the (r + 1)th term in the above expression.

Then, we have:

T_{r+1 }= ^{n}C_{r} x^{n-r} a^{r}

For this term to contain x^{9}, we must have

18 − 3r = 9

3r = 18 – 9

3r = 9

r = 9/3

= 3

**(v) **x^{m} in the expansion of (x + 1/x)^{n}

Given:

(x + 1/x)^{n}

If x^{m} occurs at the (r + 1)th term in the given expression.

Then, we have:

T_{r+1 }= ^{n}C_{r} x^{n-r} a^{r}

**(vi) **x in the expansion of (1 – 2x^{3} + 3x^{5}) (1 + 1/x)^{8}

Given:

(1 – 2x^{3} + 3x^{5}) (1 + 1/x)^{8}

If x occurs at the (r + 1)th term in the given expression.

Then, we have:

(1 – 2x^{3} + 3x^{5}) (1 + 1/x)^{8}** = **(1 – 2x^{3} + 3x^{5}) (^{8}C_{0} + ^{8}C_{1} (1/x) + ^{8}C_{2} (1/x)^{2} + ^{8}C_{3} (1/x)^{3} + ^{8}C_{4} (1/x)^{4} + ^{8}C_{5} (1/x)^{5} + ^{8}C_{6} (1/x)^{6} + ^{8}C_{7} (1/x)^{7} + ^{8}C_{8} (1/x)^{8})

So, ‘x’ occurs in the above expression at -2x^{3}.^{8}C_{2} (1/x^{2}) + 3x^{5}.^{8}C_{4} (1/x^{4})

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

= -56 + 210

*= 154*

**(vii) **a^{5}b^{7} in the expansion of (a – 2b)^{12}

Given:

(a – 2b)^{12}

If a^{5}b^{7} occurs at the (r + 1)th term in the given expression.

Then, we have:

T_{r+1 }= ^{n}C_{r} x^{n-r} a^{r}

**(viii) **x in the expansion of (1 – 3x + 7x^{2}) (1 – x)^{16}

Given:

(1 – 3x + 7x^{2}) (1 – x)^{16}

If x occurs at the (r + 1)th term in the given expression.

Then, we have:

(1 – 3x + 7x^{2}) (1 – x)^{16} = (1 – 3x + 7x^{2}) (^{16}C_{0} + ^{16}C_{1} (-x) + ^{16}C_{2} (-x)^{2} + ^{16}C_{3} (-x)^{3} + ^{16}C_{4} (-x)^{4} + ^{16}C_{5} (-x)^{5} + ^{16}C_{6} (-x)^{6} + ^{16}C_{7} (-x)^{7} + ^{16}C_{8} (-x)^{8} + ^{16}C_{9} (-x)^{9} + ^{16}C_{10} (-x)^{10} + ^{16}C_{11} (-x)^{11} + ^{16}C_{12} (-x)^{12} + ^{16}C_{13} (-x)^{13} + ^{16}C_{14} (-x)^{14} + ^{16}C_{15} (-x)^{15} + ^{16}C_{16} (-x)^{16})

So, ‘x’ occurs in the above expression at ^{16}C_{1} (-x) – 3x^{16}C_{0}

∴ Coefficient of x = -(16!/(1! 15!)) – 3(16!/(0! 16!))

= -16 – 3

*= -19*

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

**Solution:**

Let us consider T_{r+1 }th term in the given expansion contains x and y to one and the same power.

Then we have,

T_{r+1} = ^{n}C_{r} x^{n-r} a^{r}

**11. Does the expansion of (2x ^{2} – 1/x) contain any term involving x^{9}?**

**Solution:**

Given:

(2x^{2} – 1/x)

If x^{9} occurs at the (r + 1)th term in the given expression.

Then, we have:

T_{r+1} = ^{n}C_{r} x^{n-r} a^{r}

For this term to contain x^{9}, we must have

40 – 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 x ^{9} in the given expansion.*

**12. Show that the expansion of (x ^{2} + 1/x)^{12} does not contain any term involving x^{-1}.**

**Solution:**

Given:

(x^{2} + 1/x)^{12}

If x^{-1} occurs at the (r + 1)th term in the given expression.

Then, we have

T_{r+1} = ^{n}C_{r} x^{n-r} a^{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.*

**13. Find the middle term in the expansion of:**

**(i) (2/3x – 3/2x) ^{20} **

**(ii) (a/x + bx) ^{12} **

**(iii) (x ^{2} – 2/x)^{10} **

**(iv) (x/a – a/x) ^{10} **

**Solution:**

**(i) **(2/3x – 3/2x)^{20}

We have,

(2/3x – 3/2x)^{20} where n = 20 (even number)

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

Now,

T_{11} = T_{10+1}

= ^{20}C_{10} (2/3x)^{20-10} (3/2x)^{10}

= ^{20}C_{10} 2^{10}/3^{10} × 3^{10}/2^{10} x^{10-10}

= ^{20}C_{10}

*Hence, the middle term is ^{20}C_{10}.*

**(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., 7^{th} term

Now,

T_{7} = T_{6+1}

= 924 a^{6}b^{6}

*Hence, the middle term is 924 a ^{6}b^{6}.*

**(iii) **(x^{2} – 2/x)^{10}

We have,

(x^{2} – 2/x)^{10} where n = 10 (even number)

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

Now,

T_{6} = T_{5+1}

*Hence, the middle term is -8064x ^{5}.*

**(iv) **(x/a – a/x)^{10}

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) = 6, i.e., the 6^{th} term

Now,

T_{6} = T_{5+1}

*Hence, the middle term is -252.*

**14. Find the middle terms in the expansion of:**

**(i) (3x – x ^{3}/6)^{9}**

**(ii) (2x ^{2} – 1/x)^{7}**

**(iii) (3x – 2/x ^{2})^{15}**

**(iv) (x ^{4} – 1/x^{3})^{11} **

**Solution:**

**(i)** (3x – x^{3}/6)^{9}

We have,

(3x – x^{3}/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 5^{th} and 6^{th}.

Now,

T_{5} = T_{4+1}

*Hence, the middle terms are 189/8 x ^{17} and -21/16 x^{19}.*

**(ii) **(2x^{2} – 1/x)^{7}

We have,

(2x^{2} – 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 4^{th} and 5^{th}.

Now,

*Hence, the middle terms are -560x ^{5} and 280x^{2}.*

**(iii) **(3x – 2/x^{2})^{15}

We have,

(3x – 2/x^{2})^{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 8^{th} and 9^{th}.

Now,

*Hence, the middle terms are (-6435×3 ^{8}×2^{7})/x^{6} and (6435×3^{7}×2^{8})/x^{9}.*

**(iv) **(x^{4} – 1/x^{3})^{11}

We have,

(x^{4} – 1/x^{3})^{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 6^{th} and 7^{th}.

Now,

T_{7} = T_{6+1}

*Hence, the middle terms are -462x ^{9} and 462x^{2}.*

**15. Find the middle terms in the expansion of:**

**(i) (x – 1/x) ^{10} **

**(ii) (1 – 2x + x ^{2})^{n} **

**(iii) (1 + 3x + 3x ^{2} + x^{3})^{2n} **

**(iv) (2x – x ^{2}/4)^{9} **

**(v) (x – 1/x) ^{2n+1} **

**(vi) (x/3 + 9y) ^{10} **

**(vii) (3 – x ^{3}/6)^{7} **

**(viii) (2ax – b/x ^{2})^{12} **

**(ix) (p/x + x/p) ^{9} **

**(x) (x/a – a/x) ^{10} **

**Solution:**

**(i) **(x – 1/x)^{10}

We have,

(x – 1/x)^{10} where n = 10 (even number)

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

Now,

T_{6} = T_{5+1 }

*Hence, the middle term is -252.*

**(ii) **(1 – 2x + x^{2})^{n}

We have,

(1 – 2x + x^{2})^{n} = (1 – x)^{2n} where n is an even number.

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

Now,

T_{n} = T_{n+1 }

= ^{2n}C_{n} (-1)^{n} (x)^{n}

= (2n)!/(n!)^{2} (-1)^{n} x^{n}

*Hence, the middle term is (2n)!/(n!) ^{2} (-1)^{n} x^{n}.*

**(iii) **(1 + 3x + 3x^{2} + x^{3})^{2n}

We have,

(1 + 3x + 3x^{2} + x^{3})^{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,

T_{2n} = T_{3n+1 }

= ^{6n}C_{3n} x^{3n}

= (6n)!/(3n!)^{2} x^{3n}

*Hence, the middle term is (6n)!/(3n!) ^{2} x^{3n}.*

**(iv) **(2x – x^{2}/4)^{9}

We have,

(2x – x^{2}/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 5^{th} and 6^{th}.

Now,

T_{5} = T_{4+1}

And,

T_{6} = T_{5+1}

*Hence, the middle term is 63/4 x ^{13} and -63/32 x^{14}.*

**(v) **(x – 1/x)^{2n+1}

We have,

(x – 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,

T_{n} = T_{n+1}

And,

T_{n+2} = T_{n+1+1}

*Hence, the middle term is (-1) ^{n}.^{2n+1}C_{n} x and (-1)^{n+1}.^{2n+1}C_{n} (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., the 6th term.

Now,

T_{6} = T_{5+1 }

*Hence, the middle term is 61236x ^{5}y^{5}.*

**(vii) **(3 – x^{3}/6)^{7}

We have,

(3 – x^{3}/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 4^{th} and 5^{th}.

Now,

T_{4} = T_{3+1}

= ^{7}C_{3} (3)^{7-3} (-x^{3}/6)^{3}

= -105/8 x^{9}

And,

T_{5} = T_{4+1}

= ^{9}C_{4} (3)^{9-4} (-x^{3}/6)^{4}

*Hence, the middle terms are -105/8 x ^{9} and 35/48 x^{12}.*

**(viii) **(2ax – b/x^{2})^{12}

We have,

(2ax – b/x^{2})^{12} where n = 12 is an even number.

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

Now,

T_{7} = T_{6+1 }

*Hence, the middle term is (59136a ^{6}b^{6})/x^{6}.*

**(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 5^{th} and 6^{th}.

Now,

T_{5} = T_{4+1}

And,

T_{6} = T_{5+1}

= ^{9}C_{5} (p/x)^{9-5} (x/p)^{5}

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

**(x) **(x/a – a/x)^{10}

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) = 6, i.e., the 6^{th} term

Now,

T_{6} = T_{5+1}

*Hence, the middle term is -252.*

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

**(i) (3/2 x ^{2} – 1/3x)^{9} **

**(ii) (2x + 1/3x ^{2})^{9} **

**(iii) (2x ^{2} – 3/x^{3})^{25}**

**(iv) (3x – 2/x ^{2})^{15} **

**(v) ((√x/3) + √3/2x ^{2})^{10} **

**(vi) (x – 1/x ^{2})^{3n} **

**(vii) (1/2 x ^{1/3} + x^{-1/5})^{8} **

**(viii) (1 + x + 2x ^{3}) (3/2x^{2} – 3/3x)^{9} **

**(ix) (∛x + 1/2∛x) ^{18}, x > 0**

**(x) (3/2x ^{2} – 1/3x)^{6}**

**Solution:**

**(i) **(3/2 x^{2} – 1/3x)^{9}

Given:

(3/2 x^{2} – 1/3x)^{9}

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

Then, we have

T_{r+1} = ^{n}C_{r} x^{n-r} a^{r}

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

18 – 3r = 0

3r = 18

r = 18/3

= 6

So, the required term is the 7^{th} term.

We have,

T_{7} = T_{6+1}

= ^{9}C_{6} × (3^{9-12})/(2^{9-6})

= (9×8×7)/(3×2) × 3^{-3} × 2^{-3}

= 7/18

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

**(ii) **(2x + 1/3x^{2})^{9}

Given:

(2x + 1/3x^{2})^{9}

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

Then, we have

T_{r+1} = ^{n}C_{r} x^{n-r} a^{r}

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

9 – 3r = 0

3r = 9

r = 9/3

= 3

So, the required term is the 4^{th} term.

We have,

T_{4} = T_{3+1}

= ^{9}C_{3} × (2^{6})/(3^{3})

= ^{9}C_{3} × 64/27

*Hence, the term independent of x is ^{9}C_{3} × 64/27.*

**(iii) **(2x^{2} – 3/x^{3})^{25}

Given:

(2x^{2} – 3/x^{3})^{25}

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

Then, we have:

T_{r+1} = ^{n}C_{r} x^{n-r} a^{r}

= ^{25}C_{r} (2x^{2})^{25-r} (-3/x^{3})^{r}

= (-1)^{r} ^{25}C_{r }× 2^{25-r} × 3^{r} x^{50-2r-3r}

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

50 – 5r = 0

5r = 50

r = 50/5

= 10

So, the required term is the 11^{th} term.

We have,

T_{11} = T_{10+1}

= (-1)^{10} ^{25}C_{10} × 2^{25-10} × 3^{10}

= ^{25}C_{10} (2^{15} × 3^{10})

*Hence, the term independent of x is ^{25}C_{10} (2^{15} × 3^{10}).*

**(iv) **(3x – 2/x^{2})^{15}

Given:

(3x – 2/x^{2})^{15}

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

Then, we have:

T_{r+1} = ^{n}C_{r} x^{n-r} a^{r}

= ^{15}C_{r} (3x)^{15-r} (-2/x^{2})^{r}

= (-1)^{r} ^{15}C_{r} × 3^{15-r} × 2^{r} x^{15-r-2r}

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

15 – 3r = 0

3r = 15

r = 15/3

= 5

So, the required term is the 6^{th} term.

We have,

T_{6} = T_{5+1}

= (-1)^{5} ^{15}C_{5} × 3^{15-5} × 2^{5}

= -3003 × 3^{10} × 2^{5}

*Hence, the term independent of x is -3003 × 3 ^{10} × 2^{5}.*

**(v) **((√x/3) + √3/2x^{2})^{10}

Given:

((√x/3) + √3/2x^{2})^{10}

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

Then, we have:

T_{r+1} = ^{n}C_{r} x^{n-r} a^{r}

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

(10-r)/2 – 2r = 0

10 – 5r = 0

5r = 10

r = 10/5

= 2

So, the required term is the 3^{rd} term.

We have,

T_{3} = T_{2+1}

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

**(vi) **(x – 1/x^{2})^{3n}

Given:

(x – 1/x^{2})^{3n}

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

Then, we have:

T_{r+1} = ^{n}C_{r} x^{n-r} a^{r}

= ^{3n}C_{r} x^{3n-r} (-1/x^{2})^{r}

= (-1)^{r} ^{3n}C_{r} x^{3n-r-2r}

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

3n – 3r = 0

r = n

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

We have,

(-1)^{n} ^{3n}C_{n}

*Hence, the term independent of x is (-1) ^{n} ^{3n}C_{n}*

**(vii) **(1/2 x^{1/3} + x^{-1/5})^{8}

Given:

(1/2 x^{1/3} + x^{-1/5})^{8}

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

Then, we have:

T_{r+1} = ^{n}C_{r} x^{n-r} a^{r}

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

(8-r)/3 – r/5 = 0

(40 – 5r – 3r)/15 = 0

40 – 5r – 3r = 0

40 – 8r = 0

8r = 40

r = 40/8

= 5

So, the required term is the 6th term.

We have,

T_{6} = T_{5+1}

= ^{8}C_{5} × 1/(2^{8-5})

= (8×7×6)/(3×2×8)

= 7

*Hence, the term independent of x is 7.*

**(viii) **(1 + x + 2x^{3}) (3/2x^{2} – 3/3x)^{9}

Given:

(1 + x + 2x^{3}) (3/2x^{2} – 3/3x)^{9}

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

Then, we have:

(1 + x + 2x^{3}) (3/2x^{2} – 3/3x)^{9}** =**

= 7/18 – 2/27

= (189 – 36)/486

= 153/486 (divide by 9)

= 17/54

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

**(ix) **(∛x + 1/2∛x)^{18}, x > 0

Given:

(∛x + 1/2∛x)^{18}, x > 0

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

Then, we have

T_{r+1} = ^{n}C_{r} x^{n-r} a^{r}

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

(18-r)/3 – r/3 = 0

(18 – r – r)/3 = 0

18 – 2r = 0

2r = 18

r = 18/2

= 9

So, the required term is the 10th term.

We have,

T_{10} = T_{9+1}

= ^{18}C_{9} × 1/2^{9}

*Hence, the term independent of x is ^{18}C_{9} × 1/2^{9}.*

**(x) **(3/2x^{2} – 1/3x)^{6}

Given:

(3/2x^{2} – 1/3x)^{6}

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

Then, we have:

T_{r+1} = ^{n}C_{r} x^{n-r} a^{r}

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

12 – 3r = 0

3r = 12

r = 12/3

= 4

So, the required term is the 5th term.

We have,

T_{5} = T_{4+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 ^{n}C_{r-1}

So, the coefficients of the (2r + 4) and (r – 2) terms in the given expansion are ^{18}C_{2r+4-1} and ^{18}C_{r-2-1}

For these coefficients to be equal, we must have

^{18}C_{2r+4-1} = ^{18}C_{r-2-1}

^{18}C_{2r+3} = ^{18}C_{r-3}

2r + 3 = r – 3 (or) 2r + 3 + r – 3 = 18 [Since, ^{n}C_{r} = ^{n}C_{s} => r = s (or) r + s = n]

2r – r = -3 – 3 (or) 3r = 18 – 3 + 3

r = -6 (or) 3r = 18

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

r = -6 (or) r = 6

*∴ r = 6 [since r should be a positive integer.]*

**18. If the coefficients of (2r + 1)th term and (r + 2)th term in the expansion of (1 + x) ^{43} are equal, find r.**

**Solution:**

Given:

(1 + x)^{43}

We know the coefficient of the r term in the expansion of (1 + x)^{n} is ^{n}C_{r-1}

So, the coefficients of the (2r + 1) and (r + 2) terms in the given expansion are ^{43}C_{2r+1-1} and ^{43}C_{r+2-1}

For these coefficients to be equal, we must have

^{43}C_{2r+1-1} = ^{43}C_{r+2-1}

^{43}C_{2r} = ^{43}C_{r+1}

2r = r + 1 (or) 2r + r + 1 = 43 [Since, ^{n}C_{r} = ^{n}C_{s} => r = s (or) r + s = n]

2r – r = 1 (or) 3r + 1 = 43

r = 1 (or) 3r = 43 – 1

r = 1 (or) 3r = 42

r = 1 (or) r = 42/3

r = 1 (or) r = 14

*∴ r = 14 [since value ‘1’ gives the same term]*

**19. Prove that the coefficient of (r + 1)th term in the expansion of (1 + x) ^{n + 1} is equal to the sum of the coefficients of rth and (r + 1)th terms in the expansion of (1 + x)^{n}.**

**Solution:**

We know, the coefficients of (r + 1)th term in (1 + x)^{n+1} is ^{n+1}C_{r}

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

(1 + x)^{n} = ^{n}C_{r-1} + ^{n}C_{r}

= ^{n+1}C_{r} [since, ^{n}C_{r+1} + ^{n}C_{r} = ^{n+1}C_{r+1}]

*Hence proved.*

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

## Frequently Asked Questions on RD Sharma Solutions for Class 11 Maths Chapter 18

### How do RD Sharma Solutions for Class 11 Maths Chapter 18 help students to understand the concepts which are crucial from an exam perspective?

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### Name the concepts covered in Chapter 18 of RD Sharma Solutions for Class 11 Maths.

The concepts covered in Chapter 18 of RD Sharma Solutions for Class 11 Maths are as listed:

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

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