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

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

^{2})

^{25}.

**Solution:**

We are given, (2x â€“ 1/x

^{2})^{25}.The given expression contains 25 + 1 = 26 terms.

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

Hence, T

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

^{25}C_{15}(2^{10}) (x)^{10}(âˆ’1/x^{30})= â€“

^{25}C_{15}(2^{10}/ x^{20})Now, the 11th term from the beginning is,

T

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

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

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

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

^{2}â€“ 1/x

^{3})

^{10}.

**Solution:**

We are given, (3x

^{2}â€“ 1/x^{3})^{10}.The 7th term of the expression is given by,

T

_{7}= T_{6+1}=

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

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

= 17010/ x

^{10}

**Question 3. Find the 5th term from the end in the expansion of (3x â€“ 1/x**^{2})^{10}.

^{2})

^{10}.

**Solution:**

We are given, (3x â€“ 1/x

^{2})^{10}The 5th term from the end is the (11 â€“ 5 + 1)th = 7th term from the beginning.

So, T

_{7}= T_{6+1}=

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

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

= 17010/ x

^{8}

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

^{3/2}y

^{1/2}â€“ x

^{1/2}y

^{3/2})

^{10}.

**Solution:**

We are given, (x

^{3/2}y^{1/2}â€“ x^{1/2}y^{3/2})^{10}.The 8th term of the expression is given by,

T

_{8}= T_{7+1}=

^{10}C~~(x~~_{7}^{3/2}y^{1/2})^{10â€“7}(â€“x^{1/2 }y^{3/2})^{7}=

= â€“120 x

^{8}y^{12}

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

^{8}.

**Solution:**

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

^{8}.The 8th term of the expression is given by,

T

_{7}= T_{6+1}=

=

= 4375/ x

^{4}

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

^{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 âˆ’ 4 + 1) th term = 7th term from the beginning.

Hence, T

_{7}= T_{6+1}=^{9}C_{6}(x)^{9-6}(2/x)^{6}=

= 5376/ x

^{3}Now, the 4th term from the beginning is,

T

_{4}= T_{3+1}=^{9}C_{3}(2x)^{9-3}(2/x)^{3}=

= 672 x

^{3}

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

^{9}.

**Solution:**

We are given, (4x/5 â€“ 5/2x)

^{9}The 4th term from the end is (10 âˆ’ 4 + 1)th term = 7th term, from the beginning.

T

_{7}= T_{6+1}=

=

= 10500/ x

^{3}

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

^{2}â€“ 3/2x)

^{8}.

**Solution:**

We are given, (2x

^{2}â€“ 3/2x)^{8}The 7th term from the end is (9 âˆ’ 7 + 1)th term = 3rd term, from the beginning.

T

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

=

= 4032 x

^{10}

**Question 9. Find the coefficient of:**

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

^{10}in the expansion of (2x

^{2}â€“ 1/x)

^{20}

^{ }Solution:^{ }

We are given, (2x

^{2}â€“ 1/x)^{20}We know, the (r+1)

^{th}term of the expression is given by,T

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

^{20}C_{r}(2x^{2})^{20-r}(-1/x)^{r}= (-1)

^{r}^{20}C_{r}(2)^{20-r}x^{40-2r-r}If x

^{10}exists in the expansion, we must have,=> 40 âˆ’ 3r = 10

=> 3r = 30

=> r = 10

Coefficient of x

^{10}= (-1)^{10}^{20}C_{10}(2)^{20-10}=

^{20}C_{10}(2)^{10}

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

^{7}in the expansion of (x â€“ 1/x

^{2})

^{40}

**Solution:**

We are given, (x â€“ 1/x

^{2})^{40}We know, the (r+1)

^{th}term of the expression is given by,T

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

^{40}C_{r}(x)^{40-r}(-1/x^{2})^{r}= (-1)

^{r}^{40}C_{r}x^{40-r-2r}If x

^{7}exists in the expansion, we must have,=> 40 âˆ’ 3r = 7

=> 3r = 33

=> r = 11

Coefficient of x

^{7}= (-1)^{11}^{40}C_{11}= âˆ’

^{40}C_{11 }

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

^{-15}in the expansion of (3x

^{2}â€“ a/3x

^{3})

^{10}

**Solution:**

We are given, (3x

^{2}â€“ a/3x^{3})^{10}We know, the (r+1)

^{th}term of the expression is given by,T

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

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

^{r}^{10}C_{r}3^{10-r-r}x^{20-2r-3r}a^{r}If x

^{-15}exists in the expansion, we must have,=> 20 âˆ’ 5r = âˆ’15

=> 5r = 35

=> r = 7

Coefficient of x

^{-15}= (-1)^{7}^{10}C_{7}3^{10-14}a^{7}=

= âˆ’40a

^{7}/27

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

^{9}in the expansion of (x

^{2}â€“ 1/3x)

^{9}

**Solution:**

We are given, (x

^{2}â€“ 1/3x)^{9}We know, the (r+1)

^{th}term of the expression is given by,T

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

^{9}C_{r}(x^{2})^{9-r}(-1/3x)^{r}= (-1)

^{r}^{9}C_{r}x^{18-2r-r}(1/3)^{r}If x

^{9}exists in the expansion, we must have,=> 18 âˆ’ 3r = 9

=> 3r = 9

=> r = 3

Coefficient of x

^{9}= (-1)^{3}^{9}C_{3}(1/3)^{3}=

= âˆ’28/9

### (v) x^{m} 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,T

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

^{n}C_{r}x^{n- 2r }If x

^{m}exists in the expansion, we must have,=> n – 2r = m

=> r = (n – m)/2

Coefficient of x

^{m}=^{n}C_{(n-m)/2}=

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

**Solution:**

We are given, (1 – 2x

^{3}+ 3x^{5}) (1 + 1/x)^{8}We know, the (r + 1)

^{th}term of the expression is given by,= (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}))Here, x appear in the above expression at -2 x

^{3 8}C_{2}(1/x^{2}) + 3x^{5}.^{8}C_{4}(1/x^{4})So, coefficient of x =

= – 56 + 210

= 154

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

^{5}b

^{7}in the expansion of (a â€“ 2b)

^{12}

**Solution:**

We are given, (a â€“ 2b)

^{12}We know, the (r+1)

^{th}term of the expression is given by,T

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

^{r}^{12}C_{r }(a)^{12-r}(2b)^{r}If a

^{5}b^{7 }exists in the expansion, we must have,=> 12 âˆ’ r = 5

=> r = 7

Coefficient of a

^{5}b^{7}= (-1)^{7 12}C_{7}(2)^{7}=

= âˆ’ 101376

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

^{2}) (1 – x)

^{16}

**Solution:**

We are given,

(1 – 3x + 7 x^{2}) (1 – x)^{16}We know, the (r + 1)

^{th}term of the expression is given by,= (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})Here, x appear in the above expression at

^{16}C_{1}(-x) – 3x^{16}C_{0}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,T

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

=

^{21}C_{r}x^{7-r/2}y^{2r/3-7/2}If x and y have same power, we must have,

=> 7 âˆ’ r/2 = 2r/3 âˆ’ 7/2

=> 7r/6 = 21/2

=> r = 9

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

**Question 11. Does the expansion of (2x**^{2} â€“ 1/x)^{20} contain any term involving x^{9}?

^{2}â€“ 1/x)

^{20}contain any term involving x

^{9}?

**Solution:**

We are given, (2x

^{2}â€“ 1/x)^{20}We know, the (r+1)

^{th}term of the expression is given by,T

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

^{20}C_{r}(2x^{2})^{20-r}(1/x)^{r}=

^{20}C_{r}(2)^{20-r}x^{40-2r-r}If x

^{9}exists in the expansion, we must have,=> 40 âˆ’ 3r = 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.

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

^{2}+ 1/x)

^{12}does not contain any term involving x

^{-1}.

**Solution:**

We have, (x

^{2}+ 1/x)^{12}We know, the (r+1)

^{th}term of the expression is given by,T

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

^{12}C_{r}(x^{2})^{12-r}(1/x)^{r}=

^{12}C_{r}x^{24-2r-r}For this term to contain x

^{-1}, we must have=> 24 â€“ 3r = âˆ’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.

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

**(i) (2/3x â€“ 3/2x)**^{20}

^{20}

**Solution:**

We have,

(2/3x â€“ 3/2x)

^{20}where, n = 20 (which is an even number)So, the middle term is (n/2 + 1) = (20/2 + 1) = (10 + 1) = 11

^{th}termNow,

T

_{11}= T_{10+1}=

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

^{20}C_{10}(2^{10}/3^{10}) Ã— (310/210) x^{10-10}=

^{20}C_{10}

**(ii) (x**^{2} â€“ 2/x)^{10}

^{2}â€“ 2/x)

^{10}

**Solution:**

We have,

(x

^{2}â€“ 2/x)^{10}where, n = 10 (which is an even number)So, the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6

^{th}termNow,

T

_{6}= T_{5+1}=

^{10}C_{5}(x^{2})^{10-5}(-2/x)^{5}=

= âˆ’ 8064 x

^{5}

**(iii) (x/a â€“ a/x)**^{10}

^{10}

**Solution:**

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

^{th}termNow,

T

_{6}= T_{5+1}=

^{10}C_{5}(x/a)^{10-5}(-a/x)^{5}=

= âˆ’252

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