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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)

\frac{10×9×8×7×81}{4×3×2×x^{10}}

= 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)

\frac{10×9×8×7×81}{4×3×2×x^8}

= 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

\frac{-(10×9×8)}{3×2} x^{9/2} y^{3/2} (x^{7/2} y^{21/2})



= –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

^8C_6(\frac{4x}{5})^{8-6}(\frac{5}{2x})^6

\frac{8×7×6!}{6!2!}(\frac{16x^2}{25})(\frac{125×125}{64x^6})

= 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

\frac{9×8×7}{3×2}(x^3)(\frac{64}{x^6})

= 5376/ x3

Now, the 4th term from the beginning is,

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

\frac{9×8×7}{3×2}(x^6)(\frac{8}{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

^9C_6(\frac{4x}{5})^{9-6}(\frac{5}{2x})^6

\frac{9×8×7}{3×2}(\frac{64x^3}{125})(\frac{125×125}{64x^6})

= 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



^8C_2(2x^2)^{8-2}(\frac{-3}{2x})^2

\frac{8×7}{2×1}(64x^{12})(\frac{9}{4x^2})

= 4032 x10

Question 9. Find the coefficient of:

(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

-\frac{10×9×8}{3×2×9×9}a^7

= −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

-\frac{9×8×7}{2×9×9}

= −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

\frac{n!}{\left( \frac{n - m}{2} \right)! \left( \frac{n + m}{2} \right)!}

(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 = -2\left( \frac{8!}{2! 6!} \right) + 3\left( \frac{8!}{4! 4!} \right)

= – 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

-\frac{12×11×10×9×8×128}{5×4×3×2}

= − 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 = -\left( \frac{16!}{1! 15!} \right) - 3\left( \frac{16!}{0! 16!} \right)

= – 16 – 3 

= – 19

Question 10. Which term in the expansion of \left[\left(\frac{x}{\sqrt{y}}\right)^{\frac{1}{3}}+\left(\frac{x}{\sqrt[3]{y}}\right)^{\frac{1}{2}}\right]^{21}        contains x and y to one and the same power.

Solution:

We are given, \left[\left(\frac{x}{\sqrt{y}}\right)^{\frac{1}{3}}+\left(\frac{x}{\sqrt[3]{y}}\right)^{\frac{1}{2}}\right]^{21}

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

Tr+1 = nCr xn-r ar

^{21}C_r\left[(\frac{x}{\sqrt{y}})^{\frac{1}{3}}\right]^{21-r}\left[(\frac{y}{\sqrt[3]x})^{\frac{1}{2}}\right]^{r}

= 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.

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

(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

-\frac{10×9×8×7×6}{5×4×3×2×1}×32x^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

-\frac{10×9×8×7×6}{5×4×3×2×1}

= −252

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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