Class 11 RD Sharma Solutions – Chapter 18 Binomial Theorem- Exercise 18.2 | Set 2

Last Updated : 19 Jul, 2021

(i) (3x â€“ x3/6)9

Solution:

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

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

And, T6 = T5+1

= 9C5 (3x)9-5 (x3/6)5

(ii) (2x2 â€“ 1/x)7

Solution:

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,

T4 = T3+1

= 7C3 (2x2)7-3 (-1/x)3

= -\frac{7Ã—6Ã—5}{3Ã—2}Ã—16x^8Ã—\frac{1}{x^3}

= âˆ’ 560 x5

And, T5 = T4+1

= 7C4 (2x2)7-4 (-1/x)4

= 280 x2

(iii) (3x â€“ 2/x2)15

Solution:

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,

T8 = T7+1

= 15C7 (3x)15-7 (â€“ 2/x2)7

And, T9 = T8+1

= 15C8 (3x)15-8 (â€“ 2/x2)8

(iv) (x4 â€“ 1/x3)11

Solution:

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,

T6 = T5+1

= 11C5 (x4)11-5 (1/x3)5

= -462 x9

And, T7 = T6+1

= 11C6 (x4)11-6 (1/x3)6

= 462 x2

(i) (x â€“ 1/x)10

Solution:

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) = 6th term

Now,

T6 = T5+1

= 10C5 (x)10-5 (â€“1/x)5

= âˆ’252

(ii) (1 – 2x + x2)n

Solution:

We have, (1 – 2x + x2)n

= (1 – x)2n

Here, n is an even number.

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

Now,

Tn+1 = 2nCn (-1)n xn

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

Solution:

We have, (1 + 3x + 3 x2 + x3)2n

= (1 + x )6n

Here, n is an even number.

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

Now,

T3n+1 = 6nC3n x3n

(iv) (2x â€“ x2/4)9

Solution:

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

= 9C4 (2x)9-4 (â€“x2/4)4

And, T6 = T5+1

= 9C5 (2x)9-5 (â€“x2/4)5

(v) (x – 1/x)2n+1

Solution:

We have,  (x – 1/x)2n+1

Here, 2n + 1 is an odd number.

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

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

Now

Tn+1

=(-1)n  2n+1Cn x

And Tn+2 = Tn+1+1

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

Solution:

We have,

(x/3 + 9y)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/3)10-5 (9y)5

= 61236 x5 y5

(vii) (3 â€“ x3/6)7

Solution:

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

=

And, T5 = T4+1

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

(viii) (2ax â€“ b/x2)12

Solution:

We have,

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

So the middle terms are (n/2 + 1) = (12/2 + 1) = 7th term

Now,

T7 = T6+1

= 12C6 (2ax)12-6 (-b/x2)6

= 12C6 (2ax)6 (b/x2)6

= 12C6 (26a6x6)(b6/x12)

= 12C6 (26a6b6/x6)

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

Solution:

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

= 9C4 (p/x)9-4 (x3/p)4

= 9C4 (p/x)5 (x/p)4

= 9C4 (p/x)

And, T6 = T5+1

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

= 9C5 (p/x)4(x/p)5

= 9C5 (x/p)

(x) (x/a â€“ a/x)10

Solution:

We have,

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

So the middle terms are (n/2 + 1) = (10/2 + 1) 6th term

Now,

T6 = T5+1

= 10C5 (x/a)10-5 (-a/x)5

= –10C5 (x/a)5 (a/x)5

= –10C5

= -252

(i) (3/2 x2 â€“ 1/3x)9

Solution:

We have,

(3/2 x2 â€“ 1/3x)9

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

Tr+1 = nCr xn-r ar

= 9Cr (3/2x2)9-r (-1/3x)r

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

=> 18 â€“ 3r = 0

=> 3r = 18

=> r = 18/3

=> r = 6

So, the required term is 7th term.

So, T7 = T6+1

= 9C6 Ã— (39-12)/(29-6)

= 7/18

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

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

Solution:

We have,

(2x + 1/3x2)9

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

Tr+1 = nCr xn-r ar

= 9Cr (2x)9-r (1/3x2)r

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

=> 9 â€“ 3r = 0

=> r = 3

So, the required term is 4th term.

So, T7 = T6+1

= 9C6 Ã— (29-3)/(33)

= 5376/27

(iii) (2x2 â€“ 3/x3)25

Solution:

We have,

(2x2 â€“ 3/x3)25

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

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 = 10

So, the required term is 11th term.

So, T11 = T10+1

= (-1)10 25C10 Ã— 225-10 Ã— 310

= 25C10 (215 Ã— 310)

(iv) (3x â€“ 2/x2)15

Solution:

We have,

(3x â€“ 2/x2)15

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

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 = 5

So, the required term is 6th term.

So, T6 = T5+1

= (-1)5 15C5 Ã— 315-5 Ã— 25

= âˆ’3003 Ã— 310 Ã— 25

(v)

Solution:

We have

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

Now,

Tr+1

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

=> (10-r)/2 – 2r = 0

=> 10 – 5r = 0

=> r = 2

So, the required term is 3th term.

So, T3 = T2+1

T3

= 5/4

(vi)

Solution:

We have

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

Now,

Tr+1

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

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.

So,

Tn+1 = (-1)n 3nCn

(vii)

Solution:

We have

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

Now,

Tr+1

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

=> (8 – r)/3 – r/5 = 0

=> 40 – 5r – 3r = 0

=> 8r = 40

=> r = 5

So, the required term is 6th term.

So, T6 = T5+1

T6

= 7

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

Solution:

We have

(1 + x + 2x3) (3/2x2 â€“ 3/3x)9

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

Now,

Tr+1 = (1 + x + 2x3) [(3/2x2) – 9C1 (3/2x2)8 (1/3x) + . . . – 9C7 (3/2x2)2 (1/3x)7]

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

= 9C6 (33/23) (1/36) – 2x3 9C7 (23/33) (1/37) (1/x3)

= 7/18 â€“ 2/27

= (189 â€“ 36)/486

= 153/486

= 17/54

(ix) , x > 0

Solution:

We have

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

Now,

Tr+1

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

=> (18 – r)/3 – r/3 = 0

=> 18 – 2r = 0

=> r = 9

So, the required term is 9th term.

So, T9 = T8+1

T9 = 18C9 (1/29)

(x)

Solution:

We have

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

Now,

Tr+1

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

=> 12 – 3r = 0

=> r = 4

Hence, the required term is the 4th term.

So, T4 = T3+1

= 5/12

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

Solution:

We are given,

(1 + x)18

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

So, the coefficients of the (2r + 4)th and (r â€“ 2)th terms in the given expansion are,

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

According to the question, we have,

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

=> 18C2r+3 = 18Cr-3

We know if nCr = nCs , then r = s or r + s = n.

=> 2r + 3 = r â€“ 3 or 2r + 3 + r â€“ 3 = 18

=> 2r â€“ r = â€“3 â€“ 3 or 3r = 18 â€“ 3 + 3

=> r = â€“6 or 3r = 18

=> r = â€“6 or r = 6

Ignoring r = â€“ 6 as r cannot be negative.

Therefore, the value of r is 6.

Question 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:

We are given,

(1 + x)43

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

So, the coefficients of the (2r + 1)th and (r + 2)th terms in the given expansion are,

43C2r+1-1 and 43Cr+2-1

According to the question, we have,

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

=> 43C2r = 43Cr+1

We know if nCr = nCs , then r = s or r + s = n.

=> 2r = r + 1 or 2r + r + 1 = 43

=> r = 1 or 3r = 42

=> r = 1 or r = 14

Ignoring r = 1 as it gives the same term on both the sides.

Therefore, the value of r is 14.

Question 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+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

As, nCr+1 + nCr = n+1Cr+1

= n+1Cr

Hence proved.

Question 20. Prove that the term independent of x in the expansion of (x + 1/x)2n is .

Solution:

We have,

(x + 1/x)2n

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

Tr+1 = nCr xn-r ar

= 2nCr x2n-r (1/x)r

= 2nCr x2n-2r

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

=> 2n âˆ’ 2r = 0

=> 2n = 2r

=> r = n

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

Tn+1 = 2nCn x2n-n (1/x)n

= 2nCn

Hence proved.

Question 21. If the coefficients of 5th, 6th and 7th terms of the expansion (1 + x)n are in A.P., find n.

Solution:

We have, (1 + x)n

We know the coefficient of rth term of a binomial expression is given by nCr-1.

Coefficient of 5th term = nC5-1 = nC4

Coefficient of 6th term = nC6-1 = nC5

Coefficient of 7th term = nC7-1 = nC6

According to the question, we have,

=> 2 nC5 = nC4 + nC6

=>

=>

=>

=>

=>

=>

=> 60(nâˆ’4) = 150 + 5n2 âˆ’ 45n + 100

=> 60n âˆ’ 240 = 250 + 5n2 âˆ’ 45n

=> 5n2 âˆ’ 105n + 490 = 0

=> n2 âˆ’ 21n + 98 = 0

=> n2 âˆ’ 7n âˆ’ 14n + 98 = 0

=> n (n âˆ’ 7) âˆ’ 14 (n âˆ’ 7) = 0

=> (n âˆ’ 7) (n âˆ’ 14) = 0

=> n = 7 or n = 14

Therefore, the value of n is 7 or 14.

Question 22. If the coefficients of 2nd, 3rd, and 4th terms of the expansion (1 + x)2n are in A.P., show that 2n2 âˆ’ 9n + 7 = 0.

Solution:

We have, (1 + x)2n

We know the coefficient of rth term of a binomial expression is given by nCr-1.

Coefficient of 2nd term = 2nC2-1 = 2nC1

Coefficient of 3rd term = 2nC3-1 = 2nC2

Coefficient of 4th term = 2nC4-1 = 2nC3

According to the question, we have,

=> 2 2nC2 = 2nC1 + 2nC3

=> 2 =

=> 2 =

=>  = 2

=>  = 2

=> 4n2 âˆ’ 6n + 8 = 12n âˆ’ 6

=> 4n2 âˆ’ 18n + 14 = 0

=> 2 (2n2 âˆ’ 9n + 7) = 0

=> 2n2 âˆ’ 9n + 7 = 0

Hence proved.

Question 23. In the expansion of (1 + x)n, the binomial coefficients of three consecutive terms are respectively 220, 495, and 792, find the value of n.

Solution:

We have, (1 + x)n

Let the three consecutive terms be rth, (r+1)th and (r+2)th.

We know the coefficient of rth term of a binomial expression is given by nCr-1.

Coefficient of rth term = nCr-1 = 220

Coefficient of (r+1)th term = nCr+1-1 = nCr = 495

Coefficient of (r+2)th term = nCr+2-1 = nCr+1 = 792

Now,

=>

=>

=> 5n âˆ’ 5r = 8r + 8

=> 5n âˆ’ 13r = 8  . . . . (1)

Also,

=>

=> 4n âˆ’ 4r + 4 = 9r

=> 4n âˆ’ 13r = âˆ’4  . . . . (2)

Subtracting (2) from (1), we get,

=> n = 8 + 4

=> n = 12

Therefore, the value of n is 12.

Question 24. If the coefficients of 2nd, 3rd and 4th terms of the expansion (1 + x)n are in A.P., then find the value of n.

Solution:

We have, (1 + x)n

We know the coefficient of rth term of a binomial expression is given by nCr-1.

Coefficient of 2nd term = nC2-1 = nC1

Coefficient of 3rd term = nC3-1 = nC2

Coefficient of 4th term = nC4-1 = nC3

According to the question, we have,

=> 2 nC2 = nC1 + nC3

=>

=>

=>

=>  = 2

=> n2 âˆ’ 3n + 8 = 6 (n âˆ’ 1)

=> n2 âˆ’ 3n + 8 = 6n âˆ’ 6

=> n2 âˆ’ 9n + 14 = 0

=> n2 âˆ’ 7n âˆ’ 2n + 14 = 0

=> n (nâˆ’7) âˆ’ 2 (nâˆ’7) = 0

=> (n âˆ’ 2) (n âˆ’ 7) = 0

=> n = 2 or n = 7

Ignoring n = 2 as it does not satisfy our condition.

Therefore, the value of n is 7.

Question 25. If in the expansion of (1 + x)n, the coefficients of pth and qth terms are equal, then prove that p + q = n + 2, where p â‰  q.

Solution:

We have, (1 + x)n

We know the coefficient of rth term of a binomial expression is given by nCr-1.

Coefficient of pth term = nCp-1

Coefficient of qth term = nCq-1

According to the question, we have,

=> nCp-1 = nCq-1

=> p âˆ’ 1 = q âˆ’ 1 or p âˆ’ 1 + q âˆ’ 1 = n

=> p = q or p + q âˆ’ 2 = n

=> p + q âˆ’ 2 = n

=> p + q = n + 2

Hence proved.

Question 26.  If in the expansion of (1 + x)n, the binomial coefficients of three consecutive terms are respectively 56, 70, and 56, find n and the position of the terms of these coefficients.

Solution:

We have, (1 + x)n

Let the three consecutive terms be rth, (r+1)th and (r+2)th.

We know the coefficient of rth term of a binomial expression is given by nCr-1.

Coefficient of rth term = nCr-1 = 56

Coefficient of (r+1)th term = nCr+1-1 = nCr = 70

Coefficient of (r+2)th term = nCr+2-1 = nCr+1 = 56

Now,

=>

=>

=> 5n âˆ’ 5r = 4r + 4

=> 5n âˆ’ 9r = 4  . . . . (1)

Also,

=>

=> 4n âˆ’ 4r + 4 = 5r

=> 4n âˆ’ r = âˆ’4  . . . . (2)

Subtracting (2) from (1), we get,

=> n = 4 + 4

=> n = 8

Putting n = 8 in (1), we get,

=> 5(8) âˆ’ 9r = 4

=> 40 âˆ’ 9r = 4

=> 9r = 36

=> r = 4

Therefore, three consecutive terms are 4th, 5th and 6th terms.

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