Rd Sharma Xi (2018) Solutions for Class 12 Science Math Chapter 18 Binomial Theorem are provided here with simple step-by-step explanations. These solutions for Binomial Theorem are extremely popular among class 12 Science students for Math Binomial Theorem Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi (2018) Book of class 12 Science Math Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi (2018) Solutions. All Rd Sharma Xi (2018) Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

Page No 18.11:

Question 1:

Using binomial theorem, write down the expansions of the following:
(i) 2x+3y5

(ii) 2x-3y4

(iii) x-1x6

(iv) 1-3x7

(v) ax-bx6

(vi) xa-ax6

(vii) x3-a36

(viii) 1+2x-3x25

(ix) x+1-1x

(x) 1-2x+3x23

Answer:

(i) (2x + 3y)5

=C05(2x)5(3y)0+C15(2x)4(3y)1+C25(2x)3(3y)2+C35(2x)2(3y)3+C45(2x)1(3y)4+C55(2x)0(3y)5
=32x5+5×16x4×3y+10×8x3×9y2+10×4x2×27y3+5×2x×81y4+243y5=32x5+240x4y+720x3y2+1080x2y3+810xy4+243y5

(ii) (2x − 3y)4

=C04(2x)4(3y)0-C14(2x)3(3y)1+C24(2x)2(3y)2-C34(2x)1(3y)3+C44(2x)0(3y)4=16x4-4×8x3×3y+6×4x2×9y2-4×2x×27y3+81y4=16x4-96x3y+216x2y2-216xy3+81y4

(iii)
x-1x6=C06 x61x0-C16 x51x1+C26 x41x2-C36 x31x3+C46 x21x4-6C5 x11x5+C66 x01x6=x6-6 x5×1x+15 x4×1x2-20x3×1x3+15x2×1x4-6 x×1x5+1x6=x6-6x4+15x2-20+15x2-6x4+1x6

(iv) (1 − 3x)7
=C07(3x)0-C17(3x)1+C27(3x)2-C37(3x)3+C47(3x)4-C57(3x)5+C67(3x)6-C77(3x)7=1-7×3x+21×9x2-35×27x3+35×81x4-21×243x5+7×729x6-2187x7=1-21x+189x2-945x3+2835x4-5103x5+5103x6-2187x7

(v) 

(ax-bx)6=C06(ax)6(bx)0-C16(ax)5(bx)1+C26(ax)4(bx)2-C36(ax)3(bx)3+C46(ax)2(bx)4-C56(ax)1(bx)5+C66(ax)0(bx)6
=a6x6-6a5x5×bx+15a4x4×b2x2-20a3b3×b3x3+15a2x2×b4x4-6ax×b5x5+b6x6=a6x6-6a5x4b+15a4x2b2-20a3b3+15a2b4x2-6ab5x4+b6x6

(vi)
xa-ax6=C06xa6ax0-C16xa5ax1+C26xa4ax2-C36xa3ax3+C46xa2ax4-C56xa1ax5+C66xa0ax6=x3a3-6x2a2+15xa-20+15ax-6a2x2+a3x3

(vii)
x3-a36=C06(x3)6(a3)0-C16(x3)5(a3)1+C26(x3)4(a3)2-C36(x3)3(a3)3+C46(x3)2(a3)4-C56(x3)1(a3)5+C66(x3)0(a3)6=x2-6x5/3a1/3+15x4/3a2/3-20xa+15x2/3a4/3-6x1/3a5/3+a2


(viii)
(1+2x-3x2)5Consider 1-2x and 3x2 as two separate entities and apply the binomial theorem.Now,C05(1+2x)5(3x)0-C15(1+2x)4(3x2)1+C25(1+2x)3(3x2)2-C35(1+2x)2(3x2)3+C45(1+2x)1(3x2)4-C55(1+2x)0(3x2)5=(1+2x)5-5(1+2x)4×3x2+10×(1+2x)3×9x4-10×(1+2x)2×27x6+5(1+2x)×81x8-243x10=C05×(2x)0+C15×(2x)1+C25×(2x)2+C35×(2x)3+C45×(2x)4+C55×(2x)5-     15x2[C04(2x)0+C14(2x)1+C24(2x)2+C34(2x)3+C44(2x)4]+      90x4[1+8x3+6x+12x2]-270x6(1+4x2+4x)+405x8+810x9-243x10=1+10x+40x2+80x3+80x4+32x5-15x2-120x3-3604-480x5-240x6+      90x4+720x7+540x5+1080x6-270x6-1080x8-1080x7+405x8+810x9-243x10=1+10x+25x2-40x3-190x4+92x5+570x6-360x7-675x8+810x9-243x10


(ix)
(x+1-1x)3=C03(x+1)3(1x)0-C13(x+1)2(1x)1+C23(x+1)1(1x)2-C33(x+1)0(1x)3
=(x+1)3-3(x+1)2×1x+3x+1x2-1x3=x3+1+3x+3x2-3x2+3+6xx+3x+1x2-1x3=x3+1+3x+3x2-3x-3x-6+3x+3x2-1x3=x3+3x2-5+3x2-1x3

(x)
(1-2x+3x2)3=C03(1-2x)3+C13(1-2x)2(3x2)+C23(1-2x)(3x2)2+C33(3x2)3=(1-2x)3+9x2(1-2x)2+27x4(1-2x)+27x6=1-8x3+12x2-6x+9x2(1+4x2-4x)+27x4-54x5+27x6=1-8x3+12x2-6x+9x2+36x4-36x3+27x4-54x5+27x6=1-6x+21x2-44x3+63x4-54x5+27x6

Page No 18.11:

Question 2:

Evaluate the following:
(i) x+1+x-16+x+1-x-16

(ii) x+x2-16+x-x2-16

(iii) 1+2 x5+1-2 x5

(iv) 2+16+2-16

(v) 3+25-3-25

(vi) 2+37+2-37

(vii) 3+15-3-15

(viii) 0.995+1.015

(ix) 3+26-3-26

(x) a2+a2-14+a2-a2-14

Answer:

(i)
(x+1+x-1)6+(x+1-x-1)6=2[C06 (x+1)6(x-1)0+C26 (x+1)4(x-1)2+C46 (x+1)2(x-1)4+C66 (x+1)0(x-1)6]=2[(x+1)3+15(x+1)2(x-1)+15(x+1)(x-1)2+(x-1)3=2[x3+1+3x+3x2+15(x2+2x+1)(x-1)+15(x+1)(x2+1-2x)+x3-1+3x-3x2]=2[2x3+6x+15x3-15x2+30x2-30x+15x-15+15x3+15x2-30x2-30x+15x+15]=2[32x3-24x]=16x[4x2-3]

(ii)
(x+x2-1)6+(x-x2-1)6=2[C06x6(x2-1)0+C26x4(x2-1)2+C46x2(x2-1)4+C66x0(x2-1)6]=2[x6+15x4(x2-1)+15x2(x2-1)2+(x2-1)3]=2[x6+15x6-15x4+15x2(x4-2x2+1)+(x6-1+3x2-3x4)]=2[x6+15x6-15x4+15x6-30x4+15x2+x6-1+3x2-3x4]=64x6-96x4+36x2-2

(iii)
(1+2x)5+(1-2x)5=2[C05(2x)0+C25(2x)2+C45(2x)4]=2[1+10×4x+5×16x2]=2[1+40x+80x2]

(iv)
(2+1)6+(2-1)6=2[C06(2)6+C26(2)4+C46(2)2+C66(2)0]=2[8+15×4+15×2+1)=2×99 =198

(v)
(3+2)5-(3-2)5=2C15×34×(2)1+C35×32×(2)3+C55×30×(2)5

=2[5×81×2+10×9×22+42]=22(405+180+4)=11782

(vi)
(2+3)7+(2-3)7=2[C07×27×(3)0+C27×25×(3)2+C47×23×(3)4+C67×21×(3)6]=2[128+21×32×3+35×8×9+7×2×27]=2[128+2016+2520+378]=2×5042=10084

(vii)
(3+1)5-(3-1)5=2[C15×(3)4+C35×(3)2+C55×(3)0]=2[5×9+10×3+1]=2×76=152

(viii)
(0.99)5+(1.01)5=(1-0.01)5+(1+0.01)5=2[C05(0.01)0+C25(0.01)2+C45(0.01)4]=2[1+10×0.0001+5×0.00000001]=2×1.00100005=2.0020001

(ix)
(3+2)6-(3-2)6=2[C16(3)5(2)1+C36(3)3(2)3+C56(3)1(2)5]
=2[6×93×2+20×33×22+6×3×42]=2[6(54+120+24)]=3966

(x)
a2+a2-14+a2-a2-14=2[C04(a2)4(a2-1)0+C24(a2)2(a2-1)2+C44(a2)0(a2-1)4]=2[a8+6a4(a2-1)+(a2-1)2]=2[a8+6a6-6a4+a4+1-2a2]=2a8+12a6-10a4-4a2+2

Page No 18.11:

Question 3:

Find a+b4-a-b4. Hence, evaluate 3+24-3-24.

Answer:

The expression (a+b)4-(a-b)4 can be written as

(a+b)4-(a-b)4=2[C14a3b1+C34a1b3]                                =2[4a3b+4ab3]                               =8(a3b+ab3)

Putting a=3 and b =2, we get: (3+2)4-(3-2)4=8[(3)3×2+3×(2)3]                                               =8(36+26)                                               =406

 (3+2)4-(3-2)4=406                                               

Page No 18.11:

Question 4:

Find x+16+x-16. Hence, or otherwise evaluate 2+16+2-16.

Answer:

The expression x+16+x-16 can be written as
(x+1)6+(x-1)6=2[C06x6+C26x4+C46x2+C66x0]=2[x6+15x4+15x2+1]

By taking x=2, we get:
(2+1)6+(2-1)6=2[(2)6+15(2)4+15(2)2+1]
                                 =2[8+15×4+15×2+1]=2×(8+60+30+1)=198



Page No 18.12:

Question 5:

Using binomial theorem evaluate each of the following:
(i) (96)3
(ii) (102)5
(iii) (101)4
(iv) (98)5

Answer:

(i) (96)3
=(100-4)3=C03×1003×40-C13×1002×41+C23×1001×42-C33×1000×43=1000000-120000+4800-64=884736

(ii) (102)5
=(100+2)5=C05×1005×20+C15×1004×21+C25×1003×22+C35×1002×23+C45×1001×24+C55×1000×25=10000000000+1000000000+40000000+800000+8000+32=11040808032

(iii) (101)4
=(100+1)4=C04×1004+C14×1003+C24×1002+C34×1001+C44×1000=100000000+4000000+60000+400+1=104060401

(iv) (98)5
(100-2)5=C05×1005×20+-5C1×1004×21+C25×1003×22-C35×1002×23+C45×1001×24-C55×1000×25=10000000000-1000000000+40000000-800000+8000-32=9039207968

Page No 18.12:

Question 6:

Using binomial theorem, prove that 23n-7n-1 is divisible by 49, where n  N.

Answer:

23n-7n-1=8n-7n-1                ...(1)

Now,8n=(1+7)n    =C0n+C1n×71+C2n×72+C3n×73+C4n×74+...+Cnn×7n8n=1+7n+49[C2n+C3n×71+C4n×72+...+Cnn×7n-2]8n-1-7n=49×An integerNow, 8n-1-7n is divisible by 49Or,23n-1-7n is divisible by 49     From (1)

Page No 18.12:

Question 7:

Using binomial theorem, prove that 32n+2-8n-9 is divisible by 64, n  N.

Answer:

32n+2-8n-9=9n+1-8n-9       ...1
Consider
 9n+1=1+8n+19n+1 =C0n+1×80+C1n+1×81+C2n+1×82+C3n+1×83+...+Cn+1n+1×8n+1
9n+1=1+8(n+1)+[C2n+1×82+C3n+1×83+...+Cn+1n+1×8n+1]9n+1-8n-9=64(C2n+1+C3n+1×81+...+Cn+1n+1×8n-1]9n+1-8n-9=64×An integer9n+1-8n-9 is divisible by 64Or,32n+2-8n-9 is divisible by 64    From (1)Hence proved.

Page No 18.12:

Question 8:

If n is a positive integer, prove that 33n-26n-1 is divisible by 676.

Answer:

33n-26n-1=27n-26n-1       ...1
Now, we have:27n=(1+26)nOn expanding, we get(1+26)n=C0n×260+C1n×261+C2n×262+C3n×263+C4n×264+...Cnn×26n27n=1+26n+262[C2n+C3n×261+C4n×262+...Cnn×26n-2]27n-26n-1=676×an integer27n-26n-1 is divisible by 676Or,33n-26n-1 is divisible by 676   From (1)

Page No 18.12:

Question 9:

Using binomial theorem, indicate which is larger (1.1)10000 or 1000.

Answer:

We have:   
(1.1)10000  
=(1+0.1)10000=C010000×0.10+C110000×(0.1)1+C210000×(0.1)2+...C1000010000×(0.1)10000=1+10000×0.1+other positive terms=1+10000+other positive terms=10001+other positive terms10001>1000(1.1)10000>1000

Page No 18.12:

Question 10:

Using binomial theorem determine which number is larger (1.2)4000 or 800?

Answer:

We have:
(1.2)4000=(1+0.2)4000=C04000+C14000×(0.2)1+C24000×(0.2)2+...C40004000×(0.2)4000

              =1+4000×0.2+other positive terms=1+800+other positive terms=801+other positive terms801 >800

Hence, (1.2)4000 is greater than 800

Page No 18.12:

Question 11:

Find the value of (1.01)10 + (1 − 0.01)10 correct to 7 places of decimal.

Answer:

(1.01)10+(1-0.01)10=(1+0.01)10+(1-0.01)10=2[C010×(0.01)0+C210×(0.01)2+C410×(0.01)4+C610×(0.01)6+C810×(0.01)8+C1010×(0.01)10]=21+45×0.0001+210×0.00000001+... =21+0.0045+0.00000210+...=2.0090042+...

Hence, the value of (1.01)10 + (1 − 0.01)10 correct to 7 places of the decimal is 2.0090042

Page No 18.12:

Question 12:

Show that 24n+4-15n-16, where n ∈  is divisible by 225.

Answer:

We have,
24n+4-15n-16=24n+1-15n-16                        =16n+1-15n-16                        =1+15n+1-15n-16                        =n+1C0 150+n+1C1 151+n+1C2 152+...+n+1Cn+1 15n+1-15n-16                        =1+(n+1)15+n+1C2 152+...+n+1Cn+1 15n+1-15n-16                        =1+15n+15+n+1C2 152+...+n+1Cn+1 15n+1-15n-16                        =n+1C2 152+...+n+1Cn+1 15n+1                        =152n+1C2+...+n+1Cn+1 15n-1                        =225n+1C2+...+n+1Cn+1 15n-1

Thus, ​ 24n+4-15n-16, where n ∈  is divisible by 225.



Page No 18.37:

Question 1:

Find the 11th term from the beginning and the 11th term from the end in the expansion of 2x-1x225.

Answer:

Given:
2x-1x225
Clearly, the given expression contains 26 terms.

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

Thus, we have:
T16=T15+1=C1525(2x)25-15-1x215                    =C1525210x10-1x30=-C1525210x20
Now, we will find the 11th term from the beginning.

T11=T10+1      =C1025(2x)25-10-1x210      =C1025215x151x20      =C1025215x5

Page No 18.37:

Question 2:

Find the 7th term in the expansion of 3x2-1x310.

Answer:

We need to find the 7th term of the given expression.
Let it be T7
Now, we have
T7=T6+1
    =C610(3x2)10-6-1x36=C61034x81x18=10×9×8×7×814×3×2×x10=17010x10

Thus, the 7th term of the given expression is 17010x10

Page No 18.37:

Question 3:

Find the 5th term from the end in the expansion of 3x-1x210

Answer:

Given:
3x-1x210
Clearly, the expression has 6 terms.
The 5th term from the end is the (11 − 5 + 1)th, i.e., 7th, term from the beginning.
Thus, we have:

T7=T6+1=C610(3x)10-6-1x26=C61034x41x12=10×9×8×7×814×3×2×1×x8=17010x8

Page No 18.37:

Question 4:

Find the 8th term in the expansion of x3/2 y1/2- x1/2 y3/210.

Answer:

We need to find the 8th term in the given expression.
T8=T7+1

T8=C710(x3/2y1/2)10-7(-x1/2y3/2)7    =-10×9×83×2x9/2y3/2x7/2y21/2        =-120x8y12

Page No 18.37:

Question 5:

Find the 7th term in the expansion of 4x5+52x8.

Answer:

We need to find the 7th term in the given expression.

T7=T6+1

T7=T6+1    =C684x58-652x6   =8×7×4×4×125×1252×1×25×64 x2 1x6  =4375x4

Page No 18.37:

Question 6:

Find the 4th term from the beginning and 4th term from the end in the expansion of x+2x9.

Answer:

Let Tr+1 be the 4th term from the end.
Then, Tr+1 is (10 − 4 + 1)th, i.e., 7th, term from the beginning.

  T7=T6+1          =C69 x9-62x6         =9×8×73×2x364x6         =5376x3

4th term from the beginning = T4=T3+1
T4=C39 x9-3 2x3    =9×8×73×2x68x3    =672 x3

Page No 18.37:

Question 7:

Find the 4th term from the end in the expansion of 4x5-52x8.

Answer:

Let Tr+1 be the 4th term from the end of the given expression.
Then,
Tr+1 is (10 − 4 + 1)th term, i.e., 7th term, from the beginning.
Thus, we have:
T7=T6+1    =C69 4x59-652x6   =9×8×73×264125x3125×12564x6  =10500x3

Page No 18.37:

Question 8:

Find the 7th term from the end in the expansion of 2x2-32x8

Answer:

Let Tr+1 be the 7th term from the end in the given expression.
Then, we have:
Tr+1 = (9 − 7 + 1) =  3rd term from the beginning
Now,
T3=T2+1    =C28 (2x2)8-2 -32x2    =8×72×164x1294x2   =4032 x10

Page No 18.37:

Question 9:

Find the coefficient of:
(i) x10 in the expansion of 2x2-1x20

(ii) x7 in the expansion of x-1x240

(iii) x-15 in the expansion of 3x2-a3x310

(iv) x9 in the expansion of x2-13x9

(v) xm in the expansion of x+1xn

(vi) x in the expansion of 1-2x3+3x5 1+1x8.

(vii) a5b7 in the expansion of a-2b12.

(viii) x in the expansion of 1-3x+7x2 1-x16.

Answer:

(i) Suppose x10 occurs in the (+ 1)th term in the given expression.

Then, we have:

Tr+1=Crn xn-r ar  

Here,
Tr+1=Cr20(2x2)20-r -1xr         =(-1)r Cr20220-r x40-2r-rFor this term to contain x10, we must have:40-3r =103r=30r=10 Coefficient of x10 = (-1)10  C1020220-10 =C1020210 

(ii) Suppose x7 occurs at the (+ 1) th term in the given expression.

Then, we have:
Tr+1=Cr40 x40-r-1x2r         =(-1)r  Cr40 x40-r-2rFor this term to contain x7, we must have:40-3r=73r=40-7=33r=11Coefficient of x7 = (-1)11  C1140=-C1140

(iii)  Suppose x−15 occurs at the (+ 1)th term in the given expression.
Then, we have:
 Tr+1=Cr10 (3x2)10-r-a3x3r
Tr+1=(-1)r Cr10 310-r-rx20-2r-3rar
For this term to contain x-15 , we must have:20-5r =-155r=20+15r=7Coefficient of x-15 = (-1)7  C710 310-14 a7=-10×9×83×2×9×9a7=-4027a7

(iv) Suppose x9 occurs at the (+ 1)th term in the above expression.

Then, we have:

Tr+1=Cr9 (x2)9-r -13xr         =(-1)r Cr9 x18-2r-r 13rFor this term to contain x9, we must have:18-3r=93r=9r=3Coefficient of x9 =(-1)3 C39 133=-9×8×72×9×9=-289

(v)
Suppose xm occurs at the (+ 1)th term in the given expression.

Then, we have:

Tr+1=Crn xn-r 1xr=Crn xn-2rFor this term to contain xm, we must have:n-2r =mr=(n-m)/2Coefficient of xm = C(n-m)/2n=n!n-m2! n+m2!

(vi) Suppose x occurs at the (+ 1)th term in the given expression.

Then, we have:

(1-2x3+3x5)1+1x8=1-2x3+3x5C08+C18 1x+C28 1x2+C38 1x3+C48 1x4+C58 1x5+C68 1x6+C78 1x7+C88 1x8 x occurs in the above expresssion at -2x3.C28 1x2 +3x5.C48 1x4.Coefficient of x =-28!2! 6!+38!4! 4!=-56+210= 154
 
(vii)
Suppose a5 b7 occurs at the (r + 1)th term in the given expression.

Then, we have:

Tr+1=Cr12 a12-r (-2b)r=(-1)r Cr12 a12-r br 2rFor this term to contain a5 b7, we must have:12-r =5  r=7 Required coefficient = (-1)7 C712 27=-12×11×10×9×8×1285×4×3×2=-101376

(viii) Suppose x occurs at the (+ 1)th term in the given expression.

Then, we have:

1-3x+7x2 1-x16=1-3x+7x2C016+C116 -x+C216 -x2+C316 -x3+C416 -x4+16C5 -x5+C616 -x6+C716 -x7+C816 -x8+C916 -x9+C1016 -x10+C1116 -x11+C1216 -x12+C1316 -x13+C1416 -x14+C1516 -x15+C1616 -x16 x occurs in the above expresssion at 16C1 -x -3xC016.Coefficient of x =-16!1! 15!-316!0! 16!=-16-3= -19



Page No 18.38:

Question 10:

Which term in the expansion of xy1/3+yx1/31/221 contains x and y to one and the same power?

Answer:

Suppose Tr+1th term in the given expression contains x and y to one and the same power.
Then,

Tr+1 th term isCr21xy1/321-r yx1/31/2r=Cr21 x(21-r)/3xr/6yr/2y(21-r)/6=Cr21 x7-r/2y2r/3-7/2Now, if x and y have the same power, then7-r2=2r3-722r3+r2=7+727r6=212r=9Hence, the required term is the 10th term

Page No 18.38:

Question 11:

Does the expansion of 2x2-1x contain any term involving x9?

Answer:

Suppose x9 occurs in the given expression at the (r + 1)th term.
Then, we have:
Tr+1=Cr20 (2x2)20-r -1xr=(-1)r  Cr20220-r x40-2r-rFor this term to contain x9, we must have40-3r=93r=31r=313 It is not possible, as r is not an integer.

Hence, there is no term with x9 in the given expression.

Page No 18.38:

Question 12:

Show that the expansion of x2+1x12 does not contain any term involving x−1.

Answer:

Suppose x1 occurs at the (r + 1)th term in the given expression.
Then,
Tr+1=Cr12 (x2)12-r 1xr=Cr12 x24-2r-rFor this term to contain x-1, we must have24-3r=-13r=25r=253It is not possible, as r is not  an integer.

Hence, the expansion of x2+1x12 does not contain any term involving x−1.

Page No 18.38:

Question 13:

Find the middle term in the expansion of:
(i) 23x-32x20

(ii) ax+bx12

(iii) x2-2x10

(iv) xa-ax10

Answer:

(i) Here,
n = 20  (Even number) 
Therefore, the middle term is the n2+1th term, i.e., the 11th term.
Now,T11=T10+1=C1020  23x20-10 32x10=C1020  210310×310210x10-10=C1020

(ii) Here,
n = 12 (Even number)
Therefore, the middle term is the n2+1th  i.e. 7th term
Now,T7=T6+1=C612 ax12-6 (bx)6=C612 a6 b6 =12×11×10×9×8×76×5×4×3×2a6 b6=924 a6b6

(iii) Here,
n = 10    (Even number)
Therefore, the middle term is the n2+1th  i.e. 6th term
Now,T6=T5+1=C510 (x2)10-5 -2x5=-10×9×8×7×65×4×3×2×32x5=-8064 x5

(iv) Here,
n = 10 (Even number)
Therefore, the middle term is the n2+1th  i.e. 6th term
Now,T6=T5+1         =C510 xa10-5 -ax5         =-10×9×8×7×65×4×3×2=-252

Page No 18.38:

Question 14:

Find the middle terms in the expansion of:
(i) 3x-x369

(ii) 2x2-1x7

(iii) 3x-2x215

(iv) x4-1x311

Answer:

(i) Here, n, i.e. 9, is an odd number.
Thus, the middle terms are n+12th and n+12+1th, i.e. 5th and 6th
Now,T5=T4+1=C49 (3x)9-4 -x364=9×8×7×64×3×2×27×9×136×36x17=1898x17and,T6=T5+1=C59(3x)9-5 -x365=-9×8×7×64×3×2×81×1216×36x19=-2116x19

(ii) Here, n, i.e., 7, is an odd number.

Thus, the middle terms are 7+12th and 7+12+1th i.e. 4th and 5thNow,T4=T3+1=C37 (2x2)7-3 -1x3=-7×6×53×2×16 x8×1x3=-560 x5And,T5=T4+1=C47 (2x2)7-4 -1x4=35×8 ×x6×1x4=280 x2

(iii)
Given:n, i.e.15  is an odd number.Thus, the middle terms are 15+12th and 15+12+1th i.e. 8th and 9th.Now,T8= T7+1    =C715 (3x)15-7 -2x27    =-15×14×13×12×11×10×97×6×5×4×3×2×38×27 x8-14    =-6435×38×27x6And,T9=T8+1    =C815 (3x)15-8 -2x28    =15×14×13×12×11×10×97×6×5×4×3×2×37×28 ×x7-16    =6435×37×28x9

(iv)
Here, n, i.e., 11, is an odd number.Thus, the middle terms are 11+12th and 11+12+1th i.e. 6th and 7th.Now,T6=T5+1    =C511 (x4)11-5 -1x35    =-11×10×9×8×75×4×3×2×x24-15    =-462 x9And,T7=T6+1     =C611 (x4)11-6 -1x36     =11×10×9×8×75×4×3×2x20-18     =462 x2

Page No 18.38:

Question 15:

Find the middle terms(s) in the expansion of:
(i) x-1x10

(ii) 1-2x+x2n

(iii) 1+3x+3x2+x32n

(iv) 2x-x249

(v) x-1x2n+1

(vi) x3+9y10

(vii) 3-x367

(viii) 2ax-bx212

(ix) px+xp9

(x) xa-ax10

Answer:

(i)
x-1x10Here, n is an even number.  Middle term = 102+1th= 6th termNow, we haveT6=T5+1=C510 x10-5 -1x5=-10×9×8×7×65×4×3×2=-252

(ii)
(1-2x+x2)n=(1-x)2nn is an even number. Middle term = 2n2+1th=(n+1)th termNow, we haveTn+1=Cn2n (-1)n (x)n=(2n)!(n!)2(-1)n xn

(iii)
(1+3x+3x2+x3)2n=(1+x)6nHere, n  is an even number. Middle term = 6n2+1 th=(3n+1)th termNow, we haveT3n+1=C3n6n x3n=(6n)!(3n!)2x3n

(iv)
2x-x249Here, n is an odd number.Therefore, the middle terms are n+12th and n+12+1th, i.e. 5th and 6th terms.Now, we haveT5=T4+1=C49 (2x)9-4 -x244=9×8×7×64×3×2×25 144x5+8=634x13And,T6=T5+1=C59 (2x)9-5 -x245=-9×8×7×64×3×2×24 145x4+10=-6332x14

(v)
x-1x2n+1Here, 2n+1  is an odd number.Therefore, the middle terms are 2n+1+12th and2n+1+12+1th i.e. (n+1)th and (n+2)th terms.Now, we have:Tn+1=Cn2n+1 x2n+1-n  × (-1)nxn=(-1)n Cn2n+1 xAnd,Tn+2=Tn+1+1=Cn+12n+1 x2n+1-n-1   (-1)n+1xn+1=(-1)n+1 Cn+12n+1 ×1x

(vi)
x3+9y10Here, n is an even number.Therefore, the middle term is 102+1th, i.e., 6th term.Now, we haveT6=T5+1=C510 x310-5 (9y)5=10×9×8×7×65×4×3×2×135×95×x5 y5=61236 x5 y5

(vii)
3-x367Here, n is an odd number.Therefore, the middle terms are 7+12th and 7+12+1th, i.e., 4th and 5th terms.Now, we haveT4=T3+1=C37 37-3 -x363=-1058x9And,T5=T4+1=C49 39-4 -x364=7×6×53×2×35×164 x12=3548x12

(viii)
2ax-bx212Here, n is an even number.  Middle term = 122+1th= 7th termNow, we haveT7=T6+1=C612 2ax12-6 -bx26=12×11×10×9×8×76×5×4×3×2×1×2abx6=59136 a6b6x6

(ix)
px+xp9Here, n is an odd number.Therefore, the middle terms are 9+12th and 9+12+1th, i.e., 5th and 6th terms.Now, we haveT5=T4+1=C49 px9-4 xp4=9×8×7×64×3×2×1×px=126 pxAnd,T6=T5+1=C59 px9-5 xp5=9×8×7×64×3×2×1×xp=126 xp

(x)
xa-ax10Here, n is an even number.  Middle term = 102+1th= 6th termNow, we haveT6=T5+1=C510 xa10-5 -ax5=-10×9×8×7×65×4×3×2×1=-252



Page No 18.39:

Question 16:

Find the term independent of x in the expansion of the following expressions:
(i) 32x2-13x9

(ii) 2x+13x29

(iii) 2x2-3x325

(iv) 3x-2x215

(v) x3+32x210

(vi) x-1x23n

(vii) 12x1/3+x-1/58

(viii) 1+x+2x332x2-33x9

(ix) x3+12 x318, x > 0

(x) 32x2-13x6

Answer:

(i) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
32x2-13x9Tr+1=Cr9 32x29-r -13xr= (-1)r Cr9 .39-2r29-r× x18-2r-rFor this term to be independent of x, we must have18-3r=03r=18r=6Hence, the required term is the 7th term.Now, we haveC69×39-1229-6=9×8×73×2×3-3×2-3=718

(ii) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
2x+13x29Tr+1=Cr9 (2x)9-r13x2r=Cr9.29-r3r x9-r-2rFor this term to be independent of x, we must have9-3r=0r=3Hence, the required term is the 4th term.Now, we haveC39 2633=C39×6427

(iii) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
2x2-3x325Tr+1=Cr25 (2x2)25-r -3x3r=(-1)r  Cr25 ×225-r×3r x50-2r-3rFor this term to be independent of x, we must have:50-5r=0r=10Therefore, the required term is the 11th term.Now, we have(-1)10  C1025 ×225-10×310=C1025 (215× 310)

(iv)  Suppose the (r + 1)th term in the given expression is independent of x.
Now,
3x-2x215Tr+1=Cr15 (3x)15-r -2x2r= (-1)r  Cr15 ×315-r × 2r x15-r-2rFor this term to be independent of x, we must have15-3r=0r=5Hence, the required term is the 6th term.Now, we have:(-1)5  C515 .315-5 . 25=-3003 ×310×25

(v) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
x3+32x210Tr+1=Cr10 x310-r 32x2r=Cr10 .3r-10-r22r x10-r2-2rFor this term to be independent of x, we must have10-r2-2r=010-5r=0r=2Hence, the required term is the 3rd term.Now, we haveC210 ×32-10-2222=10×92×4×9=54

(vi)  Suppose the (r + 1)th term in the given expression is independent of x.
Now,
x-1x23nTr+1=Cr3n x3n-r -1x2r=(-1)r Cr3n  x3n-r-2rFor this term to be independent of x, we must have3n-3r=0r = nHence, the required term is the (n+1)th term.Now, we have(-1)n Cn3n


(vii)  Suppose the (r + 1)th term in the given expression is independent of x.
Now,
12x1/3 +x-1/58Tr+1=Cr8 12x1/38-r (x-1/5)r=Cr8. 128-r x8-r3-r5For this term to be independent of x, we must have8-r3-r5=040-5r-3r=08r=40r=5Hence, the required term is the 6th term.Now, we have:C58× 128-5=8×7×63×2×8=7

(ix) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
x3+12x318Tr+1=Cr18 (x1/3)18-r 12 x1/3r=Cr18×12r x18-r3-r3For this term to be independent of r, we must have18-r3-r3=018-2r=0r=9The term is C918×129

(x) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
32x2-13x6Tr+1=Cr6 32x26-r -13xr=-1r Cr6 × 36-r-r26-r x12-2r-rFor this term to be independent of x, we must have12-3r=0r=4Hence, the required term is the 4th term.C46 × 36-4-426-4=6×52×1×4×9=512

Page No 18.39:

Question 17:

If the coefficients of 2r + 4th and r-2th terms in the expansion of 1+x18 are equal, find r.

Answer:

Given:(1+x)18We know that the coefficient of the rth term in the expansion of (1+x)n  is Cr-1nTherefore, the coefficients of the (2r+4)th and (r-2)th terms in the given expansion are C2r+4-1 18and Cr-2-118For these coefficients to be equal, we must haveC2r+3 18= Cr-3182r+3=r-3  or, 2r+3+r-3=18      [ Crn=Csn r=s or r+s=n]r=-6   or,  r=6Neglecting negative value We getr=6

Page No 18.39:

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.

Answer:

Given :- (1+x)43We know that the coefficient of the rth term in the expansion of (1+x)n is Cr-1nTherefore, the coefficients of the (2r+1)th and (r+2)th terms in the given expression are C2r+1-143 and Cr+2-143For these coefficients to be equal, we must have:2r=r+1   or,  2r+r+1=43      [ Crn=Csn r=s or r+s=n] r=14    for r=1 it gives the same term

Page No 18.39:

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.

Answer:

Coefficient of the (r+1)th term in (1+x)n+1 is Crn+1Sum of the coefficients of the rth and (r+1)th terms in (1+x)n=Cr-1n +Crn                                                                                           =Crn+1           Cr+1n +Crn=Cr+1n+1  Hence proved.

Page No 18.39:

Question 20:

Prove that the term independent of x in the expansion of x+1x2n is 1·3·5 ... 2n-1n!. 2n.

Answer:

Given:x+1x2nSuppose the term independent of x is the (r+1) th term.Tr+1=Cr2n x2n-r 1xr         =Cr2n x2n-2rFor this term to be independent of x, we must have:2n-2r=0n=rRequired coefficient = Cn2n                                        =(2n)!(n!)2                                         =1·3·5...2n-32n-12·4·6...2n-22n(n!)2                                        =1·3·5...2n-32n-12nn!

Page No 18.39:

Question 21:

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

Answer:

Coefficients of the 5th, 6th and 7th terms in the given expansion are C4n , C5n and C6nThese coefficients are in AP. Thus, we have2 C5n =C4n+C6nOn dividing both sides by C5n, we get:2=C4nC5n+C6nC5n2=5n-4+n-5612n-48=30+n2-4n-5n+20n2-21n+98=0(n-14)(n-7)=0n=7 or 14

Page No 18.39:

Question 22:

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

Answer:

Given:(1+x)2nThus, we have:T2=T1+1     =C12n x1T3=T2+1    =C22n x2T4=T3+1    =C32n x3We have coefficients of the 2nd, 3rd and 4th terms in AP.2C22n =C12n+C32n 2=C12nC22n+C32nC22n                 2=22n-1+2n-2312n-6=6+4n2-4n-2n+24n2-18n+14=02n2-9n+7=0

Hence proved.

Page No 18.39:

Question 23:

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

Answer:

Coefficients of the 2nd, 3rd  and 4th terms in the given expansion are:
C1n, C2n and C3nWe have:2×C2n=C1n+C3nDividing both sides by C2n, we get:2=C1nC2n+C3nC2n2=2n-1+n-236n-6=6+n2+2-3nn2-9n+14=0 n=7  n2 as 2>3 in the 4th term

Page No 18.39:

Question 24:

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

Answer:

Coefficients of the pth and qth terms are Cp-1n and Cq-1n respectively.Thus, we have:Cp-1n= Cq-1np-1 =q-1 or, p-1+q-1=n             [ Crn=nCsr=s or, r+s=n]p=q or, p+q=n+2

If pq, then p+q=n+2
Hence proved

Page No 18.39:

Question 25:

Find a, if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.

Answer:

(3+ax)9=C09 .39. (ax)0 +C19 .38. (ax)1+C29 .37. (ax)2+C39 .36. (ax)3+...

We have Coefficient of x2 Coefficient of x3

C29 ×37 a2 =C39 ×36 a3a=C29C39×3       =9! ×3! ×6!×32! × 7! ×9!       =97

Page No 18.39:

Question 26:

Find the coefficient of a4 in the product (1 + 2a)4 (2 − a)5 using binomial theorem.

Answer:

(1+2a)4(2-a)5=[C04 (2a)0+C14 (2a)1+C24 (2a)2+C34 (2a)3+C44 (2a)4]×            [C05 (2)5 (-a)0 +C15 (2)4 (-a)1+C25 (2)3 (-a)2+C35 (2)2 (-a)3+C45 (2)1 (-a)4+C55 (2)0 (-a)5]=[1+8a+24a2+32a3+16a4]×[32-80a+80a2-40a3+10a4-a5]Coefficient of a4=10-320+1920-2560+512=-438



Page No 18.40:

Question 27:

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.

Answer:

Suppose the three consecutive terms are Tr-1, Tr and Tr+1.Coefficients of these terms are Cr-2n, Cr-1n and Crn, respectively.These coefficients are equal to 220, 495 and 792. Cr-2n Cr-1n=220495r-1n-r+2=499r-9=4n-4r+84n+17=13r      ...1Also,CrnCr-1n=792495n-r+1r=855n-5r+5=8r5n+5=13r 5n+5=4n+17         From Eqn1n=12 

Page No 18.40:

Question 28:

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

Answer:

Suppose rth, (r+1) thand (r+2)th terms are the three consecutive terms.Their respective coefficients are Cr-1n,  Crn and Cr+1n.We have: Cr-1n=Cr+1n=56r-1+r+1=n       [If Crn=Csnr=s or r+s =n]2r=nr=n2Now,Cn2n =70 and Cn2-1n=56Cn2-1n Cn2n=5670n2n2+1=8105n=4n+8n=8So, r=n2=4Thus, the required terms are 4th, 5th and 6th.

Page No 18.40:

Question 29:

If 3rd, 4th 5th and 6th terms in the expansion of (x + a)n be respectively a, b, c and d, prove that b2-acc2-bd=5a3c.

Answer:

  We have:(x+a)nThe 3rd, 4th, 5th and 6th terms are C2nxn-2 a2,  C3nxn-3 a3, C4nxn-4 a4 and C5nxn-5 a5, respectively.Now,C2nxn-2 a2=aC3nxn-3 a3 =bC4nxn-4 a4 =cC5nxn-5 a5 =dLHS =b2-acc2-bd

Page No 18.40:

Question 30:

If a, b, c and d in any binomial expansion be the 6th, 7th, 8th and 9th terms respectively, then prove that b2-acc2-bd=4a3c.

Answer:

Suppose the binomial expression is (1+x)n.Then, the 6th, 7th, 8th and 9th terms are C5n x5, C6n x6,  C7n x7 and C8n x8, respectively.Now, we have: C6 x6n C5n x5=ba,  C8 x8n C7n x7=dc and  C7 x7n C6n x6=cbn-56=ba and n-67=cbn-67n-56=cbba6n-367n-35=ca

Page No 18.40:

Question 31:

If the coefficients of three consecutive terms in the expansion of (1 + x)n be 76, 95 and 76, find n.

Answer:

Suppose r, r+1 and r+2 are three consecutive terms in the given expansion.The coefficients of these terms are Cr-1n, Crn and Cr+1n.According to the question,Cr-1n=76Crn=95Cr+1n=76Cr-1n=Cr+1nr-1+r+1=n             [If Crn=Csnr=s or r+s =n]r=n2CrnCr-1n=9576n-r+1r=9576n2 +1n2=957638n+76=95n2       19n2=76n=8

Page No 18.40:

Question 32:

If the 6th, 7th and 8th terms in the expansion of (x + a)n are respectively 112, 7 and 1/4, find x, a, n.

Answer:

The 6th, 7th and 8th terms in the expansion of (x+a)n are C5n xn-5 a5, C6n xn-6 a6 and C7n xn-7 a7.
According to the question,
C5n xn-5 a5=112C6n xn-6 a6=7C7n xn-7 a7=14Now,C6 nxn-6 a6C5n xn-5 a5=7112n-6+16x-1 a=116ax=38n-40                    ...1Also,C7 nxn-7 a7C6n xn-6 a6=1/47n-7+17x-1a=128ax=14n-24                        ...2From 1 and 2, we get: 38n-40=14n-2432n-10=1n-6n=8


Putting in eqn1 we geta=xNow, C58 x8-5 x85=11256x885=112x8=48x=4By putting the value of x and n in 1 we geta=12

a=3 and x=2

Page No 18.40:

Question 33:

If the 2nd, 3rd and 4th terms in the expansion of (x + a)n are 240, 720 and 1080 respectively, find x, a, n.

Answer:

In the expansion of x+an, the 2nd, 3rd and 4th terms are C1n xn-1 a1, C2n xn-2 a2 and C3n xn-3 a3, respectively.According to the question,C1n xn-1 a1=240        C2n xn-2 a2=720C3n xn-3 a3=1080C2n xn-2 a2C1n xn-1 a1=720240n-12xa=3ax=6n-1    ...1Also,C3n xn-3 a3C2n xn-2 a2=1080720n-23xa=32ax=92n-4    ...2Using 1 and 2 we get6n-1=92n-4n=5Putting in eqn1 we get2a=3xNow, C15 x5-1 32x=24015x5=480x5=32x=2By putting the value of x and n in 1 we geta=3

Page No 18.40:

Question 34:

Find a, b and n in the expansion of (a + b)n, if the first three terms in the expansion are 729, 7290 and 30375 respectively.

Answer:

We have:T1=729, T2=7290 and T3=30375Now,C0n an b0=729an=729an=36C1n an-1 b1=7290C2n an-2 b2=30375Also,C2n an-2 b2C1n an-1 b1=303757290n-12×ba=256              ...(i)(n-1)ba=253And,C1n an-1 b1C0n an b0=7290729nba=101          ...(ii)On dividing (ii) by (i), we getnba(n-1)ba=10×325nn-1=65n = 6Since, a6=36 Hence, a=3Now,  nba=10b=5

Page No 18.40:

Question 35:

If the term free from x in the expansion of x-kx210 is 405, find the value of k.

Answer:

Let (r + 1)th term, in the expansion of x-kx210, be free from x and be equal to Tr + 1. Then,
Tr+1=10Crx10-r-kx2r=10Cr x5-5r2-kr        ....(1)

If Tr + 1 is independent of x, then
5-5r2=0r=2

Putting r = 2 in (1), we obtain

T3=10C2 -k2=45k2

But it is given that the value of the term free from x is 405.
45k2=405k2=9k=±3

Hence, the value of k is ±3.

Page No 18.40:

Question 36:

Find the sixth term in the expansion y12+x13n, if the binomial coefficient of the third term from the end is 45.

Answer:

In the binomial expansion of y12+x13n, there are (n + 1) terms.

The third term from the end in the expansion of y12+x13n, is the third term from the beginning in the expansion of x13+y12n.

∴ The binomial coefficient of the third term from the end = nC2

If is given that the binomial coefficient of the third term from the end is 45.

 nC2=45nn-12=45n2-n-90=0n-10n+9=0n=10       n cannot be negative

Let Tbe the sixth term in the binomial expansion of y12+x13n. Then
T6=nC5y12n-5x135=10C5 y52x53=252 y52x53

Hence, the sixth term in the expansion of y12+x13n, is 252 y52x53.

Page No 18.40:

Question 37:

If p is a real number and if the middle term in the expansion of p2+28 is 1120, find p.

Answer:

In the binomial expansion of p2+28, we observe that 82+1th i.e., 5th term is the middle term.

It is given that the middle term is 1120.

 T5=11208C4 p28-424=1120p4=16p=±2

Hence, the real values of p is ±2.

Page No 18.40:

Question 38:

Find n in the binomial 23+133n, if the ratio of 7th term from the beginning to the 7th term from the end is 16.

Answer:

In the binomail expansion of 23+133nn+1-7+1th i.e., (n − 5)th term from the beginning is the 7th term from the end.

Now,
T7=nC623n-61336=nC6×2n3-2×132And,Tn-5=nCn-6236133n-6=nC6×22×13n3-2

It is given that,
T7Tn-5=16nC6×2n3-2×132nC6×22×13n3-2=162n3-2-2×3n3-2-2=16164-n3=164-n3=1n=9

Hence, the value of is 9.

Page No 18.40:

Question 39:

if the seventh term from the beginning and end in the binomial expansion of 23+133n are equal, find n.

Answer:

In the binomail expansion of 23+133nn+1-7+1th i.e., (n − 5)th term from the beginning is the 7th term from the end.

Now,
T7=nC623n-61336=nC6×2n3-2×132And,Tn-5=nCn-6236133n-6=nC6×22×13n3-2

It is given that,
T7=Tn-5nC6×2n3-2×132=nC6×22×13n3-22n3-222=323n3-26n3-2=62n3-2=2n=12

Hence, the value of is 12.



Page No 18.45:

Question 1:

Write the number of terms in the expansion of 2+3x10+2-3x10.

Answer:

Number of terms in the expansion (x+y)n+(x-y)n where n is even=n2+1Thus, we have:Number of terms in the given expansion=102+1=6

Page No 18.45:

Question 2:

Write the sum of the coefficients in the expansion of 1-3x+x2111.

Answer:

To find the sum of coefficients, we plug 1 for each variable then, we get the sum of coefficients of the given expression. Sum of coefficient =1-3x+x2111                              =1-3×1+12111                              =1-3+1111                              =1-3+1111                              =-1111                              =-1

Page No 18.45:

Question 3:

Write the number of terms in the expansion of 1-3x+3x2-x38.

Answer:

The given expression is (1-3x+3x2-x3)8. It can be written as [(1-x)3]8 i.e. (1-x)24Hence, the number of terms is 24+1 i.e. 25

Page No 18.45:

Question 4:

Write the middle term in the expansion of 2x23+32x2.

Answer:

Here, n, i.e., 10, is an even number. Middle term=102+1th term=6th termThus, we have:T6=T5+1    =C510 2x2310-5 32x25    =10×9×8×7×65×4×3×2×2535×3525    =252

Page No 18.45:

Question 5:

Which term is independent of x, in the expansion of x-13x29?

Answer:

Suppose Tr+1 is the term in the given expression that is independent of x.Thus, we have:Tr+1=Cr9 x9-r -13x2r=(-1)r Cr9  13r x9-r-2rFor this term to be independent of x, we must have9-3r=0r=3Hence, the required term is the 4th term.

Page No 18.45:

Question 6:

If a and b denote respectively the coefficients of xm and xn in the expansion of 1+xm+n, then write the relation between a and b.

Answer:

  Coefficient of xm in the given expansion = Cmm+n =aCoefficient of xn in the given expansion = Cnm+n =ba=b   Cmm+n =Cnm+n

Page No 18.45:

Question 7:

If a and b are coefficients of xn in the expansions of 1+x2n and 1+x2n-1 respectively, then write the relation between a and b.

Answer:

Coefficient of xn in the expansion (1+x)2n = Cn2n=aCoefficient of xn in the expansion (1+x)2n-1=Cn2n-1=bNow, we have: Cn2n=2n!n!. n!=2n(2n-1)!nn-1! n!    ...1 and Cn2n-1=(2n-1)!n!(n-1)!       ...2Dividing equation 1 by 2, we get Cn2nCn2n-1=2n(2n-1)! n! (n-1)!nn-1! n! (2n-1)!ab=2a=2b

Page No 18.45:

Question 8:

Write the middle term in the expansion of x+1x10.

Answer:

Here, n, i.e., 10, is an even number. Middle term=102+1th term=6th termThus, we have:T6=T5+1     =C510 x10-5×1x5     =C510

Page No 18.45:

Question 9:

If a and b denote the sum of the coefficients in the expansions of 1-3x+10x2n and 1+x2n respectively, then write the relation between a and b.

Answer:

Here,a=1-3+10=8=23b=1+1=2a=b3

Page No 18.45:

Question 10:

Write the coefficient of the middle term in the expansion of 1+x2n.

Answer:

Here, n, i.e., 2n, is an even number. Middle term=2n2+1th term=n+1th termThus, we have:Tn+1=2nCn12n-nxn          =2nCnxnHence, the coefficient of the middle term is 2nCn

Page No 18.45:

Question 11:

Write the number of terms in the expansion of 2x+y347.

Answer:

In the binomial expansion of a+bn, total number of terms will be (n + 1).

Now, 2x+y347=2x+y328

Therefore, in the expansion of 2x+y347, total number of terms will be 28 + 1 = 29.

Page No 18.45:

Question 12:

Find the sum of the coefficients of two middle terms in the binomial expansion of 1+x2n-1.

Answer:

1+x2n-1Here, n is an odd number.Therefore, the middle terms are 2n-1+12th and 2n-1+12+1th, i.e., nth and (n+1)th terms.Now, we haveTn=Tn-1+1=Cn-12n-1 xn-1And,Tn+1=Tn+1=Cn2n-1 xn the coefficients of two middle terms are Cn-12n-1 and Cn2n-1.Now,2n-1Cn-1+2n-1Cn=2nCn

Hence, the sum of the coefficients of two middle terms in the binomial expansion of 1+x2n-1 is 2nCn.

Page No 18.45:

Question 13:

Find the ratio of the coefficients of xp and xq in the expansion of 1+xp+q.

Answer:

Coefficient of xp in the expansion of 1+xp+q is p+qCp.

Coefficient of xq in the expansion of 1+xp+q is p+qCq.

Now,
p+qCpp+qCq=p+q!p!q!p+q!q!p!=1

Hence, the ratio of the coefficients of xp and xq in the expansion of 1+xp+q is 1 : 1.

Page No 18.45:

Question 14:

Write last two digits of the number 3400.

Answer:

3400=9200     =10-1200     =200C010200+200C110199-11+.....+200C198102-1198+200C199101-1199+200C200-1200     =10010198+200C110197-11+.....+200C198-1198+200101-1199+-1200     =10010198-200C110197+.....+200C198-210+1     =100(a natural number)+1

Hence, last two digits of the number 3400 is 01.

Page No 18.45:

Question 15:

Find the number of terms in the expansion of a+b+cn.

Answer:

We have,
a+b+cn=a+b+cn                 =an+nC1 an-1b+c1+nC2 an-2b+c2+...+nCnb+cn

Further, expanding each term of R.H.S., we note that
First term consists of 1 term.
Second term on simplification gives 2 terms.
Third term on expansion gives 3 terms.
Similarly, fourth term on expansion gives 4 terms and so on.

∴ The total number of terms = 1 + 2 + 3 + .... + (n + 1) = n+1n+22.

Page No 18.45:

Question 16:

If a and b are the coefficients of xn in the expansion of 1+x2n and 1+x2n-1 respectively, find ab.

Answer:

Coefficients of xn in the expansion of 1+x2n is 2nCn=a.

Coefficients of xn in the expansion of 1+x2n-1 is 2n-1Cn=b.

Now,
ab=2nCn2n-1Cn     =2n!n!n!2n-1!n!n-1!     =2nn     =2

Hence, ab=2.



Page No 18.46:

Question 17:

Write the total number of terms in the expansion of x+a100+x-a100.

Answer:

The total number of terms are 101 of which 50 terms get cancelled.

Hence, the total number of terms in the expansion of x+a100+x-a100 is 51.

Page No 18.46:

Question 18:

If 1-x+x2n=a0+a1 x+a2 x2+...+a2n x2n, find the value of a0+a2+a4+...+a2n.

Answer:

Putting x = 1 and −1 in
1-x+x2n=a0+a1 x+a2 x2+...+a2n x2n
we get,

1=a0+a1+a2+...+a2n       ...(1)
and
3n=a0-a1+a2-...+a2n       ...(2)

Adding (1) and (2), we get
3n+1=2a0+a2+...+a2n

Hence, the value of a0+a2+a4+...+a2n is 3n+12.

Page No 18.46:

Question 1:

If in the expansion of (1 + x)20, the coefficients of rth and (r + 4)th terms are equal, then r is equal to
(a) 7
(b) 8
(c) 9
(d) 10

Answer:

(c) 9
Coefficients of the rth and (r+4)th terms in the given expansion are Cr-120  and 20Cr+3.Here,Cr-120  = 20Cr+3 r-1+r+3=20       if nCx=nCy   x=y or x+y=nr=2 or 2r=18r=9  

Page No 18.46:

Question 2:

The term without x in the expansion of 2x-12x212 is
(a) 495
(b) −495
(c) −7920
(d) 7920

Answer:

(d) 7920

Suppose the (r+1)th term in the given expansion is independent of x.Then, we have:Tr+1=Cr12 (2x)12-r -12x2r=(-1)r Cr12  212-2r  x12-r-2rFor this term to be independent of x, we must have:12-3r=0r=4Required term:  (-1)4 C412  212-8=12×11×10×94×3×2×16=7920

Page No 18.46:

Question 3:

If rth term in the expansion of 2x2-1x12 is without x, then r is equal to
(a) 8
(b) 7
(c) 9
(d) 10

Answer:

(c) 9
  rth term in the given expansion is Cr-112 (2x2)12-r+1 -1xr-1=(-1)r-1 Cr-112  213-r x26-2r-r+1For this term to be independent of x, we must have:27-3r=0r=9Hence, the 9th term in the expansion is independent of x.

Page No 18.46:

Question 4:

If in the expansion of (a + b)n and (a + b)n + 3, the ratio of the coefficients of second and third terms, and third and fourth terms respectively are equal, then n is
(a) 3
(b) 4
(c) 5
(d) 6

Answer:

(c) n = 5

  Coefficients of the 2nd and 3rd terms in (a+b)n are C1n and C2nCoefficients of the 3rd and 4th terms in (a+b)n+3 are C2n+3 and C3n+3Thus, we haveC1nC2n=C2n+3C3n+32n-1=3n+12n+2=3n-3n=5

Page No 18.46:

Question 5:

If A and B are the sums of odd and even terms respectively in the expansion of (x + a)n, then (x + a)2n − (xa)2n is equal to
(a) 4 (A + B)
(b) 4 (AB)
(c) AB
(d) 4 AB

Answer:

(d) 4AB
If A and B denote respectively the sums of odd terms and even terms in the expansion (x+a)nThen , (x+a)n= A+B    ...1             (x-a)n= A-B   ...2Squaring and subtraction equation 2 from1 we get (x+a)2n -(x-a)2n= A+B2- A-B2(x+a)2n -(x-a)2n=4AB

Page No 18.46:

Question 6:

The number of irrational terms in the expansion of 41/5+71/1045 is
(a) 40
(b) 5
(c) 41
(d) none of these

Answer:

(c) 41

The general term Tr+1 in the given expansion is given byCr45 (41/5)45-r (71/10)rFor Tr+1 to be an integer, we must have r5 and r10as integers i.e. 0r45 r =0,10, 20, 30 and 40Hence, there are 5 rational and 41, i.e., 46-5, irrational terms.

Page No 18.46:

Question 7:

The coefficient of x-17 in the expansion of x4-1x315 is
(a) 1365
(b) −1365
(c) 3003
(d) −3003

Answer:

(b) −1365

Suppose the (r+1)th term in the given expansion contains the coefficient of x-17.Then, we have:Tr+1=Cr15 (x4)15-r -1x3r         =(-1)r Cr15  x60-4r-3rFor this term to contain x-17 , we must have:60-7r=-17 7r=77r=11Required coefficient=(-1)11 C1115=-15×14×13×124×3×2=-1365

Page No 18.46:

Question 8:

In the expansion of x2-13x9, the term without x is equal to
(a) 2881

(b) -28243

(c) 28243

(d) none of these

Answer:

(c) 28243

Suppose the (r + 1)th term in the given expansion is independent of x.
Then , we have:
Tr+1=Cr9 (x2)9-r -13xr         =(-1)r Cr9  13rx18-2r-rFor this term to be independent of x, we must have:18-3r=0r=6Required term=(-1)6  C69  136=9×8×73×2×136=28243

Page No 18.46:

Question 9:

If an the expansion of 1+x15, the coefficients of 2r+3th and r-1th terms are equal, then the value of r is
(a) 5
(b) 6
(c) 4
(d) 3

Answer:

(a) 5
Coefficients of (2r+3)th and (r-1)th terms in the given expansion areC2r+2 15 and Cr-2.15Thus, we haveC2r+215  = Cr-2152r+2=r-2      or 2r+2+r-2=15          if nCx=nCy   x=y or x+y=n r=-4      or r=5Neglecting the negative value, We haver=5



Page No 18.47:

Question 10:

The middle term in the expansion of 2x23+32x210 is
(a) 251
(b) 252
(c) 250
(d) none of these

Answer:

(b) 252

Here, n, i.e., 10, is an even number. Middle term=102+1th term=6th termThus, we have:T6=T5+1    =C510 2x2310-5 32x25    =10×9×8×7×65×4×3×2×2535×3525    =252

Page No 18.47:

Question 11:

If in the expansion of x4-1x315, x-17 occurs in rth term, then
(a) r = 10
(b) r = 11
(c) r = 12
(d) r = 13

Answer:

(c) r = 12

Here,
Tr=Cr-115 (x4)15-r+1 -1x3r-1=(-1)r× Cr-115  x64-4r-3r+3For this term to contain x-17, we must have:67-7r=-17r=12

Page No 18.47:

Question 12:

In the expansion of x-13x29, the term independent of x is
(a) T3
(b) T4
(c) T5
(d) none of these

Answer:

(b) T4

Suppose Tr+1 is the term in the given expression that is independent of x.Thus, we have:Tr+1=Cr9 x9-r -13x2r=(-1)r Cr9  13r x9-r-2rFor this term to be independent of x, we must have9-3r=0r=3Hence, the required term is the 4th term i.e. T4

Page No 18.47:

Question 13:

If in the expansion of (1 + y)n, the coefficients of 5th, 6th and 7th terms are in A.P., then n is equal to
(a) 7, 11
(b) 7, 14
(c) 8, 16
(d) none of these

Answer:

(b) 7, 14

Coefficients of the 5th, 6th and 7th terms in the given expansion are C4n , C5n and C6nThese coefficients are in AP. Thus, we have2 C5n =C4n+C6nOn dividing both sides by C5n, we get:2=C4nC5n+C6nC5n2=5n-4+n-5612n-48=30+n2-4n-5n+20n2-21n+98=0(n-14)(n-7)=0n=7 and 14

Page No 18.47:

Question 14:

In the expansion of 12x1/3+x-1/58, the term independent of x is
(a) T5
(b) T6
(c) T7
(d) T8

Answer:

(b) T6
Suppose the (r + 1)th term in the given expansion is independent of x.
Thus, we have:
Tr+1=Cr8 12x1/38-r (x-1/5)r=Cr8 128-r x8-r3-r5For this term to be independent of x, we must have8-r3-r5=040-5r-3r=0r=5Hence, the required term is the 6th term, i.e. T6

Page No 18.47:

Question 15:

If the sum of odd numbered terms and the sum of even numbered terms in the expansion of x+an are A and B respectively, then the value of x2-a2n is
(a) A2-B2
(b) A2+B2
(c) 4 AB
(d) none of these

Answer:

(a) A2-B2
If A and B denote respectively the sums of odd terms and even terms in the expansion (x+a)nThen , (x+a)n= A+B    ...1             (x-a)n= A-B   ...2Multplying both the equations we get (x+a)n (x-a)n=A2-B2(x2-a2)n=A2-B2  

Page No 18.47:

Question 16:

If the coefficient of x in x2+λx5 is 270, then λ=
(a) 3
(b) 4
(c) 5
(d) none of these

Answer:

(a) 3

The coefficient of x in the given expansion  where x occurs at the (r+1)th term.We haveCr5 (x2)5-r λxr=Cr5 λr x10-2r-rFor it to contain x, we must have:10-3r=1r=3                 Coefficient of x in the given expansion:C35 λ3 =10λ3Now, we have10λ3=270λ3=27λ=3

Page No 18.47:

Question 17:

The coefficient of x4 in x2-3x210 is
(a) 405256

(b) 504259

(c) 450263

(d) none of these

Answer:

(a) 405256

Suppose x4 occurs at the (r+1)th term in the given expansion.Then, we haveTr+1=Cr10(x2)10-r -32x2r    =(-1)r Cr10  3r210-rx10-r-2rFor this term to contain x4, we must have:10-3r=4r=2Required coefficient = C210 3228=10×9×92×28=405256

Page No 18.47:

Question 18:

The total number of terms in the expansion of x+a100+x-a100 after simplification is
(a) 202
(b) 51
(c) 50
(d) none of these

Answer:

(b) 51
Here, n, i.e., 100, is even.
∴ Total number of terms in the expansion =n2+1=1002+1=51 

Page No 18.47:

Question 19:

If T2/T3 in the expansion of a+bn and T3/T4 in the expansion of a+bn+3 are equal, then n =
(a) 3
(b) 4
(c) 5
(d) 6

Answer:

(c) 5
In the expansion (a+b)n, we haveT2T3=C1n an-1×b1C2n an-2×b2In the expansion (a+b)n+3, we haveT3T4=C2n+3 an+1 b2C3n+3 an b3Thus, we haveT2T3=T3T4C1 naC2n b=C2n+3 aC3n+3 b2n-1=3n+12n+2=3n-3n=5

Page No 18.47:

Question 20:

The coefficient of 1x in the expansion of 1+xn 1+1xn is
(a) n !n-1 ! n+1 !

(b) 2n !n-1 ! n+1 !

(c) 2n !2n-1 ! 2n+1 !

(d) none of these

Answer:

(b) 2n !n-1 ! n+1 !

Coefficient of 1xin the given expansion=Coefficient of 1 in (1+x)n×Coefficient of1xin 1+1xn+Coefficient of x in (1+x)n×Coefficient of1x2 in 1+1xn=C0n×C1n +C1n × C2n=n+n×n!2n-2!=n+nnn-12=

Page No 18.47:

Question 21:

If the sum of the binomial coefficients of the expansion 2x+1xn is equal to 256, then the term independent of x is
(a) 1120
(b) 1020
(c) 512
(d) none of these

Answer:

(a) 1120

Suppose (r+1)th tem in the given expansion is independent of x.Then, we haveTr+1=Crn(2x)n-r 1xr    =Crn 2n-r xn-2rFor this term to be independent of x, we must haven-2r=0r=n/2Required term = Cn/2n 2n-n/2=n!n/2!2 2n/2We know:Sum of the given expansion = 256Thus, we have2n. 1n=256n=8Required term = 8!4!  4!24=1120



Page No 18.48:

Question 22:

If the fifth term of the expansion  a2/3+a-1n does not contain 'a'. Then n is equal to
(a) 2
(b) 5
(c) 10
(d) none of these

Answer:

(c) 10

T5=T4+1=C4n (a2/3)n-4 (a-1)4=C4n a2n-83-4For this term to be independent of a, we must have2n-83-4=02n-20=0n=10

Page No 18.48:

Question 23:

The coefficient of x-3 in the expansion of x-mx11 is
(a) -924 m7
(b) -792 m5
(c) -792 m6
(d) -330 m7

Answer:

(d) -330 m7
Let x-3 occur at (r+1)th term in the given expansion.Then, we haveTr+1=Cr11 x11-r -mxr=(-1)r×Cr11  mr x11-r-rFor this term to contain x-3, we must have11-2r=-3r=7Required coefficient =(-1)7 C711  m7                                         =-11×10×9×84×3×2 m7                                         =-330 m7

Page No 18.48:

Question 24:

The coefficient of the term independent of x in the expansion of ax+bx14 is
(a) 14! a7 b7

(b) 14!7!a7 b7

(c) 14!7!2a7 b7

(d) 14!7!3a7b7

Answer:

(c) 14!7!2a7 b7

Suppose (r+1)th term in the given expansion is independent of x.Then, we haveTr+1=Cr14 (ax)14-r bxr         =Cr14 a14-r br x14-2rFor this term to be independent of x, we must have14-2r=0r=7 Required term =C714 a14-7 b7=14!(7!)2a7 b7

Page No 18.48:

Question 25:

The coefficient of x5 in the expansion of 1+x21+1+x22+...+1+x30 
(a) 51C5
(b) 9C5
(c) 31C621C6
(d) 30C5 + 20C5

Answer:

(c) 31C621C6
We have 1+x21+1+x22+...1+x30             =1+x211+x10-11+x-1             =1x1+x31-1+x21Coefficient of x5 in the given expansion = Coefficient of x5 in 1x1+x31-1+x21                                                                 = Coefficient of x6 in 1+x31-1+x21                                                                 =31C6-21C6

Page No 18.48:

Question 26:

The coefficient of x8 y10 in the expansion of (x + y)18 is
(a) 18C8
(b) 18p10
(c) 218
(d) none of these

Answer:

(a) 18C8

Suppose the (r+1)th term in the given expansion contains x8y10.Then, we haveTr+1=Cr18 x18-r yrFor the coefficient of x8y10 We haver= 10Hence, the required coefficient is C1018 or C818

Page No 18.48:

Question 27:

If the coefficients of the (n + 1)th term and the (n + 3)th term in the expansion of (1 + x)20 are equal, then the value of n is
(a) 10
(b) 8
(c) 9
(d) none of these

Answer:

(c) 9
Coefficient of (n+1)th term = Coefficient of (n+3)thWe have:Cn20=Cn+220      2n+2=20      if nCx=nCy   x=y or x+y=nn=9

Page No 18.48:

Question 28:

If the coefficients of 2nd, 3rd and 4th terms in the expansion of 1+xn, n  N are in A.P., then n =
(a) 7
(b) 14
(c) 2
(d) none of these

Answer:

(a) 7
Coefficients of the 2nd, 3rd  and 4th terms in the given expansion are:
C1n, C2n and C3nWe have:2×C2n=C1n+C3nDividing both sides by C2n, we get:2=C1nC2n+C3nC2n2=2n-1+n-236n-6=6+n2+2-3nn2-9n+14=0 n=7  n2 as 2>3 in the 4th term

Page No 18.48:

Question 29:

The middle term in the expansion of 2x3-32x22n is
(a) Cn2n

(b) -1n Cn2n x-n

(c) Cn2nx-n

(d) none of these

Answer:

(b) -1n Cn2n x-n

Here, n is evenMiddle term in the given expansion = 2n2+1th=(n+1)th term =Cn2n 2x32n-n-32x2n=(-1)n Cn2n x-n

Page No 18.48:

Question 30:

If rth term is the middle term in the expansion of x2-12x20, then r+3th term is
(a) C1420x214

(b) C1220 x22-12

(c) -C720 x, 2-13

(d) none of these

Answer:

(c) -C720 x× 2-13
Here n is even
So, The middle term in the given expansion is 202+1th=11th term
Therefore, (r + 3)th term is the 14th term.

T14=C1320 (x2)20-13 -12x13=-113 C1320 x14-13213=-C720 x2-13

Page No 18.48:

Question 31:

The number of terms with integral coefficients in the expansion of 171/3+351/2 x600 is
(a) 100
(b) 50
(c) 150
(d) 101

Answer:

(d) 101

The general term Tr+1 in the given expansion is given byCr600 (171/3)600-r (351/2x)r=Cr600 17200-r/3×35r/2 xrNow, Tr+1 is an integer if  r2 and r3 are integers for all 0r600Thus, we have r=0, 6, 12,...600  (Multiples of 6)Since, It is an A.PSo, 600=0+n-16    n=101Hence, there are 101 terms with integral coefficients.

Page No 18.48:

Question 32:

Constant term in the expansion of x-1x10 is
(a) 152
(b) −152
(c) −252
(d) 252

Answer:

(c) −252

Suppose (r + 1)th term is the constant term in the given expansion.
Then, we have:
Tr+1=Cr10 (x)10-r -1xr        =Cr10 (-1)r x10-r-rFor this term to be constant, we must have:10-2r=0r=5 Required term =-C510=-252

Page No 18.48:

Question 33:

If the coefficients of x2 and x3 in the expansion of (3 + ax)9 are the same, then the value of a is
(a) -79

(b) -97

(c) 79

(d) 97

Answer:

(d) 97
Coefficients of x2 Coefficients of x3

C29 ×39-2 a2 =C39 ×39-3 a3a=C29C39×3       =9! ×3! ×6!×32! × 7! ×9!       =97



View NCERT Solutions for all chapters of Class 12