**Page 18.11 Ex 18.1**

**Q1.**

**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

=32×5+5×16×4×3y+10×8×3×9y2+10×4×2×27y3+5×2x×81y4+243y5=32×5+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=16×4-4×8×3×3y+6×4×2×9y2-4×2x×27y3+81y4=16×4-96x3y+216x2y2-216xy3+81y4

(iii)

x-1×6=C06 x61x0-C16 x51x1+C26 x41x2-C36 x31x3+C46 x21x4-6C5 x11x5+C66 x01x6=x6-6 x5×1x+15 x4×1×2-20×3×1×3+15×2×1×4-6 x×1×5+1×6=x6-6×4+15×2-20+15×2-6×4+1×6

(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+21×9×2-35×27×3+35×81×4-21×243×5+7×729×6-2187×7=1-21x+189×2-945×3+2835×4-5103×5+5103×6-2187×7

(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=x2-6×5/3a1/3+15×4/3a2/3-20xa+15×2/3a4/3-6×1/3a5/3+a2

(viii)

(1+2x-3×2)5Consider 1-2x and 3×2 as two separate entities and apply the binomial theorem.Now,C05(1+2x)5(3x)0-C15(1+2x)4(3×2)1+C25(1+2x)3(3×2)2-C35(1+2x)2(3×2)3+C45(1+2x)1(3×2)4-C55(1+2x)0(3×2)5=(1+2x)5-5(1+2x)4×3×2+10×(1+2x)3×9×4-10×(1+2x)2×27×6+5(1+2x)×81×8-243×10=C05×(2x)0+C15×(2x)1+C25×(2x)2+C35×(2x)3+C45×(2x)4+C55×(2x)5- 15×2[C04(2x)0+C14(2x)1+C24(2x)2+C34(2x)3+C44(2x)4]+ 90×4[1+8×3+6x+12×2]-270×6(1+4×2+4x)+405×8+810×9-243×10=1+10x+40×2+80×3+80×4+32×5-15×2-120×3-3604-480×5-240×6+ 90×4+720×7+540×5+1080×6-270×6-1080×8-1080×7+405×8+810×9-243×10=1+10x+25×2-40×3-190×4+92×5+570×6-360×7-675×8+810×9-243×10

(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+1×2-1×3=x3+1+3x+3×2-3×2+3+6xx+3x+1×2-1×3=x3+1+3x+3×2-3x-3x-6+3x+3×2-1×3=x3+3×2-5+3×2-1×3

(x)

(1-2x+3×2)3=C03(1-2x)3+C13(1-2x)2(3×2)+C23(1-2x)(3×2)2+C33(3×2)3=(1-2x)3+9×2(1-2x)2+27×4(1-2x)+27×6=1-8×3+12×2-6x+9×2(1+4×2-4x)+27×4-54×5+27×6=1-8×3+12×2-6x+9×2+36×4-36×3+27×4-54×5+27×6=1-6x+21×2-44×3+63×4-54×5+27×6

**Q2.**

**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+3×2+15(x2+2x+1)(x-1)+15(x+1)(x2+1-2x)+x3-1+3x-3×2]=2[2×3+6x+15×3-15×2+30×2-30x+15x-15+15×3+15×2-30×2-30x+15x+15]=2[32×3-24x]=16x[4×2-3]

(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+15×4(x2-1)+15×2(x2-1)2+(x2-1)3]=2[x6+15×6-15×4+15×2(x4-2×2+1)+(x6-1+3×2-3×4)]=2[x6+15×6-15×4+15×6-30×4+15×2+x6-1+3×2-3×4]=64×6-96×4+36×2-2

(iii)

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

(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

**Q3.**

**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

**Q4.**

**Answer :**

The expression x+16+x-16 can be written as

(x+1)6+(x-1)6=2[C06x6+C26x4+C46x2+C66x0]=2[x6+15×4+15×2+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

**Q5.**

**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 terms∵10001>1000∴(1.1)10000>1000

**Q6.**

**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

**Q7.**

**Answer :**

23n-7n-1=8n-7n-1 …(1)

Now,8n=(1+7)n =C0n+C1n×71+C2n×72+C3n×73+C4n×74+…+Cnn×7n⇒8n=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)

**Q8.**

**Answer :**

32n+2-8n-9=9n+1-8n-9 …1

Consider

9n+1=1+8n+1⇒9n+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.

**Q9.**

**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×26n⇒27n=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)

**Q10.**

**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 terms∵801 >800

Hence, (1.2)^{4000} is greater than 800

**Q11.**

**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 18.32 Ex 18.2**

**Q1.**

**Answer :**

Given:

2x-1×225

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)25-15-1×215 =C1525210x10-1×30=-C1525210x20

Now, we will find the 11th term from the beginning.

T11=T10+1 =C1025(2x)25-10-1×210 =C1025215x151x20 =C1025215x5

**Page 18.33 Ex 18.2**

**Q2.**

**Answer :**

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

Let it be T_{7}

Now, we have

T7=T6+1

=C610(3×2)10-6-1×36=C61034x81x18=10×9×8×7×814×3×2×x10=17010×10

Thus, the 7th term of the given expression is 17010×10

**Q3.**

**Answer :**

Given:

3x-1×210

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)10-6-1×26=C61034x41x12=10×9×8×7×814×3×2×1×x8=17010×8

**Q4.**

**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 =-10×9×83×2×9/2y3/2×7/2y21/2 =-120x8y12

**Q5.**

**Answer :**

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

∵T7=T6+1

∴T7=T6+1 =C684x58-652×6 =8×7×4×4×125×1252×1×25×64 x2 1×6 =4375×4

**Q6.**

**Answer :**

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

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

∴ T7=T6+1 =C69 x9-62×6 =9×8×73×2x364x6 =5376×3

4th term from the beginning = T4=T3+1

∴T4=C39 x9-3 2×3 =9×8×73×2x68x3 =672 x3

**Q7.**

**Answer :**

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., 7th term, from the beginning.

Thus, we have:

T7=T6+1 =C69 4×59-652×6 =9×8×73×264125×3125×12564×6 =10500×3

**Q8.**

**Answer :**

Let T_{r+1} be the 7th term from the end in the given expression.

Then, we have:

T_{r+1} = (9 − 7 + 1) = 3rd term from the beginning

Now,

T3=T2+1 =C28 (2×2)8-2 -32×2 =8×72×164x1294x2 =4032 x10

**Q9.**

**Answer :**

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

Then, we have:

Tr+1=Crn xn-r ar

Here,

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

(ii) Suppose x7 occurs at the (r + 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=7⇒3r=40-7=33⇒r=11∴Coefficient of x7 = (-1)11 C1140=-C1140

(iii) Suppose x−15 occurs at the (r + 1)th term in the given expression.

Then, we have:

Tr+1=Cr10 (3×2)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 =-15⇒5r=20+15⇒r=7∴Coefficient of x-15 = (-1)7 C710 310-14 a7=-10×9×83×2×9×9a7=-4027a7

(iv) Suppose x9 occurs at the (r + 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=9⇒3r=9⇒r=3∴Coefficient of x9 =(-1)3 C39 133=-9×8×72×9×9=-289

(v)

Suppose xm occurs at the (r + 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 =m⇒r=(n-m)/2∴Coefficient of xm = C(n-m)/2n=n!n-m2! n+m2!

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

Then, we have:

(1-2×3+3×5)1+1×8=1-2×3+3x5C08+C18 1x+C28 1×2+C38 1×3+C48 1×4+C58 1×5+C68 1×6+C78 1×7+C88 1×8 x occurs in the above expresssion at -2×3.C28 1×2 +3×5.C48 1×4.∴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

**Q10.**

**Answer :**

Suppose T_{r+1}th 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=2r3-72⇒2r3+r2=7+72⇒7r6=212⇒r=9Hence, the required term is the 10th term

**Q11.**

**Answer :**

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

Then, we have:

Tr+1=Cr20 (2×2)20-r -1xr=(-1)r Cr20220-r x40-2r-rFor this term to contain x9, we must have40-3r=9⇒3r=31⇒r=313 It is not possible, as r is not an integer.

Hence, there is no term with x^{9} in the given expression.

**Q12.**

**Answer :**

Suppose x^{-1} 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=-1⇒3r=25⇒r=253It is not possible, as r is not an integer.

Hence, the expansion of x2+1×12 does not contain any term involving x^{-1}.

**Q13.**

**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 23×20-10 32×10=C1020 210310×310210×10-10=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 -2×5=-10×9×8×7×65×4×3×2×32×5=-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 18.34 Ex 18.2**

**Q14.**

**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=9×8×7×64×3×2×27×9×136×36×17=1898x17and,T6=T5+1=C59(3x)9-5 -x365=-9×8×7×64×3×2×81×1216×36×19=-2116×19

(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 (2×2)7-3 -1×3=-7×6×53×2×16 x8×1×3=-560 x5And,T5=T4+1=C47 (2×2)7-4 -1×4=35×8 ×x6×1×4=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 -2×27 =-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 -2×28 =15×14×13×12×11×10×97×6×5×4×3×2×37×28 ×x7-16 =6435×37×28×9

(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 -1×35 =-11×10×9×8×75×4×3×2×x24-15 =-462 x9And,T7=T6+1 =C611 (x4)11-6 -1×36 =11×10×9×8×75×4×3×2×20-18 =462 x2

**Q15.**

**Answer :**

(i)

x-1x10Here, n is an even number. ∴ Middle term = 102+1th= 6th termNow, we haveT6=T5+1=C510 x10-5 -1×5=-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+3×2+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=9×8×7×64×3×2×25 144×5+8=634x13And,T6=T5+1=C59 (2x)9-5 -x245=-9×8×7×64×3×2×24 145×4+10=-6332×14

(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=3548×12

**Q16.**

**Answer :**

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

Now,

32×2-13x9Tr+1=Cr9 32×29-r -13xr= (-1)r Cr9 .39-2r29-r× x18-2r-rFor this term to be independent of x, we must have18-3r=0⇒3r=18⇒r=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=0⇒r=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,

2×2-3x325Tr+1=Cr25 (2×2)25-r -3x3r=(-1)r Cr25 ×225-r×3r x50-2r-3rFor this term to be independent of x, we must have:50-5r=0⇒r=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=0⇒r=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=0⇒10-5r=0⇒r=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=0⇒r = 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,

12×1/3 +x-1/58Tr+1=Cr8 12×1/38-r (x-1/5)r=Cr8. 128-r x8-r3-r5For this term to be independent of x, we must have8-r3-r5=0⇒40-5r-3r=0⇒8r=40⇒r=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=0⇒18-2r=0⇒r=9The term is C918×129

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

Now,

32×2-13x6Tr+1=Cr6 32×26-r -13xr=-1r Cr6 × 36-r-r26-r x12-2r-rFor this term to be independent of x, we must have12-3r=0⇒r=4Hence, the required term is the 4th term.C46 × 36-4-426-4=6×52×1×4×9=512

**Q17.**

**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-318⇒2r+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

**Q18.**

**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-318⇒2r+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

**Q19.**

**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.

**Q20.**

**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=0⇒n=r∴Required coefficient = Cn2n =(2n)!(n!)2 =1·3·5…2n-32n-12·4·6…2n-22n(n!)2 =1·3·5…2n-32n-12nn!

**Q21.**

**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+C6nC5n⇒2=5n-4+n-56⇒12n-48=30+n2-4n-5n+20⇒n2-21n+98=0⇒(n-14)(n-7)=0⇒n=7 or 14

**Q22.**

**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-23⇒12n-6=6+4n2-4n-2n+2⇒4n2-18n+14=0⇒2n2-9n+7=0

Hence proved.

**Q23.**

**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=220495⇒r-1n-r+2=49⇒9r-9=4n-4r+8⇒4n+17=13r …1Also,CrnCr-1n=792495⇒n-r+1r=85⇒5n-5r+5=8r⇒5n+5=13r⇒ 5n+5=4n+17 From Eqn1⇒n=12

**Q24.**

**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+C3nC2n⇒2=2n-1+n-23⇒6n-6=6+n2+2-3n⇒n2-9n+14=0⇒ n=7 ∵ n≠2 as 2>3 in the 4th term

**Q25.**

**Answer :**

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

If p≠q, then p+q=n+2

Hence proved

**Page 18.35 Ex 18.2**

**Q26.**

**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=56⇒r-1+r+1=n [If Crn=Csn⇒r=s or r+s =n]⇒2r=n⇒r=n2Now,Cn2n =70 and Cn2-1n=56⇒Cn2-1n Cn2n=5670⇒n2n2+1=810⇒5n=4n+8⇒n=8So, r=n2=4Thus, the required terms are 4th, 5th and 6th.

**Q27.**

**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

**Q28.**

**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=cb⇒n-56=ba and n-67=cb⇒n-67n-56=cbba⇒6n-367n-35=ca

**Q29.**

**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=76⇒Cr-1n=Cr+1n⇒r-1+r+1=n [If Crn=Csn⇒r=s or r+s =n]⇒r=n2∴CrnCr-1n=9576⇒n-r+1r=9576⇒n2 +1n2=9576⇒38n+76=95n2 ⇒19n2=76⇒n=8

**Q30.**

**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=7112⇒n-6+16x-1 a=116⇒ax=38n-40 …1Also,C7 nxn-7 a7C6n xn-6 a6=1/47⇒n-7+17x-1a=128⇒ax=14n-24 …2From 1 and 2, we get: 38n-40=14n-24⇒32n-10=1n-6⇒n=8

Putting in eqn1 we get⇒a=xNow, C58 x8-5 x85=112⇒56×885=112⇒x8=48⇒x=4By putting the value of x and n in 1 we geta=12

a=3 and x=2

**Q31.**

**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=1080⇒C2n xn-2 a2C1n xn-1 a1=720240⇒n-12xa=3⇒ax=6n-1 …1Also,C3n xn-3 a3C2n xn-2 a2=1080720⇒n-23xa=32⇒ax=92n-4 …2Using 1 and 2 we get6n-1=92n-4⇒n=5Putting in eqn1 we get⇒2a=3xNow, C15 x5-1 32x=240⇒15×5=480⇒x5=32⇒x=2By putting the value of x and n in 1 we geta=3

**Q32.**

**Answer :**

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

**Q33.**

**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 a3⇒a=C29C39×3 =9! ×3! ×6!×32! × 7! ×9! =97

**Q34.**

**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 18.37 (Very Short Answers)**

**Q1.**

**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

**Q2.**

**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

**Q3.**

**Answer :**

The given expression is (1-3x+3×2-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

**Q4.**

**Answer :**

Here, n, i.e., 10, is an even number. ∴Middle term=102+1th term=6th termThus, we have:T6=T5+1 =C510 2×2310-5 32×25 =10×9×8×7×65×4×3×2×2535×3525 =252

**Q5.**

**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=0⇒r=3Hence, the required term is the 4th term.

**Q6.**

**Answer :**

Coefficient of xm in the given expansion = Cmm+n =aCoefficient of xn in the given expansion = Cnm+n =b∴a=b ∵Cmm+n =Cnm+n

**Q7.**

**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=2⇒a=2b

**Q8.**

**Answer :**

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

**Q9.**

**Answer :**

Here,a=1-3+10=8=23b=1+1=2⇒a=b3

**Q10.**

**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 18.37 (Multiple Choice Questions)**

**Q1.**

**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=n⇒r=2 or 2r=18⇒r=9

**Q2.**

**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=0⇒r=4∴Required term: (-1)4 C412 212-8=12×11×10×94×3×2×16=7920

**Q3.**

**Answer :**

(c) 9

rth term in the given expansion is Cr-112 (2×2)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=0⇒r=9Hence, the 9th term in the expansion is independent of x.

**Q4.**

**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+3⇒2n-1=3n+1⇒2n+2=3n-3⇒n=5

**Page 18.38 (Multiple Choice Questions)**

**Q5.**

**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

**Q6.**

**Answer :**

(a) 7920

Suppose the (r+1)th term is independent of x in the given expansion.Then, we have:Tr+1=Cr12 (2×2)12-r -1xr=(-1)r Cr12 212-r x24-2r-rFor this term to be independent of x, we must have:24-3r=0⇒r=8∴Required term=(-1)8 C812 212-8=12×11×10×94×3×2×24=7920

**Q7.**

**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=77⇒r=11∴Required coefficient=(-1)11 C1115=-15×14×13×124×3×2=-1365

**Q8.**

**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=0⇒r=6∴Required term=(-1)6 C69 136=9×8×73×2×136=28243

**Q9.**

**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-215⇒2r+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

**Q10.**

**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 2×2310-5 32×25 =10×9×8×7×65×4×3×2×2535×3525 =252

**Q11.**

**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=-17⇒r=12

**Q12.**

**Answer :**

(b) T_{4}

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=0⇒r=3Hence, the required term is the 4th term i.e. T4

**Q13.**

**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+C6nC5n⇒2=5n-4+n-56⇒12n-48=30+n2-4n-5n+20⇒n2-21n+98=0⇒(n-14)(n-7)=0⇒n=7 and 14

**Q14.**

**Answer :**

(b) T_{6}

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

Thus, we have:

Tr+1=Cr8 12×1/38-r (x-1/5)r=Cr8 128-r x8-r3-r5For this term to be independent of x, we must have8-r3-r5=0⇒40-5r-3r=0⇒r=5Hence, the required term is the 6th term, i.e. T6

**Q15.**

**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

**Q16.**

**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=1⇒r=3 ∴Coefficient of x in the given expansion:C35 λ3 =10λ3Now, we have10λ3=270⇒λ3=27⇒λ=3

**Page 18.39 (Multiple Choice Questions)**

**Q17.**

**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=4⇒r=2∴Required coefficient = C210 3228=10×9×92×28=405256

**Q18.**

**Answer :**

(b) 51

Here, n, i.e., 100, is even.

∴ Total number of terms in the expansion =n2+1=1002+1=51

**Q19.**

**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=T3T4⇒C1 naC2n b=C2n+3 aC3n+3 b⇒2n-1=3n+1⇒2n+2=3n-3⇒n=5

**Q20.**

**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=

**Q21.**

**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=0⇒r=n/2∴Required term = Cn/2n 2n-n/2=n!n/2!2 2n/2We know:Sum of the given expansion = 256Thus, we have2n. 1n=256⇒n=8∴Required term = 8!4! 4!24=1120

**Q22.**

**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=0⇒2n-20=0⇒n=10

**Q23.**

**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=-3⇒r=7∴Required coefficient =(-1)7 C711 m7 =-11×10×9×84×3×2 m7 =-330 m7

**Q24.**

**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=0⇒r=7∴ Required term =C714 a14-7 b7=14!(7!)2a7 b7

**Q25.**

**Answer :**

(c) ^{31}C_{6} − ^{21}C_{6}

We have 1+x21+1+x22+…1+x30 =1+x211+x10-11+x-1 =1×1+x31-1+x21Coefficient of x5 in the given expansion = Coefficient of x5 in 1×1+x31-1+x21 = Coefficient of x6 in 1+x31-1+x21 =31C6-21C6

**Q26.**

**Answer :**

(a) ^{18}C_{8}

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

**Q27.**

**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=n⇒n=9

**Q28.**

**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+C3nC2n⇒2=2n-1+n-23⇒6n-6=6+n2+2-3n⇒n2-9n+14=0⇒ n=7 ∵ n≠2 as 2>3 in the 4th term

**Page 18.40 (Multiple Choice Questions)**

**Q29.**

**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

**Q30.**

**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 -12×13=-113 C1320 x14-13213=-C720 x2-13

**Q31.**

**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 0≤r≤600Thus, 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.

**Q32.**

**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=0⇒r=5∴ Required term =-C510=-252

**Q33.**

**Answer :**

(d) 97

Coefficients of x<sup>2</sup> = Coefficients of x<sup>3</sup>

C29 ×39-2 a2 =C39 ×39-3 a3⇒a=C29C39×3 =9! ×3! ×6!×32! × 7! ×9! =97

**Q34.**

**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. 0≤r≤45∴ r =0,10, 20, 30 and 40Hence, there are 5 rational and 41, i.e., 46-5, irrational terms.