Q 31 :

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (24+134)n is 6:1, then the third term from the beginning is

  • 302

     

  • 603

     

  • 602

     

  • 303

     

(2)

 



Q 32 :

If the coefficients of three consecutive terms in the expansion of (1+x)n are in the ratio 1 : 5 : 20, then the coefficient of the fourth term is

  • 1827

     

  • 5481

     

  • 2436

     

  • 3654

     

(4)

Given coefficients of three consecutive terms are in the ratio 1:5:20

i.e.,  Cr-1n:Crn:Cr+1n=1:5:20

Now,  Cr-1nCrn=15⇒n=6r-1                      ...(i)

Also,  CrnCr+1n=520⇒n=5r+4                   ...(ii)

From (i) and (ii), we get

6r-1=5r+4⇒r=5⇒n=25+4=29

∴  The coefficient of 4th term=C329=3654



Q 33 :

If the term without x in the expansion of  (x23+αx3)22 is 7315, then |α| is equal to ________.



(1)

Tr+1=Cr22(x23)22-r·(αx3)r

=Cr22·(x443-2r3-3r)·(αr)

For the term without 'x', 443-2r3-3r=0⇒443=11r3

∴  r=4

So, the term without x is C422α4=7315

⇒  22×21×20×1924·α4=7315

∴  |α|=1



Q 34 :

The coefficient of x18 in the expansion of (x4-1x3)15 is _________.



(5005)

Given,  (x4-1x3)15

Tr+1=Cr15(x4)15-r(-1x3)r

=Cr15(x)60-4r·(-1)rx-3r=Cr15(x)60-7r·(-1)r

For coefficient of x18, put 60-7r=18⇒r=427=6

Hence, coefficient of x18 is C615(-1)6=5005



Q 35 :

If the coefficient of x9 in (αx3+1βx)11 and the coefficient of x-9 in (αx-1βx3)11 are equal, then (αβ)2 is equal to _______.



(1)

(r+1)th term in (αx3+1βx)11=Cr11α11-r·β-r·x33-4r

Here,  33-4r=9⇒r=6

∴  Coefficient of x9=C611α5β-6

(r+1)th term in (αx-1βx3)11=Cr11α11-r(-1β)r·x11-4r

Here,  11-4r=-9⇒r=5

∴  Coefficient of x-9=-C511α6β-5

Now,  C611α5β6=-C511α6β5    [∵ Given]

⇒  1β=-α⇒(αβ)2=1



Q 36 :

The coefficient of x-6 in the expansion of (4x5+52x2)9 is _________.



(5040)

Here, Tr+1=Cr9(4x5)9-r(52x2)r

=Cr9(45)9-r(52)rx9-r-2r

For coefficient of x-6, we have 9-r-2r=-6⇒3r=15

⇒  r=5

So, coefficient of x-6=C59(45)4(52)5

=9!5! 4!(4×4×4×45×5×5×5)(5×5×5×5×52×2×2×2×2)

=9·8·7·64·3·2(8)(5)=126×40

∴  126×40·x-6=5040



Q 37 :

If k is the coefficient of x4 in the expansion of (1+2x+3x2)6, then the sum of the digits of k is _______.



(6)

The general term is 6!α!β!γ!1α(2x)β·(3x2)γ

β-2γ=4,  α+β+γ=6⇒α=2-3γ

β=2γ+4  ∴  (α,β,γ)=(2,4,0)

The desired coefficient is

6!2! 4!·24=15×16=240



Q 38 :

If the term independent of x in the expansion of (x-λx2)10  is 405, then the value of |4λ| is _________.



(12)

Let (r+1)th term is independent of x.

We have, Tr+1=Cr10(x)10-r·(-λx2)r

                               =Cr10·x 5-r2-2r·(-λ)r

The term will be independent of x if 5-r2-2r=0

⇒  r=2

Then,  T2+1=C210·(-λ)2=45λ2=405    (Given)

∴  λ2=9⇒λ=±3⇒|4λ|=12



Q 39 :

If the magnitude of the coefficient of x7 in the expansion of (ax2+1bx)8, where a,b are positive numbers, is equal to the magnitude of the coefficient of x-7 in the expansion of (ax+1bx2)8, then |ab| is equal to _________.



(1)

In the expansion (ax2+1bx)8, (r+1)th term is

Tr+1=Cr8(ax2)8-r(1bx)r

Let the (r+1)th term contain x7.

∴  x16-3r=x7⇒r=3

∴  Coefficient of x7=C38 a5b-3                      ...(i)

In the expansion (ax+1bx2)8, (r+1)th term is

Tr+1=Cr8(ax)8-r(1bx2)r=Cr8 a8-r(b)-rx8-3r

Let the (r+1)th term contain x-7.

∴  x8-3r=x-7⇒r=5

∴  Coefficient of x-7=C58 a3b-5                  ...(ii)

From (i) and (ii), we have  |C38a5b-3|=|C58a3b-5|

⇒  a2b2=1⇒ab=1



Q 40 :

Coefficient of x99 in x100+2x99(1+x)+3(1+x)2x98+⋯+101(1+x)100 is ____

  •  C99101- C98101

     

  • 100(C99101)-C98101

     

  •  C98-101+101(C99101)

     

  • 101(C98101)-C99101

     

(3)

Required=coefficient of x99 in x100(1+2(1+xx)+.....101(1+xx)100)



Q 41 :

Statement-1: The sum of rational terms in (2+33+56)10 is 1262

Statement-2: The number of irrational terms in the expansion of (51/6+21/8)100 is 97

The correct statements among the following are

  • Statement-1 is true, statement-2 is true

     

  • Statement-1 is true, statement-2 is false

     

  • Statement-1 is false, statement-2 is true

     

  • Statement-1 is false, statement-2 is false

     

(3)

Statement-1 General term=10!α!β!γ! 2α/23β/35γ/6    for rational terms

α=0,2,4,6,8,10,  β=0,3,6,  γ=0,6

Hence possible sets=(4,6,0),(4,0,6),(10,0,0)

Hence, there are 5 rational terms.

∴  required=10!4!6! 22410!10! 25=12632

Statement-3  tr+1, the (r+1)th in the expansion of (51/6+21/8)10 is  tr+1=Cr10(5)100-r6(21/8)r

As 5 and 2 are relatively prime, tr+1 will be rational if 100-r6 and r8 are both integers.

i.e. If 100-r is a multiple of 6 and r is a multiple of 8.

As 0≤r≤100, multiple of 8 upto 100 and corresponding value of 100-r

r=0,8,16,24,…,88,96

100-r=100,92,84,76,…,12,4

Out of 100-r, multiple of 6 are 84,60,36,12.

∴ There are just four rational terms

⇒Number of irrational terms=101-4=97



Q 42 :

The remainder when the number 3223-(322)3 is divide by 8, is ______



(0)

3256-312=312×(3256-12-1)=(1+8)6((1+8)112-1)=(1+8λ)(1+8μ-1)

=8μ(1+8λ), which is divisible by 8. Hence, remainder is zero.



Q 43 :

Coefficient of x4 in 5x2(1+x2)11 is

  • 22

     

  • 33

     

  • 44

     

  • 55

     

(4)

5x2(1+C111x2+⋯)



Q 44 :

If 26(233(C212)+255(C412)+277(C612)+...+21313(C1212))=313–α, then α is equal to:          [2026]

  • 45

     

  • 48

     

  • 51

     

  • 54

     

(3)

26[233(C212)+255(C412)+277(C612)+...+21313(C1212)]=313–α

=2[23·C313+25·C513+27·C713+...+213·C1313]

=2[(1+2)13–(1–2)132–26]

=313+1–52=313–51      ∴ α=51