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=15n=6r-1                      ...(i)

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

From (i) and (ii), we get

6r-1=5r+4r=5n=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=0443=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=18r=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=9r=6

  Coefficient of x9=C611α5β-6

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

Here,  11-4r=-9r=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=-63r=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=6α=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·x5-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=9λ=±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=x7r=3

  Coefficient of x7=C38a5b-3                      ...(i)

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

Tr+1=Cr8(ax)8-r(1bx2)r=Cr8a8-r(b)-rx8-3r

Let the (r+1)th term contain x-7.

  x8-3r=x-7r=5

  Coefficient of x-7=C58a3b-5                  ...(ii)

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

  a2b2=1ab=1