Q 1 :    

Let (a+bx+cx2)10=i=020pixi,a,b,c. If p1=20 and p2=210, then 2(a+b+c) is equal to              [2023]

  • 8

     

  • 6

     

  • 12

     

  • 15

     

(3)

We have, (a+bx+cx2)10=i=020pixi, a,b,cp1=20 and p2=210

Now (a+bx+cx2)10=p0+p1x+p2x2+

C110a9·b=p1 and C210a8·b2+C110·a9·c=p2

10ba9=20 and 45b2a8+10c·a9=210ba9=2

b=2,  a=1                                                                 [a,b]

Now, 45×(2)2+10·c=210

10c=210-180c=3

   2(a+b+c)=2(1+2+3)=2×6=12



Q 2 :    

The coefficient of x7 in (1-x+2x3)10 is _________ .             [2023]



(960)

General term is given by

10!r1!r2!r3!(1)r1(-x)r2(2x3)r3 i.e., 10!r1!·r2!·r3!(-1)r2·(2)r3xr2+3r3

where r1+r2+r3=10 and r2+3r3=7

Possible values of (r1,r2,r3) are (3, 7, 0), (5, 4, 1) and (7, 1, 2)

Required coefficient

=10!3!7!(-1)7+10!5!4!(-1)4(2)+10!7!2!(-1)1(2)2

=-120+2520-1440=960



Q 3 :    

If A denotes the sum of all the coefficients in the expansion of (1-3x+10x2)n and B denotes the sum of all the coefficients in the expansion of (1+x2)n, then                    [2024]

  • B=A3

     

  • 3A=B

     

  • A=3B

     

  • A=B3

     

(4)

Putting x=1 in (1-3x+10x2)n, we get

A=(1-3+10)n=8n                                               ...(i)

Putting x=1 in (1+x2)n, we get

B=(1+1)n=2n                                                          ...(ii)

From (i) and (ii), we get A=B3

 



Q 4 :    

The sum of all rational terms in the expansion of (1+21/3+31/2)6 is equal to __________.          [2025]



(612)

Given, (1+21/3+31/2)6

   The general term of multinomial expansion is 6!x!y!z!(1)x(21/3)y(31/2)z

For terms to be rational 3 divides y and 2 divides z.

   

x y z
6 0 0
4 0 2
2 0 4
0 0 6
3 3 0
1 3 2
0 6 0

   Required sum = 6!6!0!0!+6!4!0!2!×3+6!2!0!4!×(3)2+6!0!0!6!(3)3+6!3!3!0!(2)+6!1!3!2!(2)×(3)+6!0!6!0!(2)2

= 1 + 45 + 135 + 27 + 40 + 360 + 4 = 612



Q 5 :    

Let (1+x+x2)10=a0+a1x+a2x2+...+a20x20. If (a1+a3+a5+...+a19)11a2=121k, then k is equal to __________.          [2025]



(239)

(1+x+x2)10=a0+a1x+a2x2+...+a20x20

Put x = 1, we have

310=a0+a1+a2+.....+a20          ... (i)

Put x = –1, we have

1=a0a1+a2+.....+a20          ... (ii)

Subtracting equation (ii) from (i),

a1+a3+...+a19=31012=29524

Also, {1+x(1+x)}10=1+C110x(1+x)+C210x2(1+x)2+...

  a2=C110+C210=55

  k=(a1+a3+...+a19)11a2121=29524605121=239.