F.4 A.maths........HELP!!

1:

(1+x)n(1-2x)4

= (1 + nC1x + nC2x2 + .....)[1 - 4C1(2x) + 4C2(2x)2 - .....]

= (1 + nx + n(n-1)x2/2 + ......)(1 - 8x + 24x2 - ......)

= 1(1 - 8x + 24x2 + ......) + nx(1 - 8x + ......) + n(n-1)x2/2(1 + .....) + ......

= 1 - 8x + 24x2 + nx - 8nx2 + n(n-1)x2/2 + ......

= 1 + (n-8)x + [24 - 8n + n(n-1)/2]x2 + ......

= 1 + (n-8)x + [48 - 16n + n(n-1)]x2/2 + ......

= 1 + (n-8)x + [48 - 16n + n2 - n]x2/2 + ......

= 1 + (n-8)x + [n2 - 17n + 48]x2/2 + ......



Coefficient of x2:

[n2 - 17n + 48]/2 = 54

n2 - 17n + 48 = 108

n2 - 17n - 60 = 0

(n - 20)(n + 3) = 0

n = 20 ooro n = -3 (rejected)



Coefficient of x

= 20 - 8

= 12

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2:

[x - (2/x)]6

= Σ6Crxr(2/x)6-r



The x2 term:

xr(1/x)6-r = x2

xr(x)-(6-r) = x2

r - (6 - r) = 2

r - 6 + r = 2

2r = 8

r = 4



When r = 4, the term:

= 6C4x4(2/x)6-4

= 15x4(2/x)2

= 15x4(4/x2)

= 60x2

The coefficient of x2 = 60

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3a:

(1+2x)n

= 1 + nC1(2x) + nC2(2x)2 + nC3(2x)3 + ......

= 1 + n(2x) + [n(n-1)/2](4x2) + [n(n-1)(n-2)/6](8x3) + ......

= 1 + 2nx + 2n(n-1)x2 + 4n(n-1)(n-2)x3/3 + ......



3b:

[x - (3/x)]2

= x2 - 2x(3/x) + (3/x)2

= x2 - 6 + (9/x2)



[x - (3/x)]2[1 + 2x]n

= [x2 - 2x(3/x) + (3/x)2][1 + 2nx + 2n(n-1)x2 + ......]

= [x2 - 6 + (9/x2)][1 + 2nx + 2n(n-1)x2 + ......]

= 1[x2 - 6 + (9/x2)] + 2nx[x2 - 6 + (9/x2)] + 2n(n-1)x2[x2 - 6 + (9/x2)] + ......



The constant term:

1(-6) + 2n(n-1)(9) = 210

-6 + 18n2 - 18n = 210

18n2 - 18n - 216 = 0

n2 - n - 12 = 0

(n - 4)(n + 3) = 0

n = 4 ooro n = -3 (rejected)

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4:

[1 + 2x]7[2 - x]2

= [1 + 7C1(2x) + 7C2(2x)2 + ......](22 - 2(2)x + x2)

= [1 + 14x + 84x2 + ......](4 - 4x + x2)

= 1(4 - 4x + x2) + 14x(4 - 4x + x2) + 84x2(4 - 4x + x2) + ......

= 4 - 4x + x2 + 56x - 56x2 + 336x2 + ......

= 4 + 52x + 281x2 + ......

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5:

[2x3 + (1/x)]8

= Σ6Cr(2x3)r(1/x)8-r



The constant term:

(x3)r(1/x)8-r = x0

x3rŸx-(8-r) = x0

x3r-(8-r) = x0

3r - (8 - r) = 0

3r - 8 + r = 0

r = 2



When r = 2, the term

= 6C2(2x3)2(1/x)8-2

= 15(4x6)(1/x)6

= 60
=

1:Expand (1+x)^n(1-2x)^4 in ascending powers of x up to the term x^2, where n is a positive integer.
If the coefficient of x^2 is 54, find the coefficient of x.

Ans:

(1+x)^n(1-2x)^4
=[1+nx+(1/2)(n)(n-1)x^2+....](1-8x+24x^2+...)

the coefficient of x^2 = 54
(1/2)(n)(n-1)-8n+24=54
n^2-17n-60=0
(n-20)(n+3)=0
n=20 or n=-3 rejected

so the coefficient of x = n-8=20-8=12//

2:Find the coefficient of x^2 in the expansion of (x-(2 over x))^6.

(x-2/x)^6
=...+(x^4)(-2/x)^2+...
=...+4x^2+...

3a:Expand (1+2x)^n in ascending powers of x up to the term x^3, where n is a positive integer.

(1+2x)^n
=1+2nx+2n(n-1)x^2+(4/3)(n)(n-1)(n-2)x^3+...


3b:In the expansion of (x-(3 over x))^2(1+2x)^n, the constant term is 210. Find the value of n.

(x-3/x)^2(1+2x)^n
=[1+2nx+2n(n-1)x^2+(4/3)(n)(n-1)(n-2)x^3+...][x^2-6+9/x^2]

constant term =210
9*2n(n-1)-6=210
n^2-n-12=0
(n-4)(n+3)=0
n=4 or n=-3 (rejected)

4:Expand (1+2x)^7(2-x)^2 in ascending powers of x up to the term x^2.

(1+2x)^7(2-x)^2
=(1+14x+196x^2+...)(x^2-4x+4)
=4+(-4+4*14)x+(4*196-4*16+1)x^2+...
=4+52x+613x^2+...