Form 4 A-Maths(好難呀!高手請入) 40分

1.
(1 + x + x2)n
= [(1 + x) + x2)]n
= (1+x)n + nC1(1 + x)n-1x2 + nC2(1 + x)n-2(x2)2 + ......
Only the first two terms in expansion can give x2.

(1 + x + x2)n
= [(1 + x) + x2)]n
= (1+x)n + nC1(1 + x)n-1x2 + ......
= (1 + nC1x + nC2x2 + ......) + nC1(1 + n-1C1x.......)x2 + ......

Coefficient of x2 in expansion:
nC2 + nC1 = 21
(n!/(n-2)!2!) + [n!/(n-1)!1!] = 21
[n(n - 1)/2] + n = 21
Multiply both sides by 2:
n(n - 1) + 2n = 42
n2 - n + 2n = 42
n2 + n - 42 = 0
(n - 6)(n + 7) = 0
n = 6 ooro n = -7 (rejected)

Ans: n = 6

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2.
(1 + 3x)4
= 1 + 4C1(3x) + 4C2(3x)2 + .....
= 1 + 4(3x) + 6(9x2) + ......
= 1 + 12x + 54x2 + ......

(1 - 2x)5
= 1 - 5C1(2x) + 5C2(2x)2 + ......
= 1 - 5(2x) + 10(4x2) + ......
= 1 - 10x + 40x2 + ......

(1 + 3x)4(1 - 2x)5 = 1 + ax + bx2 + .....
(1 + 12x + 54x2 + ......)( 1 - 10x + 40x2 + ......) = 1 + ax + bx2 + .....
[1(1) - 1(10x) + 1(40x2)] + [12x(1) + 12x(-10x)] + 54x2(1) + ... = 1 + ax + bx2 + ...
1 - 10x + 40x2 + 12x - 120x2 + 54x2 + ..... = 1 + ax + bx2 + ......
1 + 2x - 26x2 + ...... = 1 + ax + bx2 + ......

Compare the x terms: a = 2

Compare the x2 terms: b = -26

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3.
[1 - (2/x)]3
= 1 - 3C1(2/x) + 3C2(2/x)2 - (2/x)3
= 1 - 3(2/x) + 3(4/x2) - (8/x3)
= 1 - (6/x) + (12/x2) - (8/x3)

(1 + x)10[1 - (2/x)]3
= (1 + 10C1x + 10C2x2 + 10c3x3 + ...)[1 - 3C1(2/x) + 3C2(2/x)2 - (2/x)3]
= (1 + 10x + 45x2 + 120x3 + .....)[ 1 - 3(2/x) + 3(4/x2) - (8/x3)]
= (1 + 10x + 45x2 + 120x3 + .....)[ 1 - (6/x) + (12/x2) - (8/x3)]

The constant term in the expansion
= 1(1) + 10x(-6/x) + 45x2(12/x2) - 120x3(8/x3)
= 1 - 60 + 540 - 960
= -479
=