超急呀~~~~~f.4 a math~~20/10/06 6:00am前~~~please~~~

1-px+qx^2)^8

= [1 - x(p - qx)]^8

= 1 - (8C1)(x)(p - qx) + (8C2)(x^2)(p - qx)^2 + (8C3)(x^3)(p - qx)^3 +.....

= 1 - 8x(p - qx) + 28(x^2)(p^2 - 2pqx + q^2 x^2) + (56p^3)x^3 + .....

= 1 - 8px + 8qx^2 + (28p^2)x^2 - 56pqx^3 + (56p^3)x^3 + .....

= 1 - 8px + (28p^2 + 8q)x^2 + (56p^3 - 56pq)x^3 + .....

= 1 - 16x + 88x^2 + ......

then 8p = 16

p = 2

28p^2 + 8q = 88

28(4) + 8q = 88

q = -3

hence the coefficient of x^3 = 56p^3 - 56pq

= 56(2)^3 - 56(2)(-3)

= 56(8) + 56(6)

= 784
1 - 8px (28p^2 8q)x^2 (- 56p^3 - 56pq)x^3 .....


coefficient of x^3 = - 56p^3 - 56pq

= -56(8) 56(6)

= - 112

[1+x(qx-p)]
= 1 + 8x(qx-p) + 28x^2(qx-p)^2 + 56x^3(qx-p)^3 + ...
= -8px + 8qx^2 + 28(px)^2 - 56pqx^3 - 56(px)^3 + ...
= -8px + (28p^2 + 8q)x^2 - 56p(q + p^2)x^3 + ...

-8p = -16
p = 2

28 * (2)^2 + 8q = 88
8q = -24
q = -3

coeff. of x^3 = -56(2)( -3 + 4)
= -112

(1-px+qx^2)^8

= [1 - x(p - qx)]^8

= 1 - (8C1)(x)(p - qx) + (8C2)(x^2)(p - qx)^2 + (8C3)(x^3)(p - qx)^3 +.....

= 1 - 8x(p - qx) + 28(x^2)(p^2 - 2pqx + q^2 x^2) + (56p^3)x^3 + .....

= 1 - 8px + 8qx^2 + (28p^2)x^2 - 56pqx^3 + (56p^3)x^3 + .....

= 1 - 8px + (28p^2 + 8q)x^2 + (56p^3 - 56pq)x^3 + .....

= 1 - 16x + 88x^2 + ......

then 8p = 16

p = 2

28p^2 + 8q = 88

28(4) + 8q = 88

q = -3

hence the coefficient of x^3 = 56p^3 - 56pq

= 56(2)^3 - 56(2)(-3)

= 56(8) + 56(6)

= 784

2006-10-19 23:36:59 補充：
1 - 8px (28p^2 8q)x^2 (- 56p^3 - 56pq)x^3 .....coefficient of x^3 = - 56p^3 - 56pq= -56(8) 56(6) = - 112

p= 2, q= 4, coefficient of x^3= 896


(1-px+qx^2)^8
= 1 + 8C1( -px+qx^2)+ 8C2 (-px+qx^2)^2+ 8C3( -px+qx^2)^3+...
= 1-8px+8qx^2+ 28p^2 x^2+ -56pqx^3+ 56p^3x^3+...

SO: -8p= -16 p=2
8q+56= 88 q=4
coefficient of x^3: -56pq- 56p^3= 896