Binomial Theorem

For Q1, where is n ?

2009-09-28 16:52:11 補充：
(1) Suppose that your question is (1 - 1/x)n (1 + 2x)5, to find the expression for x3 and x4, we have:
x3 has The following combinations:
Taking constant from (1 - 1/x)n and x3 from (1 + 2x)5, giving a coefficient = 80
Taking 1/x from (1 - 1/x)n and x4 from (1 + 2x)5, giving a coefficient = - 80n
Taking 1/x2 from (1 - 1/x)n and x5 from (1 + 2x)5, giving a coefficient = 32 nC2 = 16n(n - 1)
So total = 80 - 80n + 16n(n - 1) = 16n2 - 96n + 80
x4 has The following combinations:
Taking constant from (1 - 1/x)n and x4 from (1 + 2x)5, giving a coefficient = 80
Taking 1/x from (1 - 1/x)n and x5 from (1 + 2x)5, giving a coefficient = - 32n
So total = 80 - 32n
So from the given:
(16n2 - 96n + 80)/(80 - 32n) = 4
(n2 - 6n + 5)/(5 - 2n) = 4
n2 - 6n + 5 = 20 - 8n
n2 + 2n - 15 = 0
n = 3 or -5 (rejected)
(2a) [1 - 1/(5x)]5 (1 + ax)n = [1 - 1/x + 2/(5x2) - 2/(25x3) + 1/(125x4) - 1/(3125x5)] (1 + nax + nC2 a2x2 + ...)
Coefficient of 1/x5 = - 1/3125
Coefficient of 1/x4 = 1/125 - na/3125 = (25 - na)/3125
Coefficient of 1/x3 = -2/25 + na/125 - nC2 a2/3125 = (- 250 + 25na - nC2 a2)/3125
Thus, [1 - 1/(5x)]5 (1 + ax)n = - 1/(3125x5) + (25 - na)/(3125x4) + (- 250 + 25na - nC2 a2)/(3125x3) + ...
(b) (25 - na)/3125 = 1/25
25 - na = 125
na = - 100 ... (1)
(- 250 + 25na - nC2 a2)/3125 = - 52/25
- 250 - 2500 - nC2 a2 = - 6500
nC2 (- 100/n)2 = 3750
5000 n(n - 1)/n2 = 3750
4(n - 1) = 3n
n = 4
a = - 25

(1-x^-1)^n (1+2x)^5 ??