(a)
If m is a positive integer, prove that (1-3x)^m = 1 - 3mx + (9/2)m(m-1)(x^2) + terms involving higher powers of x.
(b)
It is given that the coefficients of x^2 in the expansion of (1-3x)^m [1+2x+(x^2)] is 61. Find the value of m.
F.4 m2 binomial theorem
2009-10-18 7:54 pm
回答 (2)
2009-10-18 8:15 pm
✔ 最佳答案
(a) (1 - 3x)m = mC0 x 1m x (-3x)0 + mC1 x 1m-1 x (-3x)1 + mC2 x 1m-2 x (-3x)2 + ...= 1 - 3mx + mC2 (9x2) + ...
= 1 - 3mx + 9m(m - 1)x2/2 + ...
(b) (1 - 3x)m (1 + 2x + x2) = [1 - 3mx + 9m(m - 1)x/2 + ...] (1 + 2x + x2)
Coefficient of x2 = 1 - 6m + 9m(m - 1)/2
So,
1 - 6m + 9m(m - 1)/2 = 61
2 - 12m + 9m2 - 9m = 122
9m2 - 21m - 120 = 0
3m2 - 7m - 40 = 0
(3m + 8)(m - 5) = 0
m = -8/3 (rejected) or 5
2009-10-18 15:11:39 補充：
有的. 讀微積分. 相當於今日中五附加數
參考: Myself
2009-10-18 8:56 pm
(題外話)f4有開m2架咩
收錄日期: 2021-04-13 16:54:20
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