Expand (1+kx-3x^2)^8 in ascending order of x as far as the term in x^3
If the coefficient of the x^2 term in (a) is equal to 88 and k<0 ,find
the value of k and the x^3 term.
數學難題(Expansion)
2007-11-21 7:53 am
回答 (1)
2007-11-21 8:24 am
✔ 最佳答案
a)(1+kx-3x^2)^8
= 1 + 8kx + (28)(k^2)(x^2)+8(-3x^2)+(28+28)(-1)(3kx^3)+(56)(k^3)(x^3)
= 1 + 8kx + (28k^2-24)x^2 + (k^3-168k)x^3
b)
If the coefficient of the x^2 term in (a) is equal to 88 and k<0
28k^2-24 = 88
k=2
the coefficient of the x^3 term is
(2)^3 - 168 * 2 =-328
(Remark: 28 = 8C2 = 8!/(2!6!), 56 = 8C3)
收錄日期: 2021-05-01 22:06:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071120000051KK04515