求(1+x)^10(1-2over x)^3展開式中的常數項.
thz!!要步驟,,
f.4 a.maths
2007-12-02 12:59 am
回答 (2)
2007-12-02 1:12 am
✔ 最佳答案
爆開佢先(1+x)^10(1- 2/x)^3
= (1+10x+45x^2+120x^3+210x^4+252x^5+210x^6+120x^7+45x^8+10x^9+x^10)
(1 -6/x +12/x^2 -8/x^3)
=1 -60 + 540-960+...(將以上同色的數字相乘）
=-479
2007-12-02 1:17 am
(1+10x+45x^2+120x^3+...)(1-2/x+4/x^2-8/x^3)
the constant term:
1X1+10X(-2)+45X4-120X(-8)
=-799
2007-12-01 17:21:28 補充：
(1+10x+45x^2+120x^3+...)(1-6/x+12/x^2-8/x^3)10(-6)+45(12)+120(-8)=480
the constant term:
1X1+10X(-2)+45X4-120X(-8)
=-799
2007-12-01 17:21:28 補充：
(1+10x+45x^2+120x^3+...)(1-6/x+12/x^2-8/x^3)10(-6)+45(12)+120(-8)=480
收錄日期: 2021-04-23 22:28:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071201000051KK02636