（３ｘ＋１／ｘ）^6的常數項
我有步不明,,有冇人可以指點下我
二次式定理求常數項!!
2008-01-13 6:34 am
回答 (4)
2008-01-13 7:03 am
✔ 最佳答案
（３ｘ＋１／ｘ）^6General term
=6Cr [(3x)^(6-r)] x^(-r)
=6Cr 3^(6-r) x^(6-r) x^(-r)
=6Cr 3^(6-r) x^(6-2r)
For a 常數項 (constant term), power of x is 0.
6-2r=0
r=3
Constant term=6C3 3^(6-3)
=20 (27)
=540
參考: Me
2008-01-15 2:02 am
ans:540
2008-01-13 6:54 am
(r+1)th term= 6Cr (3x)^6-r x (1/x)^r
=6Cr (3^6-r )(x^6-r-r)
=6Cr x^6-2r
constant term=6-2r =0
r =3
constant term = 6C3 (3)^6-3=540
=6Cr (3^6-r )(x^6-r-r)
=6Cr x^6-2r
constant term=6-2r =0
r =3
constant term = 6C3 (3)^6-3=540
參考: me
2008-01-13 6:42 am
(3x + 1/x)^6
=(3x)^6 + 6C1 (3x)^5 (1/x) + 6C2 (3x)^4 (1/x)^2 + 6C3 (3x)^3 (1/x)^3 + ... ...
∴常數項 = 6C3 (3)^3 (1)^3 = 540
=(3x)^6 + 6C1 (3x)^5 (1/x) + 6C2 (3x)^4 (1/x)^2 + 6C3 (3x)^3 (1/x)^3 + ... ...
∴常數項 = 6C3 (3)^3 (1)^3 = 540
收錄日期: 2021-04-23 21:53:20
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080112000051KK04662