更新1:
To: fai109 Can you explain your answer why you have to put a=4?
Binomial
2010-04-18 8:00 pm
Given that in the expansion of [x^2 - 2/x]^n in descending powers of x, where n is a positive integer, the fifth term is the constant term. Find the constant term.
回答 (3)
2010-04-18 8:41 pm
✔ 最佳答案
(a+1)th term=[nCa](x²)ⁿ⁻ª(−2/x)ª
=[nCa](−2)ª(x²ⁿ⁻³ª)
For 5th term: a=4
Set 2n−3a=0,
2n=3(4)
∴n=6
The constant term
=[6C4](64)
=960
2010-04-18 18:39:33 補充：
如果展開 (x+1)ⁿ,
(x+1)ⁿ=[nC0]xⁿ+[nC1]xⁿ⁻¹+[nC2]xⁿ⁻²+[nC3]xⁿ⁻³+..........
睇下個coefficient!
(1st term)⇒nC0 ; (2nd term)⇒nC1 ; (3rd term)⇒nC2 ; (4th term)⇒nC3 ; .............
∴k th term ⇒ n C (k−1)
參考: , 希望你明我嘅講法
2010-04-18 8:19 pm
=nCr(x^2)^r(-2/x)^(n-r)
=nCr(x)^(3r-n)(-2)^(n-r)
You mean general term? -->
Why is not "nCr(x^2)^(n-r) (-2/x)^r"
But "nCr(x^2)^r(-2/x)^(n-r)"
=nCr(x)^(3r-n)(-2)^(n-r)
You mean general term? -->
Why is not "nCr(x^2)^(n-r) (-2/x)^r"
But "nCr(x^2)^r(-2/x)^(n-r)"
2010-04-18 8:14 pm
[x^2 - 2/x]^n
=nCr(x^2)^r(-2/x)^(n-r)
=nCr(x)^(3r-n)(-2)^(n-r)
fifth term is the constant term
So r=5
3r-n=0
n=15
constant term=nCr(-2)^(n-r)
sub n=15 r=5
constant term=15C5(-2)^10
=3075702
=nCr(x^2)^r(-2/x)^(n-r)
=nCr(x)^(3r-n)(-2)^(n-r)
fifth term is the constant term
So r=5
3r-n=0
n=15
constant term=nCr(-2)^(n-r)
sub n=15 r=5
constant term=15C5(-2)^10
=3075702
收錄日期: 2021-04-23 18:26:08
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100418000051KK00576