complex number

2009-02-22 8:31 pm
By expanding (1+i/3^1/2)^2n , find the values of
summation from k = 0 to n , C from 2k to 2n (-1/3)^k and
summation from k=0 to n-1 C from 2k+1 to 2n (-1/3)^k
更新1:

2^2n/3^n cosn兀/3 , 2^2n。3^1/2/3^nsinn兀/3 P.s. 。 = multiply

更新2:

I don't know the expansion of binomial.... seems like u put k = 2k into the 2nd step in 1st term , and k= 2n-k in 2nd term@@ totally confused

回答 (2)

2009-02-22 10:13 pm
✔ 最佳答案
Please kindly go to the following link for your answer.

http://i726.photobucket.com/albums/ww265/physicsworld2010/physicsworld05Feb221412.jpg

2009-02-22 20:32:04 補充:
I just collect the odd terms in one group, then the even terms in the other group. Try it
參考: Physics king
2009-02-22 10:09 pm
睇唔明條題目添.... =_=
你試下用 SUM[k= 4,10] { C( 2n+1, k) } 之類表示。
次方要用 ( 1 + i/3 ) ^ (1/2),多D括弧!


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