When (x+y+z)^1999 is expanded and like terms are collected, how many terms will there be??? 就是將(x+y+z)^1999 化開後，然後再將有相同xyz的簡化。例如 (x+y)^2 的答案是3，因為 x^2 + 2xy + y^2 -------> 3 個 terms
這題目是在 partitions & bijections 裏面的，應該會用到 combinatorics。
奧數題目，高手來試試看
2012-05-15 4:36 am
回答 (2)
2012-05-15 5:18 am
✔ 最佳答案
您好！我是 lop，高興能解答您的問題。Let the terms be (x^a)(y^b)[z^(1999-a-b)]. Note that a,b are possitive integers
How many terms there are ?
When a = 0,
b can be 0 ~ 1999, total 2000 different terms.
When a = 1,
b can be 0 ~ 1998, total 1999 different terms.
...
When a = 1999,
b can only be 0, total 1 term.
So the answer will be:
2000 + 1999 + 1998 + ... + 2 + 1
= (2000+1) × 2000 ÷ 2
= 2001000
Answer : There will be 2001000 terms.
2012-05-14 21:20:25 補充：
Corrections:
Note that a,b are integers not negative.
2012-05-14 21:26:24 補充：
Note that a,b are integers that not negative.
參考: My knowledge
2012-05-15 6:59 pm
根據 lop知識長的計法, 但a, b 應該由1開始計算, 不會是0啊.
收錄日期: 2021-04-20 12:44:54
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120514000051KK00659