maths

2008-09-06 2:58 am
Prove,by inducation,that for all positive integers n.

1^3 + 2^3 + 3^3 + ...+ (2n)^3 = n^2 (2n+1)^2

回答 (3)

2008-09-07 6:55 pm
✔ 最佳答案
when n=1 ,
LHS=1^3+(2*1)^3=9=(1)^2(2*1+1)^2=RHS
So it is true for n=1 .

Assume it is true for a positive integer k ,
i.e. 1^3+2^3+...(2k)^3=(k)^2(2k+1)^2

Consider , when n= k+1 ,

1^3+2^3+...+2k^3+(2k+1)^3+(2k+2)^3
=k^2 (2k+1)^2+(2k+1)^3+(2k+2)^3
=4k^4+20k^3+37k^2+30k+9
=(k+1)^2+(2k+3)^2
=RHS

so it is true for n=k+1 ,
By Mathematical induction , it is true for all positive integers n.



2008-09-07 10:56:20 補充:
=(k+1)^2 (2k+3)^2
=RHS
2008-09-06 9:37 pm
1^3 + 2^3 + 3^3 + ...+ (n)^3 = {[n(n+1)]/2}^2 ?!
2008-09-06 5:01 am
http://hk.knowledge.yahoo.com/question/question?qid=7008050501024
Think twice before you answer this question!

2008-09-06 18:54:47 補充:
題目絕對冇錯, 只係當中有一個陷阱, 發問者仲未領略到......
不過以發問者咁「了得」嘅數學基礎, 解決此題只係時間嘅問題!


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