Show that 1^3 + 2^3 + + ... + n^3 = [n(n+1)/2]^2 Whenever n is a positive integer.
=[k(k+1)/2]^2 + (k+1)^3
=[k(k+1)/2]^2 + (k+1)(k+1)^2
做到咁就唔識做了
更新1:
原來係咁.......我唔可以問多D架....... or,....我add你...得唔得呀??? 因為真係好似好多唔識
更新2:
你起碼叻過我呀......=.=
真係唔係好識計 A-math
2007-11-10 5:04 am
條問題係咁的
Show that 1^3 + 2^3 + + ... + n^3 = [n(n+1)/2]^2 Whenever n is a positive integer.
=[k(k+1)/2]^2 + (k+1)^3
=[k(k+1)/2]^2 + (k+1)(k+1)^2
做到咁就唔識做了
Show that 1^3 + 2^3 + + ... + n^3 = [n(n+1)/2]^2 Whenever n is a positive integer.
=[k(k+1)/2]^2 + (k+1)^3
=[k(k+1)/2]^2 + (k+1)(k+1)^2
做到咁就唔識做了
回答 (2)
2007-11-10 5:11 am
✔ 最佳答案
=[k(k+1)/2]² + (k+1)³=[k(k+1)/2]² + (k+1)(k+1)²
=k²(k+1)²/4+ (k+1)(k+1)²
=(k+1)²(k²/4+k+1)
=(k+1)²(k²+4k+4)/4
=(k+1)²(k+2)²/4
=[(k+1)(k+2)/2]²
2007-11-09 21:13:18 補充：
=R.H.S.So, P(k+1)is true.By the principle of M.I.,P(n) is true for al lpositive integer n.
2007-11-09 21:59:49 補充：
其實我對d amath係一知半解..因為我未讀過amath(我f.1)我未必幫到你~~
參考: ME
2007-11-10 7:10 am
Let P(n) be the statement 1^3 + 2^3 + .....+n^3=[n(n+1)/2]^2
For P(1)
L.H.S.=1^3
=1
R.H.S = [1(1+1)/2]^2
=1
L.H.S =R.H.S
P(1) is true.
Assume P(k) is true for a positive integer k,
1^3 + 2^3+....+k^3 =[k(k+1)/2]^2
For P(k+1),
1^3+2^3+.....k^3+(k+1)^3 =[k(k+1)/2]^2+(k+1)^3
=[k^2(k+1)^2 + 2^2 (k+1)^3 ]/2^2
=(k+1)^2(k+4k+4) / 2^2
=(k+1)^2(k+2)^2/2*2
=[(k+1)(k+1+1)/2]^2
P(k+1) is also true.
so P(n) is true for all positive integer n
For P(1)
L.H.S.=1^3
=1
R.H.S = [1(1+1)/2]^2
=1
L.H.S =R.H.S
P(1) is true.
Assume P(k) is true for a positive integer k,
1^3 + 2^3+....+k^3 =[k(k+1)/2]^2
For P(k+1),
1^3+2^3+.....k^3+(k+1)^3 =[k(k+1)/2]^2+(k+1)^3
=[k^2(k+1)^2 + 2^2 (k+1)^3 ]/2^2
=(k+1)^2(k+4k+4) / 2^2
=(k+1)^2(k+2)^2/2*2
=[(k+1)(k+1+1)/2]^2
P(k+1) is also true.
so P(n) is true for all positive integer n
收錄日期: 2021-04-24 09:43:38
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071109000051KK03361