歸納法 (基本問題)

👨🏻‍🏫 學術台數學

[匿名]

2011-04-06 5:53 am

1的3次方+2的3次方+3的3次方 + ... +k的3次方 = [n(n+1)/2] 的2次方

當n= k +1時,

1的3次方 +2的3次方+3的3次方+...+k的3次方+(k+1)的3次方

= [k(k+1)/2]的2次方 +(k+1)的3次方

=(k+1)的2次方 /4 [k 的2次方 +4(k+1)]

and so on

....
請問為何= [k(k+1)/2]的2次方 +(k+1)的3次方
會變成 =(k+1)的2次方 /4 [k 的2次方 +4(k+1)]呢?
謝謝

更新1:

1^3 +2^3 +3^3 + ...+ k ^3 +( k+1) ^3 =[k(k+1) /2 ]^2 + (k+1 )^3 = (k+1)^2 /4 [ k^2 + 4(k +1)] =(k+1)^2 (k+2)^2 /4 ={ (k +1) [ ( k+1) +1 ] /2 }^2 為何=[k(k+1) /2 ]^2 + (k+1 )^3 會變成= (k+1)^2 /4 [ k^2 + 4(k +1)] 呢? 謝謝

回答 (2)

匿名

2011-04-06 5:58 am

✔ 最佳答案

P(n):1^3 + 2^3 + 3^3 + ... + n^3 = [n(n+1)/2]^2
Assume P(k) is true, i.e. 1^3 + 2^3 + 3^3 + ... + k^3 = [k(k+1)/2]^2
When n = k+1,
L.H.S. = 1^3 + 2^3 + 3^3 + ... + k^3 + (k+1)^3
= [k(k+1)/2]^2 + (k+1)^3
= (k+1)^2 [k^2 /4 + k + 1]
= (k+1)^2 (k/2 + 1)^2
= (k+1)^2 (k+2)^2 / 2^2
= [(k+1)(k+2)/2]^2
= R.H.S.
...

2011-04-05 22:19:12 補充：
[k(k+1)/2]^2 + (k+1)^3 <-------- 2個term 都有(k+1)^2
然後抽(k+1)^2 出來
(k+1)^2 [(k/2)^2 + k + 1]
= (k+1)^2 (k^2/4 + k + 1)
= (k+1)^2 (k/2 + 1)^2
= (k+1)^2 (k+2)^2 /4

參考: Hope the solution can help you^^”

👍 0 👎 0

myisland8132

2011-04-06 6:02 am

[k(k + 1)/2]^2 + (k + 1)^3

= (k + 1)^2 [ k^2/4 + (k + 1)]

= (k + 1)^2/4 [k^2 + 4(k + 1)]

= (k + 1)^2/4 * (k + 2)^2

= [(k + 1)(k + 2)/2]^2

👍 0 👎 0

收錄日期: 2021-04-26 14:56:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110405000051KK01286

檢視 Wayback Machine 備份