when n is odd,
a^n+b^n=(a+b)(a^(n-1)-a^(n-2) b+a^(n-3) b^2-…-ab^(n-2)+b^(n-1))
when n is even,
a^n-b^n=(a+b)(a^(n-1)-a^(n-2) b+a^(n-3) b^2-…+ab^(n-2)-b^(n-1))
更新1:
sorry我應該問呢兩條係咪identity= =
prove identity
2014-05-07 2:16 am
請問下面兩條identity啱唔啱??
when n is odd,
a^n+b^n=(a+b)(a^(n-1)-a^(n-2) b+a^(n-3) b^2-…-ab^(n-2)+b^(n-1))
when n is even,
a^n-b^n=(a+b)(a^(n-1)-a^(n-2) b+a^(n-3) b^2-…+ab^(n-2)-b^(n-1))
when n is odd,
a^n+b^n=(a+b)(a^(n-1)-a^(n-2) b+a^(n-3) b^2-…-ab^(n-2)+b^(n-1))
when n is even,
a^n-b^n=(a+b)(a^(n-1)-a^(n-2) b+a^(n-3) b^2-…+ab^(n-2)-b^(n-1))
回答 (3)
2014-05-07 4:01 pm
✔ 最佳答案
you may expand it from RHS(a + b)(a^(n-1) - a^(n-2) b + a^(n-3) b^2 - … - a b^(n-2) + b^(n-1))
= a(a^(n-1) - a^(n-2) b + a^(n-3) b^2 - … - a b^(n-2) + b^(n-1))
+ b(a^(n-1) - a^(n-2) b + a^(n-3) b^2 - … - a b^(n-2) + b^(n-1))
= a^n - a^(n-1) b + a^(n-2) b^2 - … - a^2 b^(n-2) + ab^(n-1)
+ a^(n-1) b - a^(n-2) b^2 + a^(n-3) b^3 - … - a b^(n-1) + b^n
= a^n + b^n
參考: knowledge
2014-05-07 3:09 am
thxthxthx.
2014-05-07 2:36 am
其實 for odd n,
aⁿ + bⁿ 抽出 (a + b) 之後，餘下的會一加一減～
至於 for even n,
aⁿ - bⁿ 其實可抽 (a + b) 也可抽 (a - b)
抽 (a + b) 餘下的會一加一減～
抽 (a - b) 餘下的全部加～
2014-05-06 19:27:50 補充：
請看三個例子：
http://www.wolframalpha.com/input/?i=(a%5E7%2Bb%5E7)%2F(a%2Bb)&t=crmtb01
http://www.wolframalpha.com/input/?i=%28a%5E8-b%5E8%29%2F%28a%2Bb%29
http://www.wolframalpha.com/input/?i=%28a%5E8-b%5E8%29%2F%28a-b%29
aⁿ + bⁿ 抽出 (a + b) 之後，餘下的會一加一減～
至於 for even n,
aⁿ - bⁿ 其實可抽 (a + b) 也可抽 (a - b)
抽 (a + b) 餘下的會一加一減～
抽 (a - b) 餘下的全部加～
2014-05-06 19:27:50 補充：
請看三個例子：
http://www.wolframalpha.com/input/?i=(a%5E7%2Bb%5E7)%2F(a%2Bb)&t=crmtb01
http://www.wolframalpha.com/input/?i=%28a%5E8-b%5E8%29%2F%28a%2Bb%29
http://www.wolframalpha.com/input/?i=%28a%5E8-b%5E8%29%2F%28a-b%29
收錄日期: 2021-04-21 00:10:05
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140506000051KK00142