推 thisday :這神探柯南^^ 04/03 15:25
推 amy29585028 :感謝推 04/03 16:27
※ 引述《amy29585028 (Amy是男是女都不重要)》之銘言：
: 試證明對所有的正整數n：
: 1^2013+2^2013+...+n^2013
: 皆可以被1+2+...+n整除
: 小弟我推不出結論，不知道是不是有什麼技巧？懇請高手指點，謝謝。
In this proposition, 2013 can be replaced by any positive odd number.
We show the case n=2k,
A = 1^t + 2^t + ... + (2k)^t, where t is odd
S = 1+2+...+2k = k(2k+1)
(i) k and 2k+1 are coprime
(ii) A = 0 mod k
proof. A = 1^t + 2^t + ... + k^t + 1^t + 2^t + ... + k^t mod k
= 2(1^t + 2^t +...+ k^t) mod k
However, by Mathematical Induction
1^t + 2^t + ... + k^t = k(k+1)/2 * Q
Hence, A = k(k+1) = 0 mod k
(iii) A = 0 mod 2k+1
proof. A = 1^t + 2^t + ... + k^t + (-k)^t + ... + (-1)^t
= 0 mod 2k+1
done.
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The case n=2k+1 is similar, and left to you.
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※ 編輯: JohnMash 來自: 123.194.96.239 (04/03 13:18)