YLL討論網

由 **lskuo** 於星期日三月 18, 2012 9:50 pm

(n+1)^m = n^m + C(m,1)n^m-1 + C(m,2)n^m-2 + ... + C(m,m-1)n + 1
這裡
C(m,k) = m!/k!/(m-k)!

1^m = 1
2^m = 1^m + C(m,1)2^m-1 + C(m,2)2^m-2 + ... + C(m,m-1)2 + 1
...
(n+1)^m = n^m + C(m,1)n^m-1 + C(m,2)n^m-2 + ... + C(m,m-1)n + 1

所以, 將上面 n+1個式子左右分別相加, 消去左右相同的項,
(n+1)^m -1 = C(m,1)S_m-1 + C(m,2)S_m-2 + ... + C(m,m-1)S₁ + n

C(m,1)S_m-1 = (n+1)^m -(n+1) - C(m,2)S_m-2 - ... - C(m,m-1)S₁

例如, 當 m = 4,
C(4,1) = 4, C(4,2)=6, C(4,3)=4

4S₃ = (n+1)⁴ -(n+1) -6S₂ - 4S₁

由 **1234door** 於星期日三月 18, 2012 12:15 am

lskuo 寫到:
1234door 寫到:

其實不難
Hint:
S₁= 1 + 2 + 3 + .. + n = n(n+1)/2
S₂ = 1²+ 2² + .. + n²= n(n+1)(2n+1)/6 = n(n+1)/2 * [(2n+1)/3] = S₁(2n+1)/3
S₃ = 1³+ 2³ + .. + n³= [(n+1)⁴-(n+1)]/4 -3S₂/2 -S₁ = n(n+1)(n²+3n+3)/4-3S₂/2 -S₁ = S₁(n²+3n+3)/2-3S₂/2 -S₁ = S₁的倍數
...
S_m = [(n+1)^m+1-(n+1)] / (m+1) + S₁的倍數

證明 [(n+1)^m+1-(n+1)] / (m+1)是S₁的倍數, 就留給其他人完成吧.

S₃ = 1³+ 2³ + .. + n³= [(n+1)⁴-(n+1)]/4 -3S₂/2 -S₁ = n(n+1)(n²+3n+3)/4-3S₂/2 -S₁ = S₁(n²+3n+3)/2-3S₂/2 -S₁ = S₁的倍數

可以請教一下，上面的式子如何計算出的？

由 **lskuo** 於星期六三月 17, 2012 9:45 pm

1234door 寫到:

其實不難
Hint:
S₁= 1 + 2 + 3 + .. + n = n(n+1)/2
S₂ = 1²+ 2² + .. + n²= n(n+1)(2n+1)/6 = n(n+1)/2 * [(2n+1)/3] = S₁(2n+1)/3
S₃ = 1³+ 2³ + .. + n³= [(n+1)⁴-(n+1)]/4 -3S₂/2 -S₁ = n(n+1)(n²+3n+3)/4-3S₂/2 -S₁ = S₁(n²+3n+3)/2-3S₂/2 -S₁ = S₁的倍數
...
S_m = [(n+1)^m+1-(n+1)] / (m+1) + S₁的倍數

證明 [(n+1)^m+1-(n+1)] / (m+1)是S₁的倍數, 就留給其他人完成吧.

由 **1234door** 於星期六三月 17, 2012 1:07 pm

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[quote="lskuo"][quote="1234door"][img]http://www.yll.url.tw/richedit/upload/2k799ec71cd2.jpg[/img][/quote] 其實不難 Hint: S[SUB]1 [/SUB]= 1 + 2 + 3 + .. + n = n(n+1)/2 S[SUB]2[/SUB] = 1[SUP]2[/SUP]+ 2[SUP]2[/SUP] + .. + n[SUP]2 [/SUP]= n(n+1)(2n+1)/6 = n(n+1)/2 * [(2n+1)/3] = S[SUB]1[/SUB](2n+1)/3 S[SUB]3[/SUB] = 1[SUP]3[/SUP]+ 2[SUP]3[/SUP] + .. + n[SUP]3 [/SUP]= [(n+1)[SUP]4[/SUP]-(n+1)]/4 -3S[SUB]2[/SUB]/2 -S[SUB]1[/SUB] = n(n+1)(n[SUP]2[/SUP]+3n+3)/4-3S[SUB]2[/SUB]/2 -S[SUB]1[/SUB] = S[SUB]1[/SUB](n[SUP]2[/SUP]+3n+3)/2-3S[SUB]2[/SUB]/2 -S[SUB]1[/SUB] = S[SUB]1[/SUB]的倍數 ... S[SUB]m[/SUB] = [(n+1)[SUP]m+1[/SUP]-(n+1)] / (m+1) + S[SUB]1[/SUB]的倍數證明 [(n+1)[SUP]m+1[/SUP]-(n+1)] / (m+1)是S[SUB]1[/SUB]的倍數, 就留給其他人完成吧.[/quote]