1^4+2^4+3^4+...+n^4 等於多少(數學)?

(k+1)^5
= Σ C(5,m)*1^(5-m)*k^m , from m = 0 to m = 5 ( 二項展開式 )
= Σ C(5,m)*k^m
= C(5,0)*k^0 + C(5,1)*k^1 + C(5,2)*k^2 + C(5,3)*k^3 + C(5,4)*k^4 + C(5,5)*k^5
= 1 + 5k + 10k^2 + 10k^3 + 5k^4 + k^5

故得 :
(k+1)^5 - k^5 = 5k^4 + 10k^3 + 10k^2 + 5k + 1

分別取 k = n , n-1 , ..... , 2 , 1 可得:
(n+1)^5 - n^5 = 5n^4 + 10n^3 + 10n^2 + 5n + 1
n^5 - (n-1)^5 = 5(n-1)^4 + 10(n-1)^3 + 10(n-1)^2 + 5(n-1) + 1
...................
3^5 - 2^5 = 5*2^4 + 10*2^3 + 10*2^2 + 5*2 + 1
2^5 - 1^5 = 5*1^4 + 10*1^3 + 10*1^2 + 5*1 + 1
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以上 n 個式子加總可得 :
(n+1)^5 - 1^5 = 5*Σ k^4 + 10*Σ k^3 + 10*Σ k^2 + 5*Σ k + n , 其中各個 Σ 計算範圍皆為 1 ~ n

令 S = Σ k^4 = 1^4 + 2^4 + ..... + n^4
(n+1)^5 = 5S + 10*[ n(n+1)/2 ]^2 + 10*n(n+1)(2n+1)/6 + 5n(n+1)/2 + n + 1
(n+1)^4 = 5S/(n+1) + (5/2)n^2(n+1) + (5/3)n(2n+1) + (5/2)n + 1
6(n+1)^4 = 30S/(n+1) + 15n^2(n+1) + 10n(2n+1) + 15n + 6
6n^4 + 24n^3 + 36n^2 + 24n + 6 = 30S/(n+1) + 15n^3 + 15n^2 + 20n^2 + 10n + 15n + 6
30S/(n+1) = 6n^4 + 9n^3 + n^2 - n

S
= (1/30)(n+1)( 6n^4 + 9n^3 + n^2 - n )
= (1/30) n (n+1)( 6n^3 + 9n^2 + n - 1 )
= (1/30) n (n+1)(n+0.5)( 6n^2 + 6n - 2 )
= (1/30) n (n+1)(2n+1)( 3n^2 + 3n - 1 ) ..... Ans

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以下驗算與證明皆為補充說明.

驗算 :
以 n = 10 為例:
1^4 + 2^4 + ..... + 10^4 = 25333
(1/30) 10 (10+1)(2*10+1)( 3*10^2 + 3*10 - 1 ) = 25333
所以公式基本上沒錯, 但還需要以下的數學證明.

claim : 1^4 + 2^4 + ..... + n^4 = (1/30) n (n+1)(2n+1)( 3n^2 + 3n - 1 )
pf :
(1)
n = 1 時
1^4 = 1
(1/30) 1 (1+1)(2*1+1)( 3*1^2 + 3*1 - 1 ) = (1/30)*2*3*5 = 1
故 n = 1 時此式成立
(2)
設 n = k 時, 此式成立, 即:
1^4 + 2^4 + ..... + k^4 = (1/30) k (k+1)(2k+1)( 3k^2 + 3k - 1 )
當 n = k+1 時
1^4 + 2^4 + ..... + k^4 + (k+1)^4
= (1/30)k(k+1)(2k+1)(3k^2+3k-1) + (k+1)^4
= (1/30)(k+1) [ k(2k+1)(3k^2+3k-1) + 30(k+1)^3 ]
= (1/30)(k+1)( 6k^4 + 9k^3 + k^2 - k + 30k^3 + 90k^2 + 90k + 30 )
= (1/30)(k+1)( 6k^4 + 39k^3 + 91k^2 + 89k + 30 )
= (1/30)(k+1)(k+2)( 6k^3 + 27k^2 + 37k + 15 )
= (1/30)(k+1)(k+2)(k+1.5)( 6k^2 + 18k + 10 )
= (1/30)(k+1)(k+2)(2k+3)( 3k^2 + 9k + 5 )
= (1/30)(k+1) * [ (k+1) + 1 ] * [ 2(k+1) + 1 ] * [ 3(k+1)^2 + 3(k+1) - 1 ]
故 n = k+1 時亦成立.
由數學歸納法, 此式對所有自然數 n 皆成立, 故得證.

(1+2+3+...+n)*4

你只有給我這個式子只能解到這樣喔 因為有未知數，所以無法得到標準答案，只能得到這樣的一個式子，不知道還有沒有其他條件呢@##@