Summation:::::::::::::::::::

答案是 -19/2 沒錯!

2012-07-03 22:05:16 補充：
可以代公式 或者 要直接計算出來呢?

2012-07-03 22:50:39 補充：
預備公式:Bernoulli polynomial
(d/dx)[Bn(x)]=n B(n-1)(x),∫(0~1) Bn(x) dx = 0, n=1,2,3,...
B0(x)=1
B1(x)=x -1/2
B2(x)=x² - x + 1/6
B3(x)=x^3 - (3/2)x² + x/2
B4(x)=x^4 - 2x^3 + x² - 1/30
...
B1(1)= 1/2, B1(0)= -1/2
B2(1)= 1/6 = B2(0)
B3(1)= 0 = B3(0)
B4(1)= -1/30 = B4(0)
B6(1)=B6(0)= 1/42
...
考慮:∫(0~1) f(k+x) dx, 其中 f(x)=x^20
=∫(0~1) f(k+x)B0(x) dx (by integration by parts)
= f(x+k)B1(x) |(0~1) - ∫(0~1) f'(x+k)B1(x)/1! dx
=[ f(x+k)B1(x)/1!- f'(x+k)B2(x)/2!]|(0~1) + ∫(0~1) f"(x+k)B2(x)/2! dx
=...= Σ(n=1~21) [ (-d/dx)^(n-1) f(x+k)] Bn(x)/n! ]_(0~1) (因 (d/dx)^21 f(x) =0 )
= [(1+k)^20+k^20]/2 -20[(1+k)^19-k^19]/12 +20x19x18[(1+k)^17-k^17]/(30*4!)+ ...
=[(1+k)^20+k^20]/2 -(5/3) [(1+k)^19-k^19] + (19/2)[(1+k)^17-k^17]+ ...
又∫(0~1) f(k+x) dx=∫(0~1) (x+k)^20 dx = [(k+1)^21 - k^21]/21
故 [(k+1)^21- k^21]/21
=[(1+k)^20+k^20]/2 -(5/3)[(1+k)^19-k^19] +(19/2)[(1+k)^17- k^17] +...
Taking summation for k=0 to n-1 , then
(n^21)/21= -(n^20)/2 +[1^20+2^20+...+n^20 ]-(5/3)(n^19) +(19/2)(n^17)+...(n^15次方以下)
故 1^20+2^20+...+n^20
=n^21 /21 +(n^20)/2 +(5/3) n^19 -(19/2) n^17 + ...(n^15次方以下)

法二: Euler-Maclaurin summation formula
Σ(k=0~n) f(k)
= ∫(0~n) f(x)dx+[f(0)+f(n)]/2 +B2(0)/2![f'(n)-f'(0)]+B4(0)/4![f'''(n)-f'''(0)] + ...
代 f(x)=x^20, 得
Σ(k=0~n) k^20 = 1^20 + 2^20 + ...+ n^20
= ∫(0~n) x^20dx+n^20/2+(1/12)*20n^19 -[1/(30*4!)] [20*19*18n^17]+..(n ^15次方以下)
= (1/21)n^21+(1/2)n^20+(5/3)n^19 - (19/2) n^17+...(n^15次方以下)
故
n^17 係數 = - 19/2

please check the answer


圖片參考：http://imgcld.yimg.com/8/n/HA00108663/o/701206280066113873411880.jpg


2012-07-03 11:27:08 補充：
i think it just consider the case of k = 4