Σ k=1到n k^p=?
p=1時,原式=(1/2)n+(1/2)n^2
p=2時,原式=(1/6)n+(1/2)n^2+(1/3)n^3
p=3時,原式=(0/1)n+(1/4)n^2+(1/2)n^3+(1/4)n^4
p=4時,原式=(-1/30)n+(0/1)n^2+(1/3)n^3+(1/2)n^4+(1/5)n^5
p=5時,原式=(0/1)n+(-1/12)n^2+(0/1)n^3+(5/12)n^4+(1/2)n^5+(1/6)n^6
p=6時,原式=(1/42)n+(0/1)n^2+(-1/6)n^3+(0/1)n^4+(1/2)n^5+(1/2)n^6+(1/7)n^7
............
找不到Σ k=1到n k^p=?的規律==||
p=1時,原式=(1/2)n+(1/2)n^2
p=2時,原式=(1/6)n+(1/2)n^2+(1/3)n^3
p=3時,原式=(0/1)n+(1/4)n^2+(1/2)n^3+(1/4)n^4
p=4時,原式=(-1/30)n+(0/1)n^2+(1/3)n^3+(1/2)n^4+(1/5)n^5
p=5時,原式=(0/1)n+(-1/12)n^2+(0/1)n^3+(5/12)n^4+(1/2)n^5+(1/6)n^6
p=6時,原式=(1/42)n+(0/1)n^2+(-1/6)n^3+(0/1)n^4+(1/2)n^5+(1/2)n^6+(1/7)n^7
............
找不到Σ k=1到n k^p=?的規律==||
回應
cuteconnie0703
Re: 尋找規律
2006-01-31 01:34訪客 (未驗證)
Re: 尋找規律
2006-01-31 11:31n
Σ k = (n)(n+1) / 2
k=1
n
Σ k(k+1) = (n)(n+1)(n+2) / 3
k=1
n
Σ k(k+1)(k+2) = (n)(n+1)(n+2)(n+3) / 4
k=1
......
so
n......p.........p+1
Σ ∏ (k+i) = (∏ (k+i)) / (p+2)
k=1..i=0.......i=0
.............n
Let ap = Σ k^p (p belong to N)
.............k=1
so
p+1....................n
∏ (k+i)) / (p+2) =Σ(k^p+1 + (c1)k^p + (c2)k^p-1+...)
i=0......................k=0
=a(p+1)+c1*ap:c2*a(p-1)+...
you can infer ap (very troublesome) by dynamic programming (bottom to up)
Gauss
Re: 尋找規律
2006-02-02 23:06