■ベキ和と未定係数法(その27)

S1=Σk=n(n+1)/2

S2=Σk^2=n(n+1)(2n+1)/6

S3=Σk^3=n^2(n+1)^2/4

が多くの読者にとってお馴染みの公式であろう.さらに,

S4=Σk^4=n(n+1)(2n+1)(3n^2+3n−1)/30

S5=Σk^5=n^2(n+1)^2(2n^2+2n−1)/12

S6=Σk^6=n(n+1)(2n+1)(3n^4+6n^3−3n+1)/42

S7=Σk^7=n^2(n+1)^2(3n^4+6n^3−n^2−4n+2)/24

S8=Σk^8=n(n+1)(2n+1)(5n^6+15n^5+5n^4−15n^3−n^2+9n−3)/90

と続く.

 Sk〜n^(k+1)/(k+1)

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この近似式は

nΣ(1,n)i^(k-1)-Σ(1,n)i^k=Σ(p=1,n-1){Σ(1,p)i^(k-1)}

(n+1)Σ(1,n)i^k=Σ(1,n)i^(k+1)+Σ(p=1,n){Σ(1,p)i^(k)}

に基づいているが、4乗ベキに対しても

Σi^4=(n/5+1/5)n(n+1/2){(n+1)n-1/3}

へと翻訳できるものである

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S3=Σk^3=n^2(n+1)^2/4

1^3+2^3+3^3+4^3+・・・=1/4・n^4+a・n^3+b・n^2+c・n

1=1/4+a+b+c

9=4+8a+4b+2c

36=81/4+27a+9b+3

a=1/2,b=1/4,c=0

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