とりわけネタもないので7乗和、８乗和まで計算してみたい。

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６乗和を求めてみよう

Σ (n+1)^7-Σn^7=Σ1+7Σn+21Σn^2+35Σn^3+35Σn^4+21Σn^5+7Σn^6

(n+1)^7-1=n+7n(n+1)/2+21n(n+1)(2n+1)/6+35n^2(n+1)^2/4+35n(n+1){6n^3+9n^2+n-1}/30+21n^2(n+1)^2(2n^2+2n-1)/12+7Σn^6

(n+1){(n+1)^6-1}=7n(n+1)/2+7n(n+1)(2n+1)/2+35n^2(n+1)^2/4+7n(n+1){6n^3+9n^2+n-1}/6+7n^2(n+1)^2(2n^2+2n-1)/4+7Σn^6

12(n+1){(n+1)^6-1}=42n(n+1)+42n(n+1)(2n+1)+105n^2(n+1)^2+14n(n+1){6n^3+9n^2+n-1}+21n^2(n+1)^2(2n^2+2n-1)+84Σn^6

12n(n+1){n^5+6n^4+15n^3+20n^2+15n+6}=42n(n+1)+42n(n+1)(2n+1)+105n^2(n+1)^2+14n(n+1){6n^3+9n^2+n-1}+21n^2(n+1)^2(2n^2+2n-1)+84Σn^6

12{n^5+6n^4+15n^3+20n^2+15n+6}=42+42(2n+1)+105n(n+1)+14{6n^3+9n^2+n-1}+21n(n+1)(2n^2+2n-1)+84Σn^6/n(n+1)

12{n^5+6n^4+15n^3+20n^2+15n+6}=42+42(2n+1)+105n(n+1)+14{6n^3+9n^2+n-1}+21{2n^4+4n^3+n^2-n}+84Σn^6/n(n+1)

12n^5+30n^4+12n^3-12n^2-2n+2=84Σn^6/n(n+1)

(6n^5+15n^4+6n^3-6n^2-n+1)=42Σn^6/n(n+1)

一致した。さほど面倒ではないようだ。

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7乗和を求めてみよう

Σ (n+1)^8-Σn^8=Σ1+8Σn+28Σn^2+56Σn^3+70Σn^4+56Σn^5+28Σn^6+8Σn^7

(n+1)^8-1=n+4n(n+1)+14n(n+1)(2n+1)/3+14n^2(n+1)^2+7n(n+1){6n^3+9n^2+n-1}/3+14n^2(n+1)^2(2n^2+2n-1)/3+2n(n+1)(6n^5+15n^4+6n^3-6n^2-n+1)/3+8Σn^7

(n+1){(n+1)^7-1}=4n(n+1)+14n(n+1)(2n+1)/3+14n^2(n+1)^2+7n(n+1){6n^3+9n^2+n-1}/3+14n^2(n+1)^2(2n^2+2n-1)/3+2n(n+1)(6n^5+15n^4+6n^3-6n^2-n+1)/3+8Σn^7

3(n+1){(n+1)^7-1}=12n(n+1)+14n(n+1)(2n+1)+42n^2(n+1)^2+7n(n+1){6n^3+9n^2+n-1}+14n^2(n+1)^2(2n^2+2n-1)+2n(n+1)(6n^5+15n^4+6n^3-6n^2-n+1)+24Σn^7

3n(n+1){n^6+7n^5+21n^4+35n^3+35n^2+21n+7}=12n(n+1)+14n(n+1)(2n+1)+42n^2(n+1)^2+7n(n+1){6n^3+9n^2+n-1}+14n^2(n+1)^2(2n^2+2n-1)+2n(n+1)(6n^5+15n^4+6n^3-6n^2-n+1)+24Σn^7

3{n^6+7n^5+21n^4+35n^3+35n^2+21n+7}=12+14(2n+1)+42n(n+1)+7{6n^3+9n^2+n-1}+14{2n^4+4n^3+n^2-n}+2(6n^5+15n^4+6n^3-6n^2-n+1)+24Σn^7/n(n+1)

3n^6+9n^5+5n^4-5n^3-2n^2+2n=24Σn^7/n(n+1)

n(3n^5+9n^4+5n^3-5n^2-2n+2)=24Σn^7/n(n+1)

n(n+1)(3n^4+6n^3-n^2-4n+2)=24Σn^7/n(n+1)

一致した。さほど面倒ではないようだ。

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