(x+1)^n-x^n=Σ(n,k)x^k

(n+1)^5-n^5=(5,0)+(5,1)n+(5,2)n^2+(5,3)n^3+(5,4)n^4=1+5n+10n^2+10n^3+5n^4

Σ (n+1)^5-Σn^5=Σ1+5Σn+10Σn^2+10Σn^3+5Σn^4

(n+1)^5-1=n+5n(n+1)/2+10n(n+1)(2n+1)/6+10n^2(n+1)^2/4+5Σn^4

5Σn^4=(n+1)^5-(n+1)-5n(n+1)/2-10n(n+1)(2n+1)/6-10n^2(n+1)^2/4

5Σn^4=(n+1){(n+1)^4-1-5n/2-5n(2n+1)/3-5n^2(n+1)/2}

5Σn^4=(n+1){n^4+4n^3+6n^2+4n-5n/2-5n(2n+1)/3-5n^2(n+1)/2}

5Σn^4=n(n+1){n^3+4n^2+6n+4-5/2-5(2n+1)/3-5n(n+1)/2}

5Σn^4=n(n+1){n^3+4n^2+6n+4-5/2-10n/3-5/3-5n^2/2-5n/2}

5Σn^4=n(n+1){n^3+3n^2/2+n/6-1/6}

5Σn^4=n(n+1)/6{6n^3+9n^2+n-1}

Σn^4を求めるのは面倒である。

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

5Σn^4=n^2(n+1)^3-2Σn^3-Σn^2までわかっているとしたらどうだろうか？

30Σn^4=6n^2(n+1)^3-12Σn^3-6Σn^2

30Σn^4=6n^2(n+1)^3-3n^2(n+1)^2-n(n+1)(2n+1)

30Σn^4=n(n+1){6n(n+1)^2-3n(n+1)-(2n+1)}

30Σn^4=n(n+1){6n^3+12n^2+6n-3n^2-3n-2n-1}

30Σn^4=n(n+1){6n^3+9n^2+n-1}

大差なしといったところである

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

５乗和を求めてみよう

Σ (n+1)^6-Σn^6=Σ1+6Σn+15Σn^2+20Σn^3+15Σn^4+6Σn^5

(n+1)^6-1=n+3n(n+1)+5n(n+1)(2n+1)/2+5n^2(n+1)^2+n(n+1){6n^3+9n^2+n-1}/2+6Σn^5

(n+1){(n+1)^5-1}=3n(n+1)+5n(n+1)(2n+1)/2+5n^2(n+1)^2+n(n+1){6n^3+9n^2+n-1}/2+6Σn^5

2(n+1){(n+1)^5-1}=6n(n+1)+5n(n+1)(2n+1)+10n^2(n+1)^2+n(n+1){6n^3+9n^2+n-1}+12Σn^5

2n(n+1){n^4+5n^3+10n^2+10n+5}=6n(n+1)+5n(n+1)(2n+1)+10n^2(n+1)^2+n(n+1){6n^3+9n^2+n-1}+12Σn^5

2{n^4+5n^3+10n^2+10n+5}=6+5(2n+1)+10n(n+1)+{6n^3+9n^2+n-1}+12Σn^5/n(n+1)

2n^4+4n^3+n^2-n=12Σn^5/n(n+1)

n(2n^3+4n^2+n-1)=12Σn^5/n(n+1)

n(n+1)(2n^2+2n-1)=12Σn^5/n(n+1)

Σｋ^5＝ｎ^2（ｎ＋１）^2（２ｎ^2＋２ｎ−１）／１２と一致した。さほど面倒ではないようだ。

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

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。

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝