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累乗の和 - 八乗和の公式

 

累乗の和 - 八乗和の公式

八乗和の公式の導出です。
 \displaystyle \sum_{k=1}^{n} { k ^ 9 } - \sum_{k=1}^{n} { ( k - 1 )^9 } = 1^9 + 2^9 + \cdots + ( n - 1 )^9 + n^9 - \left\{ 0^9 + 1^9 + \cdots + ( n - 2 )^9 + ( n - 1 )^9 \right\} = n^9 \\ \displaystyle \Leftrightarrow \sum_{k=1}^{n} { k ^ 9 } = n^9 + \sum_{k=1}^{n} { ( k - 1 )^9 } \\ \displaystyle = n^9 + \sum_{k=1}^{n} { \left( k^9 - 9 k^8 + 36 k^7 -84 k^6 + 126 k^5 - 126 k^4 + 84 k^3 - 36 k^2 + 9 k - 1 \right) } \\ \displaystyle = n^9 + \sum_{k=1}^{n} { k^9 } - 9 \sum_{k=1}^{n} { k^8 } + 36 \sum_{k=1}^{n} { k^7 } - 84 \sum_{k=1}^{n} { k^6 } + 126 \sum_{k=1}^{n} { k^5 } - 126 \sum_{k=1}^{n} { k^4 } + 84 \sum_{k=1}^{n} { k^3 } - 36 \sum_{k=1}^{n} { k^2 } \\ \displaystyle + 9 \sum_{k=1}^{n} { k } - n \\ \displaystyle \Leftrightarrow 9 \sum_{k=1}^{n} { k^8 } - n ( n + 1 ) \left( n^7 - n^6 + n^5 - n^4 + n^3 - n^2 + n - 1 \right)
 \displaystyle =  6 \sum_{k=1}^{n} { k^7 } - 84 \sum_{k=1}^{n} { k^6 } + 126 \sum_{k=1}^{n} { k^5 } - 126 \sum_{k=1}^{n} { k^4 } + 84 \sum_{k=1}^{n} { k^3 } - 36 \sum_{k=1}^{n} { k^2 } + 9 \sum_{k=1}^{n} { k } \tag{1}

更に(1)と、七乗和、六乗和、五乗和、四乗和、三乗和、二乗和の公式から、

 \displaystyle 9 \sum_{k=1}^{n} { k^8 } - n ( n + 1 ) \left( n^7 - n^6 + n^5 - n^4 + n^3 - n^2 + n - 1 \right) \\ \displaystyle = 36 \left\{ \frac{1}{24} n^2 ( n + 1 )^2 \left( 3 n^4 + 6 n^3 - n ^ 2 - 4 n + 2 \right) \right\} - 84 \left\{ \frac{1}{42} n ( n + 1 ) ( 2 n + 1 ) \left( 3 n ^ 4 + 6 n ^ 3 - 3 n + 1 \right) \right\} \\ \displaystyle + 126 \left\{ \frac{ 1 }{ 12 } n^2 ( n + 1 )^2 \left( 2 n^2 + 2 n - 1 \right) \right\} - 126 \left\{ \frac{1}{30} n ( n + 1 ) ( 2 n + 1 ) \left( 3 n ^ 2 + 3 n - 1 \right) \right\} \\ \displaystyle + 84 \left\{ \frac{ 1 }{ 4 } n^2 ( n + 1 )^2 \right\} -36 \left\{ \frac{ 1 }{ 6 } n ( n + 1 ) ( 2 n + 1 ) \right\} + 9 \left\{ \frac{ 1 }{ 2 } n ( n + 1 ) \right\} \\ \displaystyle = \frac{ 3 }{ 2 } n^2 ( n + 1 )^2 \left( 3 n^4 + 6 n^3 - n ^ 2 - 4 n + 2 \right) - 2 n ( n + 1 ) ( 2 n + 1 ) \left( 3 n ^ 4 + 6 n ^ 3 - 3 n + 1 \right) \\ \displaystyle + \frac{ 21 }{ 2 } n^2 ( n + 1 )^2 \left( 2 n^2 + 2 n - 1 \right) - \frac{ 21 }{ 5 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n ^ 2 + 3 n - 1 \right) + 21 n^2 ( n + 1 )^2 \\ \displaystyle - 6 n ( n + 1 ) ( 2 n + 1 ) + \frac{ 9 }{ 2 } n ( n + 1 ) \\ \displaystyle \Leftrightarrow 9 \sum_{k=1}^{n} { k^8 } = \frac{ 1 }{ 2 } n^2 ( n + 1 )^2 \left\{ 3 \left( 3 n^4 + 6 n^3 - n ^ 2 - 4 n + 2 \right) + 21 \left( 2 n^2 + 2 n - 1 \right) +42 \right\} \\ \displaystyle - \frac{ 1 }{ 5 } n ( n + 1 ) ( 2 n + 1 ) \left\{ 10 \left( 3 n ^ 4 + 6 n ^ 3 - 3 n + 1 \right) + 21 \left( 3 n ^ 2 + 3 n - 1 \right) + 30 \right\} \\ \displaystyle + n ( n + 1 ) \left( n^7 - n^6 + n^5 - n^4 + n^3 - n^2 + n - 1 \right) + \frac{ 9 }{ 2 } n ( n + 1 ) \\ \displaystyle = \frac{ 1 }{ 2 } n^2 ( n + 1 )^2 \left( 9 n^4 + 18 n^3 -3 n^2 - 12 n + 6 + 42 n^2 + 42 n - 21 + 42 \right)  \\ \displaystyle - \frac{ 1 }{ 5 } n ( n + 1 ) ( 2 n + 1 ) \left( 30 n^4 + 60 n^3 - 30 n + 10 + 63 n^2 + 63 n - 21 + 30 \right)  \\ \displaystyle + \frac{ 1 }{ 2 } n ( n + 1 ) \left\{ 2 \left( n^7 - n^6 + n^5 - n^4 + n^3 - n^2 + n - 1 \right) + 9 \right\} \\ \displaystyle = \frac{ 1 }{ 2 } n^2 ( n + 1 )^2 \left( 9 n^4 + 18 n^3 + 39 n^2 + 30 n + 27 \right)  \\ \displaystyle - \frac{ 1 }{ 5 } n ( n + 1 ) ( 2 n + 1 ) \left( 30 n^4 + 60 n^3 + 63 n^2 + 33 n + 19 \right)  \\ \displaystyle + \frac{ 1 }{ 2 } n ( n + 1 ) \left( 2 n^7 - 2 n^6 + 2 n^5 - 2 n^4 + 2 n^3 - 2 n^2 + 2 n - 2 + 9 \right) \\ \displaystyle \Leftrightarrow 9 \sum_{k=1}^{n} { k^8 } + \frac{ 1 }{ 5 } n ( n + 1 ) ( 2 n + 1 ) \left( 30 n^4 + 60 n^3 + 63 n^2 + 33 n + 19 \right) \\ \displaystyle = \frac{ 1 }{ 2 } n ( n + 1 ) \left\{ n ( n + 1 ) \left( 9 n^4 + 18 n^3 + 39 n^2 + 30 n + 27 \right) \right\} \\ \displaystyle + \frac{ 1 }{ 2 } n ( n + 1 ) \left( 2 n^7 - 2 n^6 + 2 n^5 - 2 n^4 + 2 n^3 - 2 n^2 + 2 n + 7 \right) \\ \displaystyle = \frac{ 1 }{ 2 } n ( n + 1 ) \left( 9 n^6 + 27 n^5 + 57 n^4 + 69 n^3 + 57 n^2 + 27 n \right) \\ \displaystyle + \frac{ 1 }{ 2 } n ( n + 1 ) \left( 2 n^7 - 2 n^6 + 2 n^5 - 2 n^4 + 2 n^3 - 2 n^2 + 2 n + 7 \right) \\ \displaystyle = \frac{ 1 }{ 2 } n ( n + 1 ) \left( 9 n^6 + 27 n^5 + 57 n^4 + 69 n^3 + 57 n^2 + 27 n + 2 n^7 - 2 n^6 + 2 n^5 - 2 n^4 + 2 n^3 - 2 n^2 + 2 n + 7 \right) \\ \displaystyle = \frac{ 1 }{ 2 } n ( n + 1 ) \left( 2 n^7 + 7 n^6 + 29 n^5 + 55 n^4 + 71 n^3 + 55 n^2 + 29 n + 7 \right) \\ \displaystyle = \frac{ 1 }{ 2 } n ( n + 1 ) ( 2 n + 1 ) \left( n^6 + 3 n^5 + 13 n^4 + 21 n^3 + 25 n^2 + 15 n + 7 \right) \\ \displaystyle \Leftrightarrow 9 \sum_{k=1}^{n} { k^8 } = \frac{ 1 }{ 2 } n ( n + 1 ) ( 2 n + 1 ) \left( n^6 + 3 n^5 + 13 n^4 + 21 n^3 + 25 n^2 + 15 n + 7 \right) \\ \displaystyle - \frac{ 1 }{ 5 } n ( n + 1 ) ( 2 n + 1 ) \left( 30 n^4 + 60 n^3 + 63 n^2 + 33 n + 19 \right) \\ \small \displaystyle = \frac{ 1 }{ 10 } n ( n + 1 ) ( 2 n + 1 ) \left\{ 5 \left( n^6 + 3 n^5 + 13 n^4 + 21 n^3 + 25 n^2 + 15 n + 7 \right) - 2 \left( 30 n^4 + 60 n^3 + 63 n^2 + 33 n + 19 \right) \right\} \\ \displaystyle = \frac{ 1 }{ 10 } n ( n + 1 ) ( 2 n + 1 ) \left( 5 n^6 + 15 n^5 + 65 n^4 + 105 n^3 + 125 n^2 + 75 n + 35 - 60 n^4 - 120 n^3 - 126 n^2 - 66 n - 38 \right) \\ \displaystyle = \frac{ 1 }{ 10 } n ( n + 1 ) ( 2 n + 1 ) \left( 5 n^6 + 15 n^5 + 5 n^4 - 15 n^3 - n^2 + 9 n - 3 \right)
 \displaystyle \Leftrightarrow \sum_{k=1}^{n} { k^8 } = \frac{ 1 }{ 90 } n ( n + 1 ) ( 2 n + 1 ) \left( 5 n^6 + 15 n^5 + 5 n^4 - 15 n^3 - n^2 + 9 n - 3 \right) \tag{2}