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累乗の和 - 六乗和の公式

イメージ 1

六乗和の公式の導出です。
$\displaystyle \sum_{k=1}^{n} { k ^ 7 } - \sum_{k=1}^{n} { ( k - 1 )^7 } = 1^7 + 2^7 + \cdots + ( n - 1 )^7 + n^7 - \left\{ 0^7 + 1^7 + \cdots + ( n - 2 )^7 + ( n - 1 )^7 \right\} = n^7 \\ \displaystyle \Leftrightarrow \sum_{k=1}^{n} { k ^ 7 } = n^7 + \sum_{k=1}^{n} { ( k - 1 )^7 } = n^7 + \sum_{k=1}^{n} { \left( k^7 - 7 k^6 + 21 k^5 - 35 k^4 + 35 k^3 - 21 k^2 + 7 k - 1 \right) } \\ \displaystyle = n^7 + \sum_{k=1}^{n} { k ^ 7 } - 7 \sum_{k=1}^{n} { k ^ 6 } + 21 \sum_{k=1}^{n} { k ^ 5 } - 35 \sum_{k=1}^{n} { k ^ 4 } + 35 \sum_{k=1}^{n} { k ^ 3 } - 21 \sum_{k=1}^{n} { k ^ 2 } + 7 \sum_{k=1}^{n} { k } - n \\ \displaystyle \Leftrightarrow 7 \sum_{k=1}^{n} { k ^ 6 } = n^7 - n + 21 \sum_{k=1}^{n} { k ^ 5 } - 35 \sum_{k=1}^{n} { k ^ 4 } + 35 \sum_{k=1}^{n} { k ^ 3 } - 21 \sum_{k=1}^{n} { k ^ 2 } + 7 \sum_{k=1}^{n} { k }$
$\displaystyle = n ( n + 1 ) \left( n^5 - n^4 + n^3 - n^2 + n - 1 \right) + 21 \sum_{k=1}^{n} { k ^ 5 } - 35 \sum_{k=1}^{n} { k ^ 4 } + 35 \sum_{k=1}^{n} { k ^ 3 } - 21 \sum_{k=1}^{n} { k ^ 2 } + 7 \sum_{k=1}^{n} { k } \tag{12}$

更に(12)と、五乗和、四乗和、三乗和、二乗和の公式から、

$\displaystyle 7 \sum_{k=1}^{n} { k ^ 6 } - n ( n + 1 ) \left( n^5 - n^4 + n^3 - n^2 + n - 1 \right) = 21 \sum_{k=1}^{n} { k ^ 5 } - 35 \sum_{k=1}^{n} { k ^ 4 } + 35 \sum_{k=1}^{n} { k ^ 3 } - 21 \sum_{k=1}^{n} { k ^ 2 } + 7 \sum_{k=1}^{n} { k } \\ \displaystyle \Leftrightarrow \sum_{k=1}^{n} { k ^ 6 } - \frac{ 1 }{ 7 } n ( n + 1 ) \left( n^5 - n^4 + n^3 - n^2 + n - 1 \right) = 3 \sum_{k=1}^{n} { k ^ 5 } - 5 \sum_{k=1}^{n} { k ^ 4 } + 5 \sum_{k=1}^{n} { k ^ 3 } - 3 \sum_{k=1}^{n} { k ^ 2 } + \sum_{k=1}^{n} { k } \\ \displaystyle = 3 \left\{ \frac{ 1 }{ 12 } n^2 ( n + 1 )^2 \left( 2 n^2 + 2 n - 1 \right) \right\} - 5 \left\{ \frac{ 1 }{ 30 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n^2 + 3 n - 1 \right) \right\} + 5 \left\{ \frac{ 1 }{ 4 } n^2 ( n + 1 )^2 \right\} \\ \displaystyle - 3 \left\{ \frac{ 1 }{ 6 } n ( n + 1 ) ( 2 n + 1 ) \right\} + \frac{ 1 }{ 2 } n ( n + 1 ) \\ \displaystyle = \frac{ 1 }{ 4 } n^2 ( n + 1 )^2 \left( 2 n^2 + 2 n - 1 \right) - \frac{ 1 }{ 6 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n^2 + 3 n - 1 \right) + \frac{ 5 }{ 4 } n^2 ( n + 1 )^2 - \frac{ 1 }{ 2 } n ( n + 1 ) ( 2 n + 1 ) \\ \displaystyle + \frac{ 1 }{ 2 } n ( n + 1 ) \\ \displaystyle = \frac{ 1 }{ 4 } n^2 ( n + 1 )^2 \left( 2 n^2 + 2 n - 1 + 5 \right) - \frac{ 1 }{ 6 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n^2 + 3 n - 1 + 3 \right) + \frac{ 1 }{ 2 } n ( n + 1 ) \\ \displaystyle = \frac{ 1 }{ 4 } n^2 ( n + 1 )^2 \left( 2 n^2 + 2 n + 4 \right) - \frac{ 1 }{ 6 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n^2 + 3 n + 2 \right) + \frac{ 1 }{ 2 } n ( n + 1 ) \\ \displaystyle = \frac{ 1 }{ 2 } n^2 ( n + 1 )^2 \left( n^2 + n + 2 \right) - \frac{ 1 }{ 6 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n^2 + 3 n + 2 \right) + \frac{ 1 }{ 2 } n ( n + 1 )$

$\displaystyle \Leftrightarrow \sum_{k=1}^{n} { k ^ 6 } - \frac{ 1 }{ 2 } n^2 ( n + 1 )^2 \left( n^2 + n + 2 \right) + \frac{ 1 }{ 6 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n^2 + 3 n + 2 \right) \\ \displaystyle = \frac{ 1 }{ 7 } n ( n + 1 ) \left( n^5 - n^4 + n^3 - n^2 + n - 1 \right) + \frac{ 1 }{ 2 } n ( n + 1 ) \\ \displaystyle = \frac{ 1 }{ 14 } n ( n + 1 ) \left\{ 2 \left( n^5 - n^4 + n^3 - n^2 + n - 1 \right) + 7 \right\} \\ \displaystyle = \frac{ 1 }{ 14 } n ( n + 1 ) \left( 2 n^5 - 2 n^4 + 2 n^3 - 2 n^2 + 2 n - 2 + 7 \right) \\ \displaystyle = \frac{ 1 }{ 14 } n ( n + 1 ) \left( 2 n^5 - 2 n^4 + 2 n^3 - 2 n^2 + 2 n + 5 \right)$

$\displaystyle \Leftrightarrow \sum_{k=1}^{n} { k ^ 6 } + \frac{ 1 }{ 6 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n^2 + 3 n + 2 \right) \\ \displaystyle = \frac{ 1 }{ 2 } n^2 ( n + 1 )^2 \left( n^2 + n + 2 \right) + \frac{ 1 }{ 14 } n ( n + 1 ) \left( 2 n^5 - 2 n^4 + 2 n^3 - 2 n^2 + 2 n + 5 \right) \\ \displaystyle = \frac{ 1 }{ 14 } n ( n + 1 ) \left\{ 7 n ( n + 1 ) \left( n^2 + n + 2 \right) + 2 n^5 - 2 n^4 + 2 n^3 - 2 n^2 + 2 n + 5 \right\} \\ \displaystyle = \frac{ 1 }{ 14 } n ( n + 1 ) \left( 7 n^4 + 14 n^3 + 21 n^2 + 14 n + 2 n^5 - 2 n^4 + 2 n^3 - 2 n^2 + 2 n + 5 \right) \\ \displaystyle = \frac{ 1 }{ 14 } n ( n + 1 ) \left( 2 n^5 + 5 n^4 + 16 n^3 + 19 n^2 + 16 n + 5 \right) \\ \displaystyle = \frac{ 1 }{ 14 } n ( n + 1 ) ( 2 n + 1 ) \left( n^4 + 2 n^3 + 7 n^2 + 6 n + 5 \right)$

$\displaystyle \Leftrightarrow \sum_{k=1}^{n} { k ^ 6 } = \frac{ 1 }{ 14 } n ( n + 1 ) ( 2 n + 1 ) \left( n^4 + 2 n^3 + 7 n^2 + 6 n + 5 \right) - \frac{ 1 }{ 6 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n^2 + 3 n + 2 \right) \\ \displaystyle = \frac{ 1 }{ 14 } n ( n + 1 ) ( 2 n + 1 ) \left\{ 3 \left( n^4 + 2 n^3 + 7 n^2 + 6 n + 5 \right) - 7 \left( 3 n^2 + 3 n + 2 \right) \right\} \\ \displaystyle = \frac{ 1 }{ 14 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n^4 + 6 n^3 + 21 n^2 + 18 n + 15 - 21 n^2 - 21 n - 14 \right) \\ \displaystyle = \frac{1}{42} n ( n + 1 ) ( 2 n + 1 ) \left( 3 n ^ 4 + 6 n ^ 3 - 3 n + 1 \right)$

$\displaystyle \therefore \sum_{k=1}^{n} { k ^ 6 } = \frac{1}{42} n ( n + 1 ) ( 2 n + 1 ) \left( 3 n ^ 4 + 6 n ^ 3 - 3 n + 1 \right) \tag{13}$

これが六乗和の公式となります。

また(12)より、
$\displaystyle \sum_{k=1}^{n} { k ^ 6 } = \frac{ 1 }{ 7 } n ( n + 1 ) \left( n^5 - n^4 + n^3 - n^2 + n - 1 \right) + 3 \sum_{k=1}^{n} { k ^ 5 } - 5 \sum_{k=1}^{n} { k ^ 4 } + 5 \sum_{k=1}^{n} { k ^ 3 } - 3 \sum_{k=1}^{n} { k ^ 2 } + \sum_{k=1}^{n} { k } \tag{14}$
の関係式も成り立ちます。

LibreOffice 数式(Math)のソース：

k ^7 - ( k - 1 ) ^7 = 7 k ^6 - 21 k ^5 + 35 k ^4 - 35 k ^3 + 21 k ^2 - 7 k + 1

n ^7 - ( n - 1 ) ^7 = 7 n ^6 - 21 n ^5 + 35 n ^4 - 35 n ^3 + 21 n ^2 - 7 n + 1
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dotsvert
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3 ^7 - 2 ^7 = 7 cdot 3 ^6 - 21 cdot 3 ^5 + 35 cdot 3 ^4 - 35 cdot 3 ^3 + 21 cdot 3 ^2 - 7 cdot 3 + 1
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2 ^7 - 1 ^7 = 7 cdot 2 ^6 - 21 cdot 2 ^5 + 35 cdot 2 ^4 - 35 cdot 2 ^3 + 21 cdot 2 ^2 - 7 cdot 2 + 1
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1 ^7 - 0 ^7 = 7 cdot 1 ^6 - 21 cdot 1 ^5 + 35 cdot 1 ^4 - 35 cdot 1 ^3 + 21 cdot 1 ^2 - 7 cdot 1 + 1

alignl n ^7 = 7 sum from { k = 1 } to { n } { k ^6 } - 21 sum from { k = 1 } to { n } { k ^5 } + 35 sum from { k = 1 } to { n } { k ^4 } - 35 sum from { k = 1 } to { n } { k ^3 } + 21 sum from { k = 1 } to { n } { k ^2 } - 7 sum from { k = 1 } to { n } { k } + n
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alignl phantom { y } dlrarrow 7 sum from { k = 1 } to { n } { k ^6 } = n ^7 + 21 sum from { k = 1 } to { n } { k ^5 } - 35 sum from { k = 1 } to { n } { k ^4 } + 35 sum from { k = 1 } to { n } { k ^3 } - 21 sum from { k = 1 } to { n } { k ^2 } + 7 sum from { k = 1 } to { n } { k } - n
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alignl phantom { yyyyy } = { alignc n ^7 + { { 21 } over { 12 } } ( 2 n ^6 + 6 n ^5 +5 n ^4 - n ^2 ) - { { 35 } over { 30 } } ( 6 n ^5 + 15 n ^4 + 10 n ^3 - n ) + { { 35 } over { 4 } } ( n ^4 + 2 n ^3 + n ^2 ) }
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alignl phantom { yyyyyyy } { alignc - { { 21 } over { 6 } } ( 2 n ^3 + 3 n ^2 + n ) + { { 7 } over { 2 } } ( n ^2 + n ) - n }
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alignl phantom { yyyyy } = { alignc n ^7 + { { 7 } over { 4 } } ( 2 n ^6 + 6 n ^5 +5 n ^4 - n ^2 ) - { { 7 } over { 6 } } ( 6 n ^5 + 15 n ^4 + 10 n ^3 - n ) + { { 35 } over { 4 } } ( n ^4 + 2 n ^3 + n ^2 ) }
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alignl phantom { yyyyyyy } { alignc - { { 7 } over { 2 } } ( 2 n ^3 + 3 n ^2 + n ) + { { 7 } over { 2 } } ( n ^2 + n ) - n }
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alignl phantom { yyyyy } = { alignc { { 1 } over { 12 } } left lbrace 12 n ^7 + 21 ( 2 n ^6 + 6 n ^5 +5 n ^4 - n ^2 ) - 14 ( 6 n ^5 + 15 n ^4 + 10 n ^3 - n ) + 105 ( n ^4 + 2 n ^3 + n ^2 ) right none }
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alignl phantom { yyyyyyy } left none - 42 ( 2 n ^3 + 3 n ^2 + n ) + 42 ( n ^2 + n ) - 12 n right rbrace
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alignl phantom { yyyyy } = { alignc { { 1 } over { 12 } } left ( 12 n ^7 + 42 n ^6 + 42 n ^5 - 14 n ^3 + 2 n right ) } = { alignc { { 1 } over { 6 } } left ( 6 n ^7 + 21 n ^6 + 21 n ^5 - 7 n ^3 + n right ) }
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alignl phantom { yyyyy } = { alignc { { 1 } over { 6 } } n ( n + 1 ) ( 2 n + 1 ) ( 3 n ^4 + 6 n ^3 - 3 n + 1 ) }
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alignl phantom { y } dlrarrow sum from { k = 1 } to { n } { k ^6 } = { alignc { { 1 } over { 42 } } n ( n + 1 ) ( 2 n + 1 ) ( 3 n ^4 + 6 n ^3 - 3 n + 1 ) }