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累乗の和 - 五乗和の公式

イメージ 1

五乗和の公式の導出です。

$\displaystyle \sum_{k=1}^{n} { k ^ 6 } - \sum_{k=1}^{n} { ( k - 1 )^6 } = 1^6 + 2^6 + \cdots + ( n - 1 )^6 + n^6 - \left\{ 0^6 + 1^6 + \cdots + ( n - 2 )^6 + ( n - 1 )^6 \right\} = n^6 \\ \displaystyle \Leftrightarrow \sum_{k=1}^{n} { k ^ 6 } = n^6 + \sum_{k=1}^{n} { ( k - 1 )^6 } = n^6 + \sum_{k=1}^{n} { \left( k^6 - 6 k^5 + 15 k^4 - 20 k^3 + 15 k^2 -6 k + 1 \right) } \\ \displaystyle = n^6 + \sum_{k=1}^{n} { k ^ 6 } - 6 \sum_{k=1}^{n} { k ^ 5 } + 15 \sum_{k=1}^{n} { k ^ 4 } - 20 \sum_{k=1}^{n} { k ^ 3 } + 15 \sum_{k=1}^{n} { k ^ 2 } - 6 \sum_{k=1}^{n} { k } + n \\ \displaystyle \Leftrightarrow 6 \sum_{k=1}^{n} { k ^ 5 } = n^6 + n + 15 \sum_{k=1}^{n} { k ^ 4 } - 20 \sum_{k=1}^{n} { k ^ 3 } + 15 \sum_{k=1}^{n} { k ^ 2 } - 6 \sum_{k=1}^{n} { k }$

$\displaystyle = n ( n + 1 ) \left( n^4 - n^3 + n^2 - n + 1 \right) + 15 \sum_{k=1}^{n} { k ^ 4 } - 20 \sum_{k=1}^{n} { k ^ 3 } + 15 \sum_{k=1}^{n} { k ^ 2 } - 6 \sum_{k=1}^{n} { k } \tag{10}$

更に(10)と、四乗和、三乗和、二乗和の公式から、
$\displaystyle 6 \sum_{k=1}^{n} { k ^ 5 } = n ( n + 1 ) \left( n^4 - n^3 + n^2 - n + 1 \right) + 15 \sum_{k=1}^{n} { k ^ 4 } - 20 \sum_{k=1}^{n} { k ^ 3 } + 15 \sum_{k=1}^{n} { k ^ 2 } - 6 \sum_{k=1}^{n} { k } \\ \displaystyle = n ( n + 1 ) \left( n^4 - n^3 + n^2 - n + 1 \right) + 15 \left\{ \frac{ 1 }{ 30 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n^2 + 3 n - 1 \right) \right\} \\ \displaystyle - 20 \left\{ \frac{ 1 }{ 4 } n^2 ( n + 1 )^2 \right\} + 15 \left\{ \frac{ 1 }{ 6 } n ( n + 1 ) ( 2 n + 1 ) \right\} - 6 \left\{ \frac{ 1 }{ 2 } n ( n + 1 ) \right\} \\ \displaystyle = n ( n + 1 ) \left( n^4 - n^3 + n^2 - n + 1 \right) + \frac{ 1 }{ 2 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n^2 + 3 n - 1 \right) - 5 n^2 ( n + 1 )^2 \\ \displaystyle + \frac{ 5 }{ 2 } n ( n + 1 ) ( 2 n + 1 ) - 3 n ( n + 1 ) \\ \displaystyle = \frac{ 1 }{ 2 } n ( n + 1 ) \left\{ 2 \left( n^4 - n^3 + n^2 - n + 1 \right) + ( 2 n + 1 ) \left( 3 n^2 + 3 n - 1 \right) - 10 n ( n + 1 ) + 5 ( 2 n + 1 ) - 6 \right\} \\ \displaystyle = \frac{ 1 }{ 2 } n ( n + 1 ) \left\{ 2 \left( n^4 - n^3 + n^2 - n + 1 \right) + ( 2 n + 1 ) \left( 3 n^2 + 3 n - 1 + 5 \right) - 10 n ( n + 1 ) - 6 \right\} \\ \displaystyle = \frac{ 1 }{ 2 } n ( n + 1 ) \left\{ 2 \left( n^4 - n^3 + n^2 - n + 1 \right) + ( 2 n + 1 ) \left( 3 n^2 + 3 n + 4 \right) - 10 n ( n + 1 ) - 6 \right\} \\ \displaystyle = \frac{ 1 }{ 2 } n ( n + 1 ) \left\{ 2 n^4 - 2 n^3 + 2 n^2 - 2 n + 2 + 6 n^3 + 6 n^2 + 8 n + 3 n^2 + 3 n + 4 - 10 n - 10 - 6 \right\} \\ \displaystyle = \frac{ 1 }{ 2 } n ( n + 1 ) \left\{ 2 n^4 + 4 n^3 + n^2 - n \right\} = \frac{ 1 }{ 2 } n^2 ( n + 1 ) \left\{ 2 n^3 + 4 n^2 + n - 1 \right\} \\ \displaystyle = \frac{ 1 }{ 2 } n^2 ( n + 1 )^2 \left\{ 2 n^2 + 2 n - 1 \right\}$

$\displaystyle \Leftrightarrow \sum_{k=1}^{n} { k ^ 5 } = { \frac{1}{12} } { n ^ 2 } { ( n + 1 ) ^ 2 } \left( 2 n ^ 2 + 2 n - 1 \right) \tag{11}$

これが五乗和の公式となります。

また(10)より、

$\displaystyle \sum_{k=1}^{n} { k ^ 5 } = \frac{ 1 }{ 6 } n ( n + 1 ) \left( n^4 - n^3 + n^2 - n + 1 \right) + \frac{ 5 }{ 2 } \sum_{k=1}^{n} { k ^ 4 } - \frac{ 10 }{ 3 } \sum_{k=1}^{n} { k ^ 3 } + \frac{ 5 }{ 2 } \sum_{k=1}^{n} { k ^ 2 } - \sum_{k=1}^{n} { k } \tag{12}$

の関係式も成り立ちます。

LibreOffice 数式(Math)のソース：

k ^6 - ( k - 1 ) ^6 = 6 k ^5 - 15 k ^4 + 20 k ^3 - 15 k ^2 + 6 k - 1

n ^6 - ( n - 1 ) ^6 = 6 n ^5 - 15 n ^4 + 20 n ^3 - 15 n ^2 + 6 n - 1
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dotsvert
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3 ^6 - 2 ^6 = 6 cdot 3 ^5 - 15 cdot 3 ^4 + 20 cdot 3 ^3 - 15 cdot 3 ^2 + 6 cdot 3 - 1
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2 ^6 - 1 ^6 = 6 cdot 2 ^5 - 15 cdot 2 ^4 + 20 cdot 2 ^3 - 15 cdot 2 ^2 + 6 cdot 2 - 1
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1 ^6 - 0 ^6 = 6 cdot 1 ^5 - 15 cdot 1 ^4 + 20 cdot 1 ^3 - 15 cdot 1 ^2 + 6 cdot 1 - 1

alignl n ^6 = 6 sum from { k = 1 } to { n } { k ^5 } - 15 sum from { k = 1 } to { n } { k ^4 } + 20 sum from { k = 1 } to { n } { k ^3 } - 15 sum from { k = 1 } to { n } { k ^2 } + 6 sum from { k = 1 } to { n } { k } - n
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alignl phantom { y } dlrarrow 6 sum from { k = 1 } to { n } { k ^5 } = n ^6 + 15 sum from { k = 1 } to { n } { k ^4 } - 20 sum from { k = 1 } to { n } { k ^3 } + 15 sum from { k = 1 } to { n } { k ^2 } - 6 sum from { k = 1 } to { n } { k } + n
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alignl phantom { yyyyy } = { alignc n ^6 + { { 15 } over { 30 } } ( 6 n ^5 + 15 n ^4 + 10 n ^3 - n ) - { { 20 } over { 4 } } ( n ^4 + 2 n ^3 + n ^2 ) + { { 15 } over { 6 } } ( 2 n ^3 + 3 n ^2 + n ) - { { 6 } over { 2 } } ( n ^2 + n ) + 2 n }
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alignl phantom { yyyyy } = { alignc { { 1 } over { 2 } } left lbrace 2 n ^6 + ( 6 n ^5 + 15 n ^4 + 10 n ^3 - n ) -10 ( n ^4 + 2 n ^3 + n ^2 ) + 5 ( 2 n ^3 + 3 n ^2 + n ) - 6 ( n ^2 + n ) + 2 n right rbrace }
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alignl phantom { yyyyy } = { alignc { { 1 } over { 2 } } ( 2 n ^6 + 6 n ^5 + 5 n ^4 - n ^2 ) } = { alignc { { 1 } over { 2 } } n ^2 ( n + 1 ) ^2 ( 2 n ^2 + 2 n - 1 ) }
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alignl phantom { y } dlrarrow sum from { k = 1 } to { n } { k ^5 } = { alignc { { 1 } over { 12 } } n ^2 ( n + 1 ) ^2 ( 2 n ^2 + 2 n - 1 ) }