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累乗の和 - 四乗和の公式(2)

四乗和の公式の導出です。

$k ( k + 1 ) ( k + 2 ) ( k + 3 ) ( k + 4 ) - ( k - 1 ) k ( k + 1 ) ( k + 2 ) ( k + 3 ) \tag{1}$

を考えます。

(1)をそのまま展開すると、

$k ( k + 1 ) ( k + 2 ) ( k + 3 ) ( k + 4 ) - ( k - 1 ) k ( k + 1 ) ( k + 2 ) ( k + 3 ) \\ \displaystyle = k ( k + 1 ) ( k + 2 ) ( k + 3 ) \left\{ ( k + 4 ) - ( k - 1 ) \right\}$
$\displaystyle = 5 k ( k + 1 ) ( k + 2 ) ( k + 3 ) = 5 \left( k^4 + 6 k^3 + 11 k^2 + 6 k \right) \tag{2}$

(1)の和を考えると、

$\displaystyle \sum_{k=1}^{n} { \left\{ k ( k + 1 ) ( k + 2 ) ( k + 3 ) ( k + 4 ) - ( k - 1 ) k ( k + 1 ) ( k + 2 ) ( k + 3 ) \right\} } \\ \displaystyle = \sum_{k=1}^{n} { k ( k + 1 ) ( k + 2 ) ( k + 3 ) ( k + 4 ) } - \sum_{k=1}^{n} { ( k - 1 ) k ( k + 1 ) ( k + 2 ) ( k + 3 ) } \\ \displaystyle = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 + 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 + \cdots + ( n - 1 ) n ( n + 1 ) ( n + 2 ) ( n + 3 ) + n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) \\ \displaystyle - \left\{ 0 \cdot 1 \cdot 2 \cdot 3 \cdot 4 + 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 + \cdots + ( n - 2 ) ( n - 1 ) n ( n + 1 ) ( n + 2 ) + ( n - 1 ) n ( n + 1 ) ( n + 2 ) ( n + 3 ) \right\}$
$= n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) \tag{3}$
これと、(2)の和が等しいので、

$\displaystyle n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) = \sum_{k=1}^{n} { \left\{ 5 \left( k^4 + 6 k^3 + 11 k^2 + 6 k \right) \right\} } = 5 \sum_{k=1}^{n} { \left( k^4 + 6 k^3 + 11 k^2 + 6 k \right) } \\ \displaystyle \Leftrightarrow \frac{ 1 }{ 5 } n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) = \sum_{k=1}^{n} { \left( k^4 + 6 k^3 + 11 k^2 + 6 k \right) } = \sum_{k=1}^{n} { k^4 } + 6 \sum_{k=1}^{n} { k^3 } + 11 \sum_{k=1}^{n} { k^2 } + 6 \sum_{k=1}^{n} { k }$
$\displaystyle \Leftrightarrow \sum_{k=1}^{n} { k^4 } = \frac{ 1 }{ 5 } n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) - 6 \sum_{k=1}^{n} { k^3 } - 11 \sum_{k=1}^{n} { k^2 } - 6 \sum_{k=1}^{n} { k } \tag{4}$

これと、三乗和、二乗和の公式から、

$\displaystyle \sum_{k=1}^{n} { k^4 } = \frac{ 1 }{ 5 } n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) - 6 \left\{ \frac{ 1 }{ 4 } n^2 ( n + 1 )^2 \right\} - 11 \left\{ \frac{ 1 }{ 6 } n ( n + 1 ) ( 2 n + 1 ) \right\} - 6 \left\{ \frac{ 1 }{ 2 } n ( n + 1 ) \right\} \\ \displaystyle = \frac{ 1 }{ 5 } n ( n + 1 ) \left( n^3 + 9 n^2 + 26 n + 24 \right) - \frac{ 3 }{ 2 } n ( n + 1 ) \left( n^2 + n \right) - \frac{ 11 }{ 6 } n ( n + 1 ) ( 2 n + 1 ) - 3 n ( n + 1 ) \\ \displaystyle \Leftrightarrow \sum_{k=1}^{n} { k^4 } + \frac{ 11 }{ 6 } n ( n + 1 ) ( 2 n + 1 ) = \frac{ 1 }{ 10 } n ( n + 1 ) \left\{ 2 \left( n^3 + 9 n^2 + 26 n + 24 \right) - 15 \left( n^2 + n \right) - 30 \right\} \\ \displaystyle = \frac{ 1 }{ 10 } n ( n + 1 ) \left( 2 n^3 + 18 n^2 + 52 n + 48 - 15 n^2 - 15 n - 30 \right) \\ \displaystyle = \frac{ 1 }{ 10 } n ( n + 1 ) \left( 2 n^3 + 3 n^2 + 37 n + 18 \right) = \frac{ 1 }{ 10 } n ( n + 1 ) ( 2 n + 1 ) \left( n^2 + n + 18 \right) \\ \displaystyle \Leftrightarrow \sum_{k=1}^{n} { k^4 } = \frac{ 1 }{ 10 } n ( n + 1 ) ( 2 n + 1 ) \left( n^2 + n + 18 \right) - \frac{ 11 }{ 6 } n ( n + 1 ) ( 2 n + 1 ) \\ \displaystyle = \frac{ 1 }{ 30 } n ( n + 1 ) ( 2 n + 1 ) \left\{ 3 \left( n^2 + n + 18 \right) - 55 \right\} = \frac{ 1 }{ 30 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n^2 + 3 n + 54 - 55 \right) \\ \displaystyle = \frac{ 1 }{ 30 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n^2 + 3 n - 1 \right)$
$\displaystyle \therefore \sum_{k=1}^{n} { k^4 } = \frac{ 1 }{ 30 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n^2 + 3 n - 1 \right) \tag{5}$