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# $3{\sum }\limits_{ k=1 }^{n} {k}^{3}$ の計算式

$3\displaystyle{{\sum }\limits_{ k=1 }^{n} {k}^{3} ={1}^{3}+{2}^{3}+{3}^{3}+\cdot \cdot \cdot \cdot \cdot \cdot +{n}^{3}={ \{ \frac{\hspace{2} n \( n+1 \) \hspace{2}}{\hspace{2}2\hspace{2}} \} }^{2}}$

$3\displaystyle{{ \( k+1 \) }^{4}- {k}^{4}=4{k}^{3}+6{k}^{2}+4k+1}$ に順に $3\displaystyle{k=1,2,3,\cdot \cdot \cdot ,n}$ を代入し，以下のように縦にそろえて加えると，

 $3\displaystyle{{2}^{4}- {1}^{4}}$ $3\displaystyle{=}$ $3\displaystyle{4\cdot {1}^{3}}$ $3\displaystyle{+6\cdot {1}^{2}}$ $3\displaystyle{+4\cdot 1}$ $3\displaystyle{+1}$ $3\displaystyle{{3}^{4}- {2}^{4}}$ $3\displaystyle{=}$ $3\displaystyle{4\cdot {2}^{3}}$ $3\displaystyle{+6\cdot {2}^{2}}$ $3\displaystyle{+4\cdot 2}$ $3\displaystyle{+1}$ $3\displaystyle{{4}^{4}- {3}^{4}}$ $3\displaystyle{=}$ $3\displaystyle{4\cdot {3}^{3}}$ $3\displaystyle{+6\cdot {3}^{2}}$ $3\displaystyle{+4\cdot 3}$ $3\displaystyle{+1}$ $3\displaystyle{\cdot }$ $3\displaystyle{\cdot }$ $3\displaystyle{\cdot }$ $3\displaystyle{+\)}$ $3\displaystyle{{ \( n+1 \) }^{4}- {n}^{4}}$ $3\displaystyle{=}$ $3\displaystyle{4\cdot {n}^{3}}$ $3\displaystyle{+6\cdot {n}^{2}}$ $3\displaystyle{+4\cdot n}$ $3\displaystyle{+1}$ $3\displaystyle{\bar{ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}{ \( n+1 \) }^{4}- 1\text{\hspace{1} }\text{\hspace{1} }\text{\hspace{1} }\text{\hspace{1} }\text{\hspace{1} }\text{\hspace{1} }=4{\sum }\limits_{ k=1 }^{n} {k}^{3} +6{\sum }\limits_{ k=1 }^{n} {k}^{2} +4{\sum }\limits_{ k=1 }^{n}k+n }}$

$3\displaystyle{\displaystyle{{\sum }\limits_{ k=1 }^{n}{k}^{3}}}$$3\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} \{ { \( n+ 1 \) }^{4}- 1- 6{\sum }\limits_{ k=1 }^{n}{k}^{2}- 4{\sum }\limits_{ k=1 }^{n}k- n \} }$

$3\displaystyle{\displaystyle{\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} \{ { \( n+ 1 \) }^{4}- 6{\sum }\limits_{ k=1 }^{n}{k}^{2}- 4{\sum }\limits_{ k=1 }^{n}k- n- 1 \} }}}$

$3\displaystyle{\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} \{ { \( n+ 1 \) }^{4}- n \( n+ 1 \) \( 2n+ 1 \) - 2n \( n+ 1 \) - \( n+ 1 \) \} }}$

$3\displaystyle{\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} \( n+ 1 \) \{ { \( n+ 1 \) }^{3}- n \( 2n+ 1 \) - 2n- 1 \} }}$

$3\displaystyle{\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} \( n+ 1 \) \{ \( {n}^{3}+ 3{n}^{2}+ 3n+ 1 \) - \( 2{n}^{2}+ n \) - 2n- 1 \} }}$

$3\displaystyle{\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} \( n+ 1 \) \{ {n}^{3}+ \( 3- 2 \) {n}^{2}+ \( 3- 1- 2 \) n+ \( 1- 1 \) \} }}$

$3\displaystyle{\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} \( n+ 1 \) \( {n}^{3}+ {n}^{2} \) }}$

$3\displaystyle{\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} \( {n}^{4}+ {n}^{3}+ {n}^{3}+ {n}^{2} \) }}$

$3\displaystyle{\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} \( {n}^{4}+ 2{n}^{3}+ {n}^{2} \) }}$

$3\displaystyle{\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}{n}^{2} \( {n}^{2}+ 2n+ 1 \) }}$

$3\displaystyle{\displaystyle{={ \{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}n \( n+ 1 \) \} }^{2}}}$

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