Exercise:
Show that $\sum\limits_{k=1}^{n}{{{k}^{3}}={{w}^{4}}}$ where $\sum\limits_{k=1}^{n}{k}={{w}^{2}}\,\,,\,\,w\in \mathbb{N}$
Proof: we know that $\sum\limits_{k=1}^{n}{k=}1+2+3+..+n=\frac{n\left( n+1 \right)}{2}={{w}^{2}}\Leftrightarrow n\left( n+1 \right)=2{{w}^{2}}$
Also $\sum\limits_{k=1}^{n}{{{k}^{4}}-{{\left( k-1 \right)}^{4}}=\sum\limits_{k=1}^{n}{\left( {{k}^{4}}-\left( {{k}^{4}}-4{{k}^{3}}+6{{k}^{2}}-4k+1 \right) \right)}}$
$=\sum\limits_{k=1}^{n}{\left( {{k}^{4}}-{{k}^{4}}+4{{k}^{3}}-6{{k}^{2}}+4k-1 \right)}=4\sum\limits_{k=1}^{n}{{{k}^{3}}-6\sum\limits_{k=1}^{n}{{{k}^{2}}}+4\sum\limits_{k=1}^{n}{k}-\sum\limits_{k=1}^{n}{1}}$ (**)
But $\sum\limits_{k=1}^{n}{{{k}^{3}}-{{\left( k-1 \right)}^{3}}=\sum\limits_{k=1}^{n}{\left( {{k}^{3}}-\left( {{k}^{3}}-3{{k}^{2}}+3k-1 \right) \right)=3\sum\limits_{k=1}^{n}{{{k}^{2}}-3\sum\limits_{k=1}^{n}{k}+\sum\limits_{k=1}^{n}{1}}}}$ (*)
Notice that $\sum\limits_{k=1}^{n}{{{k}^{4}}-{{\left( k-1 \right)}^{4}}={{n}^{4}}}\,\,\And \,\,\,\sum\limits_{k=1}^{n}{{{k}^{3}}-{{\left( k-1 \right)}^{3}}={{n}^{3}}}$ so (*) become
${{n}^{3}}=3\sum\limits_{k=1}^{n}{{{k}^{2}}-3\sum\limits_{k=1}^{n}{k+n}\Leftrightarrow {{n}^{3}}-n=3\sum\limits_{k=1}^{n}{{{k}^{2}}-\frac{3n\left( n+1 \right)}{2}}}\Leftrightarrow {{n}^{3}}-n+\frac{3n\left( n+1 \right)}{2}=3\sum\limits_{k=1}^{n}{{{k}^{2}}}$
$\Leftrightarrow 3\sum\limits_{k=1}^{n}{{{k}^{2}}=n\left( {{n}^{2}}-1+\frac{3\left( n+1 \right)}{2} \right)=n\left( \frac{2{{n}^{2}}-2+3n+3}{2} \right)=\frac{n}{2}\left( 2{{n}^{2}}+3n+1 \right)}$
So $\sum\limits_{k=1}^{n}{{{k}^{2}}=\frac{n}{6}\left( 2{{n}^{2}}+3n+1 \right)=\frac{n\left( 2n+1 \right)\left( n+1 \right)}{6}}$ so (**) become
${{n}^{4}}=4\sum\limits_{k=1}^{n}{{{k}^{3}}-\frac{6n\left( n+1 \right)\left( 2n+1 \right)}{6}+\frac{4n\left( n+1 \right)}{2}-n}$
$\Leftrightarrow {{n}^{4}}=4\sum\limits_{k=1}^{n}{{{k}^{3}}-n\left( n+1 \right)\left( 2n+1 \right)+2n\left( n+1 \right)}-n$
$\Leftrightarrow 4\sum\limits_{k=1}^{n}{{{k}^{3}}}={{n}^{4}}+n\left( n+1 \right)\left( 2n+1 \right)-2n\left( n+1 \right)+n$
$\Leftrightarrow 4\sum\limits_{k=1}^{n}{{{k}^{3}}=n\left( {{n}^{3}}+2{{n}^{2}}+3n+1-2n-2+1 \right)}$
$\Leftrightarrow \sum\limits_{k=1}^{n}{{{k}^{3}}=\frac{n}{4}\left( {{n}^{3}}+2{{n}^{2}}+n \right)=\frac{{{n}^{2}}}{4}\left( {{n}^{2}}+2n+1 \right)=\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}={{\left( \frac{n\left( n+1 \right)}{2} \right)}^{2}}}$
Thus $\sum\limits_{k=1}^{n}{{{k}^{3}}={{\left( \frac{2{{w}^{2}}}{2} \right)}^{2}}={{w}^{4}}}$ Q.E.D
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