2018 HCI Promo Q1

Given that \(\sum\limits_{r=1}^{n}{{{r}^{3}}=\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}}\), show that\[{{2}^{3}}+{{4}^{3}}+{{6}^{3}}+...+{{\left( 2n\right)}^{3}}=2{{n}^{2}}{{\left( n+1 \right)}^{2}}\]

Hence show that \(\sum\limits_{r=1}^{n}{{{\left( 2r-1 \right)}^{3}}={{n}^{2}}\left( 2{{n}^{2}}-1 \right)}\).[3]

Using the results in part (i), find \(\sum\limits_{r=1}^{2n}{{{\left( -1 \right)}^{r}}{{\left( r \right)}^{3}}}\), simplifying your answer.[2]