<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Given that $\sum\limits_{r=1}^{n}{{{r}^{3}}=\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}}$, show that\[{{2}^{3}}+{{4}^{3}}+{{6}^{3}}+...+{{\left( 2n\right)}^{3}}=2{{n}^{2}}{{\left( n+1 \right)}^{2}}\]</span></div><div class="list-item-paragraph">Hence show that $\sum\limits_{r=1}^{n}{{{\left( 2r-1 \right)}^{3}}={{n}^{2}}\left( 2{{n}^{2}}-1 \right)}$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-content"><span class="list-item-text">Using the results in part (i), find $\sum\limits_{r=1}^{2n}{{{\left( -1 \right)}^{r}}{{\left( r \right)}^{3}}}$, simplifying your answer.</span></div><p style="text-align: right;">[2]</p></li></ol>

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 $\sum\limits_{r=1}^{n}{{{\left( 2r-1 \right)}^{3}}={{n}^{2}}\left( 2{{n}^{2}}-1 \right)}$.

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

2018 HCI Promo Q1

2018 HCI Promo Q1

Video Solution: