The sum \(\frac{{1}^{3}}{1}+\frac{{1}^{3}+{2}^{3}}{1+3}+\frac{{1}^{3}+{2}^{3}+{3}^{3}}{1+3+5}+⋯\) up to \(8\) terms, is:…

\( \mathrm{T}_{\mathrm{r}}=\frac{1^3+2^3+3^3+\ldots +\ldots r^3}{1+3+5+\ldots .+(2 \mathrm{r}-1)}=\frac{\left(\frac{\mathrm{r}(\mathrm{r}+1)}{2}\right)^2}{\mathrm{r}^2} \)
\( =\frac{\mathrm{r}^2+2 \mathrm{r}+1}{4} \)
\( \mathrm{~S}_{\mathrm{n}}=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{~T}_{\mathrm{r}} \)
\( \mathrm{~S}_{\mathrm{n}}=\frac{1}{4} \sum_{\mathrm{r}=1}^{\mathrm{n}} \left(\mathrm{r}^2+2 \mathrm{r}+1\right) \)
\( =\frac{1}{4}\left[\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}+\mathrm{n}\right] \)
\( \mathrm{S}_8=\frac{1}{4}\left[\frac{8 \times 9 \times 17}{6}+8 \times 9+8\right]\)
\( =\frac{1}{4}[204+72+8]=71\)