Question

Sol. [Key Concept: General term $T_{r} = (\frac{r + 1}{2})^{2}$ (cubes over odd-sum); add the 8 terms.]

$${T_{r} = \frac{1^{3} + 2^{3} + 3^{3} + \ldots\ldots r^{3}}{1 + 3 + 5 + \ldots.. + (2r - 1)} = \frac{\left( \frac{r(r + 1)}{2} \right)^{2}}{r^{2}} }{= \frac{r^{2} + 2r + 1}{4} }{S_{n} = \sum_{r = 1}^{n}\mspace{2mu} T_{r} }{S_{n} = \frac{1}{4}\sum\left( r^{2} + 2r + 1 \right) }{= \frac{1}{4}\left\lbrack \frac{n(n + 1)(2n + 1)}{6} + 2\frac{n(n + 1)}{2} + n \right\rbrack }{S_{8} = \frac{1}{4}\left\lbrack \frac{8 \times 9 \times 17}{6} + 8 \times 9 + 8 \right\rbrack }{= \frac{1}{4}\lbrack 204 + 72 + 8\rbrack = 71 }$$