The sum Up to 8 terms, is: [2026]

Question

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: [2026]

Smart Achievers · Accepted Answer

(2)

$T_{r} = \frac{1^{2} + 2^{3} + 3^{3} + . . . + r^{3}}{1 + 3 + 5 + . . . + (2 r - 1)}$

$\Rightarrow T_{r} = \frac{{(\frac{r (r + 1)}{2})}^{2}}{r^{2}}$

$\Rightarrow T_{r} = \frac{1}{4} [r^{2} + 2 r + 1] \Rightarrow S_{n} = \sum_{r = 1}^{n} T_{r}$

$∴ S_{n} = \frac{1}{4} \frac{n (n + 1) (2 n + 1)}{6} + \frac{1}{2} \frac{n (n + 1)}{2} + \frac{1}{4} n$

For n = 8,

$S_{8} = \frac{1}{24} (8) (9) (17) + \frac{1}{4} (8) (9) + \frac{1}{4} (8)$

= 51 + 18 + 2 = 71

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