The sum T_1+T_2++T_8, where T_r = 1^3+2^3++r^31+3+5++(2r-1), is:

Question

The sum $T_1+T_2+\cdots+T_8$, where $T_r = \dfrac{1^3+2^3+\cdots+r^3}{1+3+5+\cdots+(2r-1)}$, is:

Accepted Answer

$$T_r = \frac{\left(\dfrac{r(r+1)}{2}\right)^2}{r^2} = \frac{(r+1)^2}{4}.$$

$$\sum_{r=1}^{8} T_r = \frac14\sum_{r=1}^{8}(r+1)^2 = \frac14\sum_{k=2}^{9}k^2 = \frac14\left(\sum_{k=1}^{9}k^2 - 1\right) = \frac{285-1}{4} = 71.$$

Solutions · 1