The sum (1^3)/(1) + (1^3 + 2^3)/(1 + 3) + (1^3 + 2^3 + 3^3)/(1 + 3 + 5) + up to 8 terms, i…

Question

The sum $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \ldots$ up to 8 terms, is :

Accepted Answer

$\begin{gathered} {T}_{r}=\frac{{1}^{3}+{2}^{3}+{3}^{3}+\ldots \ldots {r}^{3}}{1+3+5+\ldots ..+(2r-1)}=\frac{{(\frac{r(r+1)}{2})}^{2}}{{r}^{2}} \ =\frac{{r}^{2}+2r+1}{4} \end{gathered}$

$\begin{gathered} {S}_{n}=\sum _{r=1}^{n}{T}_{r} \ {S}_{n}=\frac{1}{4}\sum ({r}^{2}+2r+1) \ =\frac{1}{4}[\frac{n(n+1)(2n+1)}{6}+2\frac{n(n+1)}{2}+n] \ {S}_{8}=\frac{1}{4}[\frac{8	imes 9	imes 17}{6}+8	imes 9+8] \ =\frac{1}{4}[204+72+8]=71 \end{gathered}$

Solutions · 1