Question

$\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3} +\frac{ 1^{3}+2^{3}+3^{3}}{1+3+5} +... $ to $16$ terms =

• A. $420$
• B. $416$
• C. $436$
• D. $446$

Answer 1

Answer: (D)

The $ r^{th} $ term, $t_{r}=\frac{ 1^{3}+2^{3}+...r^{3}}{1+3+...+\left(2r-1\right)}$
$= \left(\frac{r\left(r+1\right)}{2}\right)\cdot\frac{1}{r^{2}} $
$= \frac{1}{4}\left(r+1\right)^{2}$
$ \sum\limits_{r=1}^{16} t_{r} = \frac{1}{4} \left[2^{2} +3^{2} + ....+17^{2}\right]$
$ = \frac{1}{4}\left[\frac{17\times18\times35}{6}-1\right] $
$ = 446$.