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, $t_r=\frac{1^3+2^3+...r^3}{1+3+...+\left(2r-1\right)}$
$=\left(\frac{r\left(r+1\right)}{2}\right)\cdot \frac{1}{r^2}$
$=\frac{1}{4}{\left(r+1\right)}^2$
$\sum _{r=1}^{16}{t}_r=\frac{1}{4}\left[2^2+3^2+....+17^2\right]$
$=\frac{1}{4}\left[\frac{17\times 18\times 35}{6}-1\right]$
$=446$.