Are you the smartest #2

In this series, the sum of numerator of each term is $(n(n+1)/2) ^{2}$ and the sum of denominator of each term is $n^{2}.$

So each term is $(n+1)^{2}/4.$

So the sum is

$=(2^{2} +3^{2} + . . . + 17^{2})/4$

$=(1^{2} + 2^{2} +3^{2} + . . . + 17^{2} -1)/4$

$=[(17*18*35/6)-1]/4$

$= \boxed{446}$

Let $T_{r}$ denote the $r^{th}$ term of the series.

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

$=\frac{\displaystyle\sum_{m=1}^{r} m^{3}}{\displaystyle\sum_{m=1}^{r} (2m-1)}$

Denominator comes out to be $r^{2}$ .

This can be proved by sum of an AP with initial term $1$ , common difference $2$ and number of terms $r$ .

Numerator comes out to be $\bigg (\dfrac{r(r+1)}{2}\bigg )^{2}$ .

So $T_{r}=\dfrac{\Big (\frac{r(r+1)}{2}\Big )^{2}}{r^{2}}$

$=\dfrac{(r+1)^{2}}{4}$

$S_{16}=\dfrac{\displaystyle\sum_{r=1}^{16} (r+1)^{2}}{4}$

To ease up the question we can say that

$S_{16}=\dfrac{\displaystyle\sum_{r=2}^{17} r^{2}}{4}$

$\dfrac{\Big (\displaystyle\sum_{r=1}^{17} r^{2}\Big )-1^{2}}{4}$

Now by using the formula that $\displaystyle\sum_{r=1}^{n} n^{2}=\frac{n(n+1)(2n+1)}{6}$ ,

$S$ comes out to be $\dfrac{\frac{17×18×35}{6}-1}{4}=\boxed{446}$