The largest term in the sequence ${\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2}{{\mathrm{n}}^3+200}$is given by 

Consider the function $\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2}{\left({\mathrm{x}}^3+200\right)} \left(1\right)$
$\therefore {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\mathrm{x}\frac{\left(400-{\mathrm{x}}^3\right)}{{\left({\mathrm{x}}^3+200\right)}^2}=0$
When $\mathrm{x}=(400{)}^{1/3}, (\because \mathrm{x}\ne 0)$
          $\begin{array}{r}\mathrm{x}=(400{)}^{1/3}-\mathrm{h}\text{ or }{\mathrm{f}}^{\mathrm{\prime }}(\mathrm{x})>0\\ \mathrm{x}=(400{)}^{1/3}+\mathrm{h}\text{ or }{\mathrm{f}}^{\mathrm{\prime }}(\mathrm{x})<0\end{array}$
Thus, f(x) has maxima at $\mathrm{x}=(400{)}^{1/3}$.
Since $7<(400{)}^{1/3}<8$, either ${\mathrm{a}}_7\text{ or }{\mathrm{a}}_8$ is the greatest term of the sequence. Therefore,
${\mathrm{a}}_7=\frac{49}{543}\text{ and }{\mathrm{a}}_8=\frac{8}{89}\text{ and }\frac{49}{543}>\frac{8}{89}$
Thus, ${\mathrm{a}}_7=\frac{49}{543}$is the greatest term.