Mathematics - Sequence & Series Question with Solution | TestHub

By A.M. $\ge$ G.M. on $1,x,x^2,\dots ,x^{200}:\frac{1+x+\cdots +x^{200}}{201}\ge {\left(x^{0+1+\cdots +200}\right)}^{1/201}=x^{100}$. So $\frac{x^{100}}{1+x+\cdots +x^{200}}\le \frac{1}{201}$, with equality at $x=1$. Greatest value $=\frac{1}{201}$, so $p+q=1+201=202$.