Question

If$x>0$, then find greatest value of the expression $\frac{x^{100}}{1+x+x^2+x^3+....+x^{200}}$

Answer 1

$a+ax+a{x}^2+...+a{x}^{n-1}=\frac{a(x^n-1)}{x-1}$

$\Rightarrow 1+x+x^2+...+x^{200}=\frac{x^{201}-1}{x-1}$

So, $f\left(x\right)=\frac{x^{100}}{1+x+x^2+...+x^{200}}=\frac{x^{100}(x-1)}{x^{201}-1}$

When $f^{{}^{\prime }}\left(x\right)=\frac{(x^{201}-1)x^{99}(101x-100)-201x^{300}(x-1)}{(x^{201}-1{)}^2}=0$

$\Rightarrow x^{99}(x^{201}-1)(101x-100)=201x^{300}(x-1)$

$\Rightarrow 101x^{202}-100x^{201}-101x+100=201x^{202}-201x^{201}\Rightarrow 100x^{202}+101x=101x^{201}+100$

Possible only if $x=\pm 1$

As $x>0$, $\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}\frac{101x^{100}-100x^{99}}{201x^{200}}=\frac{1}{201}$

So, the maximum value of $f\left(x\right)$ is $f\left(1\right)\approx 0.005$