${\left(1+x\right)}^{101}{\left(1+x^2-x\right)}^{100}$

$=\left(1+x\right){\left(\left(1+x\right)\left(1-x+x^2\right)\right)}^{100}$

$=\left(1+x\right){\left(1+x^3\right)}^{100}$

$=1\times {\left(1+x^3\right)}^{100}+x\times {\left(1+x^3\right)}^{100}$

So, the number of terms$=$( $101$ terms of the form $x^{3k}$ ) $+$ ($101$ terms of the form $x^{3k+1}$ )
$=202$ terms

$\Rightarrow n=202$ 

$\because \frac{n}{25}=8.08$