$\frac{1+x^{2020}}{1+x^{2018}}=\frac{x^2\left(1+x^{2018}\right)+1-x^2}{1+x^{2018}}$
$=x^2+\frac{1-x^2}{1+x^{2018}}$
Put $x=2 \therefore\left[4+\frac{(-3)}{1+2^{2018}}\right]=3$
Put $x=3 \therefore\left[9-\frac{8}{1+3^{2018}}\right]=8$
Similarly for $x=4,\left[16-\frac{15}{1+4^{2018}}\right]=15$
For $x=5,\left[25-\frac{24}{1+5^{2018}}\right]=24$
For $x=6,\left[36-\frac{35}{1+6^{2018}}\right]=35$
$\therefore$ Required sum
$=3+8+15+24+35=85$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.