Question
$\left[\frac{2^{2020}+1}{2^{2018}+1}\right]+\left[\frac{3^{2020}+1}{3^{2018}+1}\right]+\left[\frac{4^{2020}+1}{4^{2018}+1}\right] +\left[\frac{5^{2020}+1}{5^{2018}+1}\right] + \left[\frac{6^{2020}+1}{6^{2018}+1}\right]$ is
$\frac{1+x^{2020}}{1+x^{2018}}=\frac{x^2\left(1+x^{2018}\right)+1-x^2}{1+x^{2018}}$
$=x^2+\frac{1-x^2}{1+x^{2018}}$
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
$=3+8+15+24+35=85$
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$g(x) = (2x + 1)(x - k) + 3,\,0 \leqslant x < \infty $ then $g(f(x)),$ will be continuous at $x = 1$ if $k$ is equal