If $\frac{d}{dx}[1+x+x^2+x^3+\cdots +x^{100}]=\frac{a\cdot x^{101}-b\cdot x^{100}+1}{{(x-1)}^2}$ then  $b-a=$

$\frac{d}{dx}\left(\frac{x^{101}-1}{x-1}\right)$

$[1+x+x^2+\cdots +x^{100}=\frac{[x^{101}-1]}{x-1}]$

$=\frac{(x-1)[101\cdot x^{100}]-[x^{101}-1]\left[1\right]}{{(x-1)}^2}$

$=\frac{101\cdot x^{101}-101\cdot x^{100}-x^{101}+1}{{(x-1)}^2}$

$=\frac{100\cdot x^{101}-101\cdot x^{100}+1}{{(x-1)}^2}$

$\therefore a=100,b=101,b-a=1$