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Let $y=\cfrac{1} {1-x}$ Expressing $x$ in terms of $y$ , we have $x =1-\cfrac{1}{y}$ .

Substituting this into our original equation, we get $(1-\cfrac{1}{y})^{2018}+(1-\cfrac{1}{y})^{2017}+...+(1-\cfrac{1}{y})^{2}+(1-\cfrac{1}{y})^{1}-1345=0$

$(y^{2018})((1-\cfrac{1}{y})^{2018}+(1-\cfrac{1}{y})^{2017}+...+(1-\cfrac{1}{y})^{2}+(1-\cfrac{1}{y})^{1}-1345) =0$

$(y-1) ^{2018} +y(y-1) ^{2017}+y^{2} (y-1) ^{2016} +... +y^{2016} (y-1) ^{2} +y^{2 017}(y-1)-1345y^{2018}=0$

In the expansion above,

Coefficient of $y^{2018}=1\cdot 2018-1345=673$

Coefficient of $y^{2017}=-2018-2017-2016-...-1=-(1+2+3+...+2017+2018) =-\cfrac{2018\cdot2019}{2}$

By Vieta's Formula, if $y_{k}$ for $k=1,2,3,...2018$ are roots for the equation above in terms of y,

We get $\sum_{k=1}^{2018} y_{k} =\cfrac{\cfrac{2018\cdot2019}{2}}{673}=3027$

Hence, returning to our original question, $\sum_{n=1}^{2018} \cfrac{1}{1-a_{n}}=3027$

Let $y = x^{2018} + x^{2017} + .... + x - 1345$

Thus, $y = (x-a_{1})(x-a_{2}).....(x-a_{2018})$ , where $a_{1}, a_{2}$ and so on are the roots.

$ln(y) = \sum_{n=1}^{2018} ln(x-a_{n})$

Differentiating: $\frac{1}{y} \frac{dy}{dx} = \sum_{n=1}^{2018} \frac{1}{x-a_{n}}$

Since we require the RHS value for x=1, all that remains to be done is finding $y(1)$ and $y'(1)$ .

$y(1) = 1*2018 - 1345 = 673$

$y'(x) = 2018x^{2017} + 2017x^{2016} + ... + 1 => y'(1) = \frac{(2018)(2019)}{2}$

Thus, the desired sum is $3027$ .