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If \(x_1,x_2,x_3, \dots,x_{2017}\) are the solution of the equation \( \displaystyle{\sum_{n=0}^{2017} {\sum_{r=0}^{n} {\left[{\binom{n}{r} \left(\frac{2017}{x}\right)^r (x)^{n} \cos(\pi r)}\right]} }} = \sum_{t=0}^{2016} {(-1)^{t+1}(t x^t)}\) and \( \displaystyle{ {\prod_{p=1}^{2018} {p \Bigg( \sum_{m=0}^{p} {\binom{p}{m} \Bigg[{\sum_{q=1}^{2017} { \Bigg( x_q+\binom{2017}{q}+1} \Bigg) \Bigg]^{p-m} } (-2016-2^{2017})^m \Bigg) ^p} } } = \Lambda } \) and \( \displaystyle{ \log_{4} \Bigg\{\sum_{\Gamma=1}^{20^{17}} {\prod_{U_\Gamma=1}^{2017} {\Big(\Lambda^{U_\Gamma}+{(U_\Gamma)}^\Lambda\Big)}}\Bigg\} = a + b\log_2{\sqrt{5}} } \) when \( a,b \in \mathbb{Z}^+ \), then what is the value of \( \displaystyle{ \Bigg | {\frac{14a^2-2b^2}{3a^2+4b^2+5ab} + \frac {34a^2-7b^2}{16a^2+18ab+20b^2}i - \frac {54a^2-12b^2}{52a^2+56ab+60b^2} - \frac {74a^2-17b^2}{144a^2+152ab+160b^2}i } }+ \dots \Bigg |^2 \)

**Note:** \(i = \sqrt{-1} \)

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