\[\displaystyle \sum_{\psi =1}^n (x+\psi-1)(x+\psi)=\phi\]

If the equation above has \(\alpha\) and \(\alpha+1\) as its roots, then we get the relation between \(n\) and \(\phi\) as below. \[n^2=1+\dfrac{\mathfrak{\color{blue}{T}}\phi}{n}\]

Then find the value of the expression below.

\[\large \sum_{i=\mathfrak{\color{blue}{\frac{\mathfrak{T}}{6}}}}^{\infty}\dfrac{40i}{i^4+4}\]

**Details**

\(\alpha\in \mathbb{R} \text{ and } n,\mathfrak{T}\in\mathbb{Z}, n>1\)

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