Keep it up-2

Inspiration $\displaystyle \sum_{\psi =1}^n (x+\psi-1)(x+\psi)=\phi$ $=\displaystyle\sum_{\psi=1}^n \left(\color{#20A900}{\psi^2}+\color{#D61F06}{\psi}(2x-1)+(x^2-x)(\color{#302B94}{1})\right)-\phi=0$ Using $\small{\color{#20A900}{\sum_{\psi=1}^n \psi^2=\dfrac{n(n+1)(2n+1)}{6}}, \color{#D61F06}{\sum_{\psi=1}^n\psi=\dfrac{n(n+1)}{2}},\color{#302B94}{\sum_{\psi=1}^n 1=n}}$ Equation simplifies to :- $3x^2+3nx+n^2-1-\dfrac{3\phi}{n}=0$ Since square of difference of roots of this equation is 1 ( $\because ( (\alpha+1)-\alpha)^2=1^2$ ) $\implies n^2-\dfrac{4(n^2-1-\dfrac{3\phi}n)}{3}=1^2$ $\implies n^2=1+\dfrac{12\phi}{n}\implies \mathfrak{T}=12$ Summation can be written as: $40 \sum_{i=2}^{\infty}\dfrac{i}{(i^2+2-2i)(i^2+2+2i)}$ $=10\sum_{i=2}^{\infty}\left(\dfrac{1}{i^2+2-2i}-\dfrac{1}{i^2+2+2i}\right)$ $\color{forestgreen}{\textbf{A Telescopic Series}}$ $=\left(\dfrac{1}{2}-\xcancel{\color{skyblue}{\dfrac{1}{10}}}\right)+\left(\dfrac{1}{5}-\xcancel{\color{magenta}{\dfrac{1}{17}}}\right)+\left(\xcancel{\color{skyblue}{\dfrac{1}{10}}}-\color{#EC7300}{\dfrac{1}{26}}\right)+ \left(\xcancel{\color{magenta}{\dfrac{1}{17}}}-\color{#D61F06}{\dfrac{1}{37}}\right)\cdots$ $\Large =10\left(\dfrac 12+\dfrac 15\right)$ $\huge=\boxed 7$

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

By solving the above expression, we'll get,

$nx^2+2\cdot \frac{n(n+1)}{2} x-nx+\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}=\phi \\ 6nx^2+6n^2x+2n^3+n-6\phi=0$

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$n=1+\frac{\mathfrak{T}\phi}{n} \Rightarrow \boxed{n^3-n=\mathfrak{T}\phi}$

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$2\alpha+1=-n \Rightarrow \alpha= -\frac{(n+1)}{2} \\ \alpha(\alpha+1)=\frac{2n^3+n-6\phi}{6n} \\ \frac{(n+1)^2}{4}-\frac{n+1}{2}=\frac{2n^3+n-6\phi}{6n} \\ 3n(n+1)^2-6n(n+1)=2(2n^3+n-6\phi)$

By solving this, we'll get,

$12\phi = n^3-n$

By comparing it with the expression in the box,

$12\phi=\mathfrak{T}\phi \Rightarrow 12=\mathfrak{T}$

Coming to the next part of the problem,

$\begin{aligned} & = 40 \displaystyle \sum^{\infty}_{i=2} \frac{i}{i^4+4} \\ & = 40 \displaystyle \sum^{\infty}_{i=2} \frac{i}{(i^2+2-2i)(i^2+2+2i)} \\ & = \frac{40}{4} \displaystyle \sum^{\infty}_{i=2}\left( \frac{1}{i^2+2-2i}-\frac{1}{i^2+2+2i}\right) \\ & = 10 \left( \frac{1}{2}-\frac{1}{10}+\frac{1}{5}-\frac{1}{17}+\frac{1}{10}-\frac{1}{26}+\frac{1}{17}-\ldots \right) \\ & = 10\left( \frac{1}{2}+\frac{1}{5} \right) =10\cdot \frac{7}{10}=\boxed{7} \end{aligned}$