<s>[INST] Calculate $\frac{1}{{1 - 100 + 500}} + \frac{{{2^2}}}{{{2^2} - 200 + 500}} + \ldots + \frac{{{{99}^2}}}{{{{99}^2} - 9900 + 500}} $ I need a hint to solve this problem : Find the sum of

$$ \frac{1}{{1 - 100 + 500}} + \frac{{{2^2}}}{{{2^2} - 200 + 500}} + \frac{{{3^2}}}{{{3^2} - 300 + 500}} +  +  \ldots  + \frac{{{{99}^2}}}{{{{99}^2} - 9900 + 500}}
$$

 [/INST] Hint:
$\frac{N^2}{N^2-100N+500} = \frac{N^2}{N^2-100N+2500-2500+500} = \frac{N^2}{(N-50)^2-(10\sqrt{20})^2)} $
$ a = (50-10\sqrt{20})$
$S = \frac{AN+B}{N-a} + \frac{C}{N+a}$
Simplifying this
$S = \sum_{N=1}^{99} 1+ a^2\frac{1}{N^2-a^2} =  \sum_{N=1}^{99}1+ a^2\frac{1}{(\frac{N}{a})^2-1}$
$S= 99+(50-10\sqrt{20})^2*(-6.67) $
I evaluated the last sum by Wolfram Alpha, if anyone has an idea of how to evaluate that will be great
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