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Express rational functions as a sum of fractions with simpler denominators. You can apply this to telescoping series to mass cancel terms in a seemingly complicated sum.

\[\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \frac{4}{5!} + \ldots = \ ?\]

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\[\large \displaystyle \sum_{n = 1}^{\infty} \dfrac {1}{n^2 + 3n + 2} = \ ? \]

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\[ \frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} + \frac{1}{(x-3)(x-4)} = \frac{1}{6} \]

What is the sum of all real values of \(x\) that satisfy the above equation?

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