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If $\lim_{x \to 1} \frac{x + x^2 + x^3 + \cdots + x^n - n}{x - 1} = 820$, $(n \in \mathbb{N})$ then the value of $n$ is equal to ______.
By L’Hospital rule: $\lim_{x\to 1} \frac{1 + 2x + 3x^2 + \cdots + nx^{n-1}}{1} = 820$. So $1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2} = 820 \Rightarrow n^2 + n - 1640 = 0 \Rightarrow n = 40$.
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