Limits, Continuity & Differentiability Question - None

Question

If $\lim_{x \to 1} \frac{x + x^2 + x^3 + \ldots + x^n - n}{x - 1} = 820$, where $n \in \mathbb{N}$, then the value of $n$ is equal to ………

Accepted Answer

As $x \to 1$, both numerator and denominator approach 0, giving the form $\frac{0}{0}$.

Numerator: $x + x^2 + \ldots + x^n - n \to 1 + 1 + \ldots + 1 - n = n - n = 0$.

Using L'Hôpital's rule: $\lim_{x \to 1} \frac{1 + 2x + 3x^2 + \ldots + nx^{n-1}}{1} = 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}$.

Setting equal to 820: $\frac{n(n+1)}{2} = 820$ gives $n(n+1) = 1640$.

Solving: $n^2 + n - 1640 = 0$ yields $n = 40$ (taking positive root).

∴ $n = 40$

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