The given mathematical expression appears to involve a limit calculation as $n \to 1$. The expression is:

$\lim_{n \to 1} \frac{\sum_{k=1}^{n} (x^k + 1) - 2n}{n - 1}$

To solve this, we need to break it down:

1. Summation term: $\sum_{k=1}^{n} (x^k + 1)$ means we sum the expression $x^k + 1$ from $k = 1$ to $k = n$.

   • The summation for $x^k$ is a geometric series: $\sum_{k=1}^{n} x^k = \frac{x(x^n - 1)}{x - 1} \text{ if } x \neq 1$
   • The sum of the constant term $1$ from $k=1$ to $n$ is simply $n$.

   Thus: $\sum_{k=1}^{n} (x^k + 1) = \frac{x(x^n - 1)}{x - 1} + n$

2. Simplification of numerator: The numerator becomes: $\frac{x(x^n - 1)}{x - 1} + n - 2n = \frac{x(x^n - 1)}{x - 1} - n$

3. Division by $n - 1$: Now we need to divide this expression by $n - 1$: $\frac{\frac{x(x^n - 1)}{x - 1} - n}{n - 1}$

4. Taking the limit as $n \to 1$: To compute the limit, we might need to apply L'Hôpital's Rule because the expression takes the indeterminate form $\frac{0}{0}$ when $n = 1$. Differentiating the numerator and denominator with respect to $n$ would help.

Would you like to proceed with L'Hôpital's Rule, or shall I simplify it another way?

Here are five related questions to deepen your understanding:

1. How does L'Hôpital's Rule apply when limits take an indeterminate form?
2. What is the formula for the sum of a geometric series, and how can it be applied in different problems?
3. How does the behavior of a function change near singular points like $n = 1$?
4. What other techniques besides L'Hôpital's Rule are helpful for solving limits?
5. What role does the summation notation play in solving limits involving series?

Tip: Whenever you encounter indeterminate forms like $0/0$ in limits, L'Hôpital's Rule is often a good method to apply.