What is the value of formula?

If we replace $x=1$ in the given expression, we will get $\frac{0}{0}$ which is an indeterminate form so we need to apply L’Hopital’s Rule

ApplyingL’Hopita’s Rule :

$\lim_{x \to 1} \frac{\frac{\partial }{\partial x}(x^n-1)}{\frac{\partial}{\partial x}(x-1)} =\lim_{x\to 1} \frac{nx^{n-1}}{1} = n$

The above numerator can be factored into $x^{n}-1 = (x-1)(x^{n-1} + x^{n-2} + ... + x + 1)$ , and the $x-1$ factor cancels with the denominator. The latter nonlinear factor contains $n$ terms which the limit equals:

$lim_{x \rightarrow 1} \Sigma_{k=0}^{n-1} x^k = \Sigma_{k=0}^{n-1} 1 = \boxed{n}.$

Let $S = 1 + x + x^2 + x^3 + \cdots + x^n \newline$ So $xS = x + x^2 + x^3 + \cdots + x^n + x^{n+1} \newline$ $xS - S = x^{n+1} - 1 \newline$ $S = \frac{x^{n+1}-1}{x-1} \newline$ When $x \to 1$ , the formula's (not $S$ ) value is $\newline$ $1 + 1 + 1^2 + 1^3 + \cdots + 1^{n-1} \newline = 1 \times n \newline = n$ . $\newline$ Choose the third.