The value of `lim_(xrarr1)(x+x^(2)+….+x^(n)-n)/(x-1)` is

Options

A. n
B. `(n+1)/(2)`
C. `(n(n+1))/(2)`
D. `(n(n-1))/(2)`

Answer

To solve the limit \( \lim_{x \to 1} \frac{x + x^2 + \ldots + x^n - n}{x - 1} \), we can follow these steps:

### Step 1: Recognize the Form
First, we note that substituting \( x = 1 \) directly into the limit gives us an indeterminate form \( \frac{0}{0} \):
\[ x + x^2 + \ldots + x^n = 1 + 1^2 + \ldots + 1^n = n \quad \text{and} \quad x - 1 = 0. \]
Thus, we have:
\[ \frac{n - n}{0} = \frac{0}{0}. \]

**Hint:** Check if direct substitution leads to an indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator:
\[ \lim_{x \to 1} \frac{f(x)}{g(x)} = \lim_{x \to 1} \frac{f'(x)}{g'(x)}. \]

**Hint:** Remember that L'Hôpital's Rule can be applied when you encounter \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).

### Step 3: Differentiate the Numerator and Denominator
The numerator is \( f(x) = x + x^2 + \ldots + x^n - n \). The derivative \( f'(x) \) is:
\[ f'(x) = 1 + 2x + 3x^2 + \ldots + nx^{n-1}. \]
The denominator is \( g(x) = x - 1 \) and its derivative \( g'(x) \) is:
\[ g'(x) = 1. \]

**Hint:** Differentiate each term carefully and remember that the derivative of a constant is zero.

### Step 4: Evaluate the New Limit
Now we can rewrite our limit using the derivatives:
\[ \lim_{x \to 1} \frac{1 + 2x + 3x^2 + \ldots + nx^{n-1}}{1}. \]
Substituting \( x = 1 \):
\[ 1 + 2(1) + 3(1^2) + \ldots + n(1^{n-1}) = 1 + 2 + 3 + \ldots + n. \]

**Hint:** Simplify the expression by substituting \( x = 1 \) after differentiating.

### Step 5: Use the Formula for the Sum of the First n Natural Numbers
The sum \( 1 + 2 + 3 + \ldots + n \) can be calculated using the formula:
\[ \frac{n(n + 1)}{2}. \]

**Hint:** Recall the formula for the sum of the first \( n \) natural numbers.

### Final Answer
Thus, the value of the limit is:
\[ \lim_{x \to 1} \frac{x + x^2 + \ldots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}. \]