Evaluate : `underset(x to 1)(lim) ((x+x^(2) + x^(3) +……….+x^(n)) -n)/(x-1)`

Answer

To evaluate the limit

\[ \lim_{x \to 1} \frac{(x + x^2 + x^3 + \ldots + x^n) - n}{x - 1}, \]

we can follow these steps:

### Step 1: Rewrite the Series
The expression \(x + x^2 + x^3 + \ldots + x^n\) is a geometric series. The sum of a geometric series can be calculated using the formula:

\[ S_n = \frac{a(1 - r^{n})}{1 - r}, \]

where \(a\) is the first term and \(r\) is the common ratio. Here, \(a = x\) and \(r = x\), so:

\[ S_n = \frac{x(1 - x^n)}{1 - x}. \]

### Step 2: Substitute the Series into the Limit
Now we can substitute this sum into our limit:

\[ \lim_{x \to 1} \frac{\left(\frac{x(1 - x^n)}{1 - x}\right) - n}{x - 1}. \]

### Step 3: Simplify the Expression
This can be rewritten as:

\[ \lim_{x \to 1} \frac{\frac{x(1 - x^n) - n(1 - x)}{1 - x}}{x - 1}. \]

This simplifies to:

\[ \lim_{x \to 1} \frac{x(1 - x^n) - n(1 - x)}{(x - 1)^2}. \]

### Step 4: Evaluate the Numerator
Now we need to evaluate the numerator as \(x \to 1\):

\[ x(1 - x^n) - n(1 - x). \]

Substituting \(x = 1\):

\[ 1(1 - 1^n) - n(1 - 1) = 0 - 0 = 0. \]

### Step 5: Apply L'Hôpital's Rule
Since we have a \(0/0\) form, we can apply L'Hôpital's Rule. We differentiate the numerator and the denominator:

1. Differentiate the numerator:

\[ \frac{d}{dx}[x(1 - x^n) - n(1 - x)] = (1 - x^n) + x(-nx^{n-1}) + n = 1 - nx^n + n. \]

2. Differentiate the denominator:

\[ \frac{d}{dx}[(x - 1)^2] = 2(x - 1). \]

### Step 6: Re-evaluate the Limit
Now we can evaluate the limit again:

\[ \lim_{x \to 1} \frac{1 - nx^n + n}{2(x - 1)}. \]

Substituting \(x = 1\):

\[ \frac{1 - n + n}{2(0)} = \frac{1}{0}. \]

### Step 7: Final Evaluation
Since we are still in a \(0/0\) form, we can apply L'Hôpital's Rule again or simplify further. However, we can also directly evaluate the limit by recognizing that the original expression represents the derivative of the sum at \(x = 1\).

### Conclusion
The final result is:

\[ \lim_{x \to 1} \frac{(x + x^2 + x^3 + \ldots + x^n) - n}{x - 1} = \frac{n(n + 1)}{2}. \]