`lim_(xto1) (sum_(r=1)^(n)x^(r)-n)/(x-1)` is equal to

Options

A. `n/2`
B. `(n(n+1))/2`
C. 1
D. None of these

Answer

To solve the limit

\[ \lim_{x \to 1} \frac{\sum_{r=1}^{n} x^r - n}{x - 1}, \]

we can follow these steps:

### Step 1: Rewrite the Summation
The summation \(\sum_{r=1}^{n} x^r\) can be expressed as a geometric series. The formula for the sum of a geometric series is:

\[ \sum_{r=1}^{n} x^r = x + x^2 + x^3 + \ldots + x^n = \frac{x(1 - x^n)}{1 - x} \quad \text{for } x \neq 1. \]

### Step 2: Substitute the Summation into the Limit
Substituting the geometric series into the limit, we have:

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

### Step 3: Simplify the Expression
This can be simplified as follows:

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

### Step 4: Evaluate the Limit
As \(x\) approaches 1, both the numerator and denominator approach 0, which indicates we can apply L'Hôpital's Rule or simplify further.

Calculating the numerator when \(x = 1\):

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

### Step 5: Differentiate the Numerator and Denominator
Applying L'Hôpital's Rule:

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

2. Differentiate the denominator:
\[ \frac{d}{dx}[(x - 1)(1 - x)] = -2(x - 1). \]

### Step 6: Evaluate the New Limit
Now we evaluate the limit again:

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

Substituting \(x = 1\):

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

### Step 7: Use Taylor Series Expansion
Instead of L'Hôpital's Rule, we can use the Taylor series expansion around \(x = 1\):

\[ x^r \approx 1 + r(x - 1) + \frac{r(r-1)}{2}(x - 1)^2 + \ldots. \]

Summing from \(r=1\) to \(n\):

\[ \sum_{r=1}^{n} x^r \approx n + \frac{n(n+1)}{2}(x - 1) + O((x - 1)^2). \]

### Step 8: Substitute Back into the Limit
Substituting this back into the limit gives:

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

### Final Result
Thus, the limit is:

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