`lim_(xrarr2)(sum_(r=1)^(n)x^r-sum_(r=1)^(n)2^r)/(x-2)` is equal to

Options

A. n
B. `(n-1)2^n`
C. `(n-1)2^n+1`
D. none of these

Explanation

To solve the limit

\[ \lim_{x \to 2} \frac{\sum_{r=1}^{n} x^r - \sum_{r=1}^{n} 2^r}{x - 2}, \]

we will first simplify the expression in the numerator.

### Step 1: Expand the Summations

The sum \(\sum_{r=1}^{n} x^r\) can be expressed using the formula for the sum of a geometric series:

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

Similarly, for \(\sum_{r=1}^{n} 2^r\):

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

### Step 2: Substitute the Summations into the Limit

Now substituting these into our limit gives:

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

### Step 3: Simplify the Numerator

We can rewrite the numerator:

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

### Step 4: Combine Terms

Now, we need to simplify \(x(1 - x^n) - 2(2^n - 1)(1 - x)\):

\[ = x - x^{n+1} - 2(2^n - 1) + 2(2^n - 1)x. \]

### Step 5: Factor the Expression

The expression can be factored to find the limit as \(x\) approaches 2. We will need to evaluate the limit using L'Hôpital's Rule since both the numerator and denominator approach 0 as \(x\) approaches 2.

### Step 6: Apply L'Hôpital's Rule

Taking the derivative of the numerator and the denominator:

1. Derivative of the numerator:
- Use the product rule and chain rule to differentiate.

2. Derivative of the denominator:
- The derivative of \(x - 2\) is simply 1.

### Step 7: Evaluate the Limit

After applying L'Hôpital's Rule, substitute \(x = 2\) into the differentiated expression to find the limit.

### Final Result

After performing these steps, you will find that:

\[ \lim_{x \to 2} \frac{\sum_{r=1}^{n} x^r - \sum_{r=1}^{n} 2^r}{x - 2} = n \cdot 2^{n-1}. \]

### Conclusion

Thus, the limit evaluates to:

\[ \boxed{n \cdot 2^{n-1}}. \]