Let ` a= min { x^2+2x+3:x in R}and b=lim_(x to0) (sin xcos x) /(e^x-e^-x)`. Then the value of `sum_(r=0)^(n) a^r,b^(n-r)`, is

Options

A. `(2^(n+1)+1)/(3.2^n)`
B. `(2^(n+1)-1)/(3.2^n)`
C. `(2^(n)-1)/(3.2^n)`
D. `(4^(n+1)-1)/(3.2^n)`

Answer

To solve the problem step by step, we need to find the values of \( a \) and \( b \) first, and then evaluate the summation.

### Step 1: Find the value of \( a \)

We need to find the minimum value of the function \( f(x) = x^2 + 2x + 3 \).

1. **Identify the coefficients**:
- Here, \( a = 1 \), \( b = 2 \), and \( c = 3 \).

2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 3 = 4 - 12 = -8 \]
Since the discriminant is negative, the quadratic has no real roots and opens upwards (as \( a > 0 \)), indicating it has a minimum.

3. **Find the vertex**:
The vertex \( x \) coordinate is given by \( x = -\frac{b}{2a} = -\frac{2}{2 \cdot 1} = -1 \).

4. **Calculate \( f(-1) \)**:
\[ f(-1) = (-1)^2 + 2(-1) + 3 = 1 - 2 + 3 = 2 \]
Thus, \( a = 2 \).

### Step 2: Find the value of \( b \)

We need to evaluate the limit:
\[ b = \lim_{x \to 0} \frac{\sin x \cos x}{e^x - e^{-x}} \]

1. **Check the form**:
As \( x \to 0 \), both the numerator and denominator approach 0, giving a \( \frac{0}{0} \) form.

2. **Use L'Hôpital's Rule**:
Differentiate the numerator and denominator:
- Derivative of the numerator: \( \frac{d}{dx}(\sin x \cos x) = \cos^2 x - \sin^2 x \).
- Derivative of the denominator: \( \frac{d}{dx}(e^x - e^{-x}) = e^x + e^{-x} \).

3. **Evaluate the limit**:
\[ b = \lim_{x \to 0} \frac{\cos^2 x - \sin^2 x}{e^x + e^{-x}} = \frac{1 - 0}{1 + 1} = \frac{1}{2} \]

### Step 3: Evaluate the summation

The summation we need to evaluate is:
\[ S = \sum_{r=0}^{n} a^r b^{n-r} \]

1. **Substitute values of \( a \) and \( b \)**:
\[ S = \sum_{r=0}^{n} 2^r \left(\frac{1}{2}\right)^{n-r} \]
This can be rewritten as:
\[ S = \sum_{r=0}^{n} 2^r \cdot \frac{1}{2^{n-r}} = \sum_{r=0}^{n} \frac{2^r}{2^{n-r}} = \sum_{r=0}^{n} \frac{2^{2r - n}}{1} \]

2. **Factor out \( \frac{1}{2^n} \)**:
\[ S = \frac{1}{2^n} \sum_{r=0}^{n} 2^{2r} \]

3. **Recognize the summation as a geometric series**:
The series \( \sum_{r=0}^{n} 2^{2r} \) is a geometric series with first term \( 1 \) and common ratio \( 4 \):
\[ \sum_{r=0}^{n} 2^{2r} = \frac{1(4^{n+1} - 1)}{4 - 1} = \frac{4^{n+1} - 1}{3} \]

4. **Combine results**:
\[ S = \frac{1}{2^n} \cdot \frac{4^{n+1} - 1}{3} = \frac{4^{n+1} - 1}{3 \cdot 2^n} \]

### Final Result
Thus, the value of the summation is:
\[ S = \frac{4^{n+1} - 1}{3 \cdot 2^n} \]