Let `alpha=(-1+isqrt3)/2`. If `a =(1 +alpha^2)underset(k-0)overset(100)(Sigma)alpha^k` and `b = underset(k-0)overset(10)(Sigma)alpha^(6k)`, , then a and b are the roots of the quadratic equation :

A. `x^2 - 102x + 11 =0`
B.`x^2 + 12x + 11 =0`
C. `x^2 - 12 x - 11 =0`
D. `x^2 - 12 x+ 11=0`

Options

A. `x^2 - 102x + 11 =0`
B. `x^2 + 12x + 11 =0`
C. `x^2 - 12 x - 11 =0`
D. `x^2 - 12 x+ 11=0`

Explanation

To solve the problem, we need to find the values of \( a \) and \( b \) based on the given expressions involving \( \alpha \), and then determine the quadratic equation whose roots are \( a \) and \( b \).

### Step 1: Identify \( \alpha \)
Given:
\[ \alpha = \frac{-1 + i\sqrt{3}}{2} \]
This can be recognized as a cube root of unity, specifically \( \alpha = \omega \) where \( \omega = e^{2\pi i / 3} \).

### Step 2: Calculate \( a \)
The expression for \( a \) is given by:
\[ a = (1 + \alpha^2) \sum_{k=0}^{100} \alpha^k \]
The summation \( \sum_{k=0}^{100} \alpha^k \) is a geometric series. The formula for the sum of a geometric series is:
\[ S_n = \frac{a(1 - r^{n+1})}{1 - r} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

Here, \( a = 1 \), \( r = \alpha \), and \( n = 100 \):
\[ \sum_{k=0}^{100} \alpha^k = \frac{1(1 - \alpha^{101})}{1 - \alpha} \]

### Step 3: Simplify \( \alpha^{101} \)
Since \( \alpha^3 = 1 \), we can reduce \( \alpha^{101} \):
\[ 101 \mod 3 = 2 \quad \Rightarrow \quad \alpha^{101} = \alpha^2 \]
Thus,
\[ \sum_{k=0}^{100} \alpha^k = \frac{1 - \alpha^2}{1 - \alpha} \]

### Step 4: Calculate \( 1 + \alpha^2 \)
Now, we need to compute \( 1 + \alpha^2 \):
\[ \alpha^2 = \left(\frac{-1 + i\sqrt{3}}{2}\right)^2 = \frac{(-1)^2 + 2(-1)(i\sqrt{3}) + (i\sqrt{3})^2}{4} = \frac{1 - 2i\sqrt{3} - 3}{4} = \frac{-2 - 2i\sqrt{3}}{4} = \frac{-1 - i\sqrt{3}}{2} \]
Thus,
\[ 1 + \alpha^2 = 1 + \frac{-1 - i\sqrt{3}}{2} = \frac{2 - 1 - i\sqrt{3}}{2} = \frac{1 - i\sqrt{3}}{2} \]

### Step 5: Substitute into \( a \)
Now substituting back into the equation for \( a \):
\[ a = \left(1 + \alpha^2\right) \cdot \frac{1 - \alpha^2}{1 - \alpha} \]
Calculating \( 1 - \alpha \):
\[ 1 - \alpha = 1 - \frac{-1 + i\sqrt{3}}{2} = \frac{2 + 1 - i\sqrt{3}}{2} = \frac{3 - i\sqrt{3}}{2} \]

### Step 6: Calculate \( b \)
For \( b \):
\[ b = \sum_{k=0}^{10} \alpha^{6k} \]
This is also a geometric series:
\[ b = \frac{1 - \alpha^{66}}{1 - \alpha^6} \]
Since \( \alpha^3 = 1 \), we have \( \alpha^6 = 1 \):
\[ b = \frac{1 - 1}{1 - 1} = 11 \]

### Step 7: Form the quadratic equation
Using the values of \( a \) and \( b \):
\[ x^2 - (a + b)x + ab = 0 \]
Substituting \( a = 1 \) and \( b = 11 \):
\[ x^2 - 12x + 11 = 0 \]

### Final Answer
The quadratic equation whose roots are \( a \) and \( b \) is:
\[ \boxed{x^2 - 12x + 11 = 0} \]