Let `alpha=(-1+isqrt(3))/2` and `a=(1+alpha)sum_(k=0)^(100) alpha^(2k),b=sum_(k=0)^(100) alpha^(3k)` . If a and b are roots of quadratic equation then quadratic equation is

Options

A. `x^2 – 102x + 101 = 0`
B. `x^2 – 101x + 100 = 0`
C. `x^2 + 101x + 100 = 0`
D. `x^2 + 102x + 100 = 0`

Explanation

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

### Step 1: Define \( \alpha \)
Given:
\[ \alpha = \frac{-1 + i\sqrt{3}}{2} \]
This represents a complex number.

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

Here, \( a = 1 \), \( r = \alpha^2 \), and there are \( 101 \) terms (from \( k=0 \) to \( k=100 \)):
\[ \sum_{k=0}^{100} \alpha^{2k} = \frac{1(1 - \alpha^{202})}{1 - \alpha^2} \]

### Step 3: Calculate \( \alpha^2 \) and \( \alpha^{202} \)
First, we find \( \alpha^2 \):
\[ \alpha^2 = \left(\frac{-1 + i\sqrt{3}}{2}\right)^2 = \frac{(-1)^2 + 2(-1)(i\sqrt{3}) + (i\sqrt{3})^2}{4} = \frac{1 - 2i\sqrt{3} - 3}{4} = \frac{-2 - 2i\sqrt{3}}{4} = \frac{-1 - i\sqrt{3}}{2} \]

Next, we find \( \alpha^{202} \). Since \( \alpha \) is a cube root of unity (specifically, \( \omega = e^{2\pi i / 3} \)), we know:
\[ \alpha^3 = 1 \implies \alpha^{202} = \alpha^{3 \cdot 67 + 1} = \alpha^1 = \alpha \]

### Step 4: Substitute back into the sum
Now substituting back:
\[ \sum_{k=0}^{100} \alpha^{2k} = \frac{1 - \alpha}{1 - \alpha^2} \]

### Step 5: Calculate \( a \)
Now substituting into \( a \):
\[ a = (1 + \alpha) \cdot \frac{1 - \alpha}{1 - \alpha^2} \]
Calculating \( 1 - \alpha^2 \):
\[ 1 - \alpha^2 = 1 - \frac{-1 - i\sqrt{3}}{2} = \frac{2 + 1 + i\sqrt{3}}{2} = \frac{3 + i\sqrt{3}}{2} \]
Thus,
\[ a = (1 + \alpha) \cdot \frac{1 - \alpha}{\frac{3 + i\sqrt{3}}{2}} = \frac{(1 + \alpha)(1 - \alpha)}{\frac{3 + i\sqrt{3}}{2}} = \frac{1 - \alpha^2}{\frac{3 + i\sqrt{3}}{2}} = \frac{2(1 - \alpha^2)}{3 + i\sqrt{3}} \]

### Step 6: Calculate \( b \)
The expression for \( b \) is:
\[ b = \sum_{k=0}^{100} \alpha^{3k} \]
This is also a geometric series:
\[ b = \frac{1(1 - \alpha^{303})}{1 - \alpha^3} = \frac{1(1 - 1)}{1 - 1} = 101 \]

### Step 7: Form the quadratic equation
The roots of the quadratic equation are \( a \) and \( b \). The quadratic equation can be expressed as:
\[ x^2 - (a + b)x + ab = 0 \]
Substituting \( a \) and \( b \):
\[ x^2 - (1 + 101)x + (1 \cdot 101) = 0 \]
This simplifies to:
\[ x^2 - 102x + 101 = 0 \]

### Final Answer
Thus, the quadratic equation is:
\[ \boxed{x^2 - 102x + 101 = 0} \]