Let $$\alpha = \frac{-1+i\sqrt{3}}{2}$$. If $$a = (1 + \alpha)\sum_{k=0}^{100} \alpha^{2k}$$ and $$b = \sum_{k=0}^{100} \alpha^{3k}$$, then $$a$$ and $$b$$ are the roots of the quadratic equation.

We are given $$\alpha=\dfrac{-1+i\sqrt{3}}{2}$$. This number is a non-real cube root of unity, so it satisfies the well-known relations $$\alpha^3=1 \quad\text{and}\quad \alpha^2+\alpha+1=0.$$

First we tackle the sum $$\displaystyle S_1=\sum_{k=0}^{100}\alpha^{2k}.$$ Notice that $$\alpha^{2k}=(\alpha^2)^k,$$ and since $$\bigl(\alpha^2\bigr)^3=\alpha^{6}=1,$$ the powers of $$\alpha^2$$ repeat every three steps. Let us examine how the exponents behave modulo 3:

For $$k=3m$$ we have $$2k=6m\equiv0\pmod3,$$ giving the term $$1.$$
For $$k=3m+1$$ we have $$2k=6m+2\equiv2\pmod3,$$ giving the term $$\alpha^2.$$
For $$k=3m+2$$ we have $$2k=6m+4\equiv1\pmod3,$$ giving the term $$\alpha.$$

Thus every block of three consecutive terms contributes $$1+\alpha^2+\alpha=0,$$ because $$1+\alpha+\alpha^2=0.$$

Among the $$101$$ indices $$k=0,1,\dots,100$$ we have exactly $$33$$ complete blocks (which account for $$99$$ terms) and two more terms for $$k=99$$ and $$k=100$$. Their contributions are

$$k=99:\;2k=198\equiv0\pmod3\implies\alpha^{198}=1,$$
$$k=100:\;2k=200\equiv2\pmod3\implies\alpha^{200}=\alpha^2.$$

Hence $$S_1 = 1+\alpha^2.$$

Now we construct $$a$$:

$$a=(1+\alpha)S_1=(1+\alpha)(1+\alpha^2).$$

We expand the product completely:

$$\begin{aligned} (1+\alpha)(1+\alpha^2)&=1+\alpha^2+\alpha+\alpha^3\\ &=1+\alpha^2+\alpha+1 \quad(\text{because }\alpha^3=1)\\ &=2+(\alpha+\alpha^2)\\ &=2-1 \quad(\text{since }\alpha+\alpha^2=-1)\\ &=1. \end{aligned}$$

Therefore $$a=1.$$

Next we evaluate $$b=\displaystyle\sum_{k=0}^{100}\alpha^{3k}.$$ Because $$\alpha^3=1,$$ each term in this sum equals $$1$$. With $$101$$ such terms, we obtain

$$b=101.$$

Thus the two roots of the required quadratic are $$a=1$$ and $$b=101$$. For a quadratic with roots $$r_1$$ and $$r_2$$ we always have the form $$x^2-(r_1+r_2)x+r_1r_2=0.$$

Here the sum of roots is $$r_1+r_2 = 1+101 = 102,$$ and the product of roots is $$r_1r_2 = 1\cdot101 = 101.$$

Substituting these values, the quadratic equation becomes $$x^2-102x+101=0.$$

Hence, the correct answer is Option B.