If $$\alpha$$ and $$\beta$$ are the roots of the equation $$375x^2 - 25x - 2 = 0$$, then $$\lim_{n \to \infty} \sum_{r=1}^{n} \alpha^r + \lim_{n \to \infty} \sum_{r=1}^{n} \beta^r$$ is equal to:

We begin with the quadratic equation $$375x^{2}-25x-2=0$$ whose roots are $$\alpha$$ and $$\beta$$.

For a quadratic $$ax^{2}+bx+c=0$$ the roots are given by the quadratic formula $$x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$ Here $$a=375,\; b=-25,\; c=-2.$$ So

$$\alpha,\beta=\dfrac{-(-25)\pm\sqrt{(-25)^{2}-4\cdot375\cdot(-2)}}{2\cdot375}.$$

First compute the discriminant: $$(-25)^{2}-4\cdot375\cdot(-2)=625+3000=3625.$$ Hence $$\sqrt{3625}=\sqrt{25\cdot145}=5\sqrt{145}.$$

Therefore $$\alpha,\beta=\dfrac{25\pm5\sqrt{145}}{750} =\dfrac{5(5\pm\sqrt{145})}{750} =\dfrac{5\pm\sqrt{145}}{150}.$$

Because $$\sqrt{145}\approx12.0416,$$ we have $$\alpha=\dfrac{5+12.0416}{150}\approx0.1136,\qquad \beta=\dfrac{5-12.0416}{150}\approx-0.0469.$$ Clearly $$|\alpha|<1$$ and $$|\beta|<1,$$ so the infinite geometric series formed by their powers will converge.

The required expression is $$\lim_{n\to\infty}\sum_{r=1}^{n}\alpha^{r}+\lim_{n\to\infty}\sum_{r=1}^{n}\beta^{r}.$$ For any number $$|x|<1,$$ the sum of an infinite geometric series is $$\sum_{r=1}^{\infty}x^{r}=\dfrac{x}{1-x}.$$ Applying this formula we get

$$\sum_{r=1}^{\infty}\alpha^{r}=\dfrac{\alpha}{1-\alpha},\qquad \sum_{r=1}^{\infty}\beta^{r}=\dfrac{\beta}{1-\beta}.$$

Hence our expression equals $$S=\dfrac{\alpha}{1-\alpha}+\dfrac{\beta}{1-\beta}.$$

To evaluate $$S$$, it is convenient to use Vieta’s relations, which state that for the roots of $$ax^{2}+bx+c=0$$

$$\alpha+\beta=-\dfrac{b}{a},\qquad \alpha\beta=\dfrac{c}{a}.$$

In our case $$\alpha+\beta=-\dfrac{-25}{375}=\dfrac{25}{375}=\dfrac{1}{15},$$ $$\alpha\beta=\dfrac{-2}{375}=-\dfrac{2}{375}.$$

Now write $$S$$ over a common denominator:

$$S=\dfrac{\alpha}{1-\alpha}+\dfrac{\beta}{1-\beta} =\dfrac{\alpha(1-\beta)+\beta(1-\alpha)}{(1-\alpha)(1-\beta)}.$$

Simplify the numerator:

$$\alpha(1-\beta)+\beta(1-\alpha)=\alpha-\alpha\beta+\beta-\alpha\beta =(\alpha+\beta)-2\alpha\beta.$$

Simplify the denominator:

$$(1-\alpha)(1-\beta)=1-\alpha-\beta+\alpha\beta =1-(\alpha+\beta)+\alpha\beta.$$

Substituting the Vieta values we obtain

Numerator $$ (\alpha+\beta)-2\alpha\beta=\dfrac{1}{15}-2\!\left(-\dfrac{2}{375}\right) =\dfrac{1}{15}+\dfrac{4}{375}.$$

Convert to a common denominator $$375$$: $$\dfrac{1}{15}=\dfrac{25}{375},$$ so the numerator becomes $$\dfrac{25}{375}+\dfrac{4}{375}=\dfrac{29}{375}.$$

Denominator $$1-(\alpha+\beta)+\alpha\beta =1-\dfrac{1}{15}-\dfrac{2}{375} =\dfrac{14}{15}-\dfrac{2}{375}.$$

Again, write everything over $$375$$: $$\dfrac{14}{15}=\dfrac{14\cdot25}{15\cdot25}=\dfrac{350}{375},$$ so the denominator is $$\dfrac{350}{375}-\dfrac{2}{375}=\dfrac{348}{375}.$$

Thus

$$S=\dfrac{29/375}{348/375}=\dfrac{29}{375}\times\dfrac{375}{348} =\dfrac{29}{348}.$$

Notice that $$348=29\times12,$$ so

$$S=\dfrac{29}{29\cdot12}=\dfrac{1}{12}.$$

Hence, the correct answer is Option A.