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If $$\alpha$$ and $$\beta$$ are the roots of the quadratic equation $$4{x}^{2}+2x-1=0$$ then the value of $$\sum _{r=1}^{\infty }\left({\alpha }^{r}+{\beta }^{r}\right)$$ is:
(a) $$2$$
(b) $$3$$
(c) $$6$$
(d) $$0$$

$$A=\sum_{r=1}^{\infty}\left( \alpha^{r}+\beta^{r}\right)\\\begin{array}{l}=\left(\alpha^{1}+\beta^{1}+\alpha^{2}+\beta^{2}+\alpha^{3}+\beta^3+\cdots \cdots\infty\right) \\=\left(\alpha^{1}+\alpha^{2}+\alpha^{3}+\cdots \infty\right)+\left(\beta^{1}+\beta^{2}+\cdots \infty\right) \\a=\alpha, \quad r=\alpha, \quad a=\beta, \quad r=\beta\end{array}$$
$$A=\frac{\alpha}{1-\alpha}+\frac{\beta}{1-\beta}=\frac{\alpha-\alpha \beta+\beta-\alpha \beta}{(1-\alpha)(1-\beta)}$$.....(i)
Now, on solving $$4 x^{2}+2 x-1=0$$
We get $$\alpha+\beta=-\frac{b}{a}=\frac{-2}{4}=\frac{-1}{2}$$
And $$\alpha \cdot \beta=\frac{c}{a}=-\frac{1}{4}$$
Putting value of $$\alpha+\beta$$ and $$\alpha \beta$$ in (i) we get
$$A=\frac{(\alpha+\beta)-2 \alpha \beta}{(1-\alpha)(1-\beta)}\\A=\frac{(-2 / 4)-2\left(-\frac{1}{4}\right)}{(1-\alpha)(1-\beta)} \\A=\frac{-2 / 4+2 / 4}{(1-\alpha)(1-\beta)} \\A=0$$
Hence, the correct option is (d).