Question

If $\alpha, \beta$ are the roots of the equation $375 x^{2}-25 x-2=0$ and $S_{n}=\alpha^{n}+\beta^{n}$, then $\underset{{n \rightarrow \infty}}{\text{Lt}} \displaystyle\sum_{r=1}^{n} S_{r}$ is

• A. $\frac{7}{12}$
• B. $\frac{1}{12}$
• C. $\frac{35}{12}$
• D. None of these

Answer 1

Answer: (B)

$\displaystyle\sum_{r=1}^{n} S_{r}=(\alpha+\beta)+\left(\alpha^{2}+\beta^{2}\right)+\ldots+\left(\alpha^{n}+\beta^{n}\right)$
$=\left(\alpha+\alpha^{2}+\ldots+\alpha^{n}\right)+\left(\beta+\beta^{2}+\ldots+\beta^{n}\right)$
$\underset{{n \rightarrow \infty}}{{\text{Lt}}} \displaystyle\sum_{r=1}^{n} S_{r}=\left(\alpha+\alpha^{2}+\ldots+\infty\right)+\left(\beta+\beta^{2}+\ldots+\infty\right)$
$=\frac{\alpha}{1-\alpha}+\frac{\beta}{1-\beta}$
$=\frac{\alpha-\alpha \beta+\beta-\alpha \beta}{1-(\alpha+\beta)+\alpha \beta}$
$=\frac{\alpha+\beta-2 \alpha \beta}{1-(\alpha+\beta)+\alpha \beta}$
$=\frac{\frac{25}{375}+\frac{4}{375}}{1-\frac{25}{375}-\frac{2}{375}}$
$=\frac{29}{348}=\frac{1}{12}$