If $\alpha ,\text{β}$  are roots of $375x^2-25x-2=0$  and ${\text{S}}_n={\alpha }^n+{\text{β}}^n$ , then $\underset{n\to \infty }{\lim }{\displaystyle \sum _{r=1}^n{\text{S}}_r}$  is

The given equation is $375x^2-25x-2=0$ 

$\alpha ,\text{β}$ are the roots of the equation, then

Sum of the roots : $\alpha +\text{β}=\frac{25}{375}=\frac{1}{15}$  and

Product of the roots : $\alpha \text{β}=\frac{-2}{375}$

Given that, ${\text{S}}_n={\alpha }^n+{\text{β}}^n$

                    $\sum _{r=1}^n{\text{S}}_r=(\alpha +{\alpha }^2+\mathrm{.....}+{\alpha }^n)+(\text{β}+{\text{β}}^2+\mathrm{....}+{\text{β}}^n)$

$\therefore$  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{r=1}^n{\text{S}}_r=\left(\alpha +{\alpha }^2+\mathrm{.....}\text{ to}\infty \right)+\left(\text{β}+{\text{β}}^2+\mathrm{.....}\text{ to }\infty \right)}$

                     $=\frac{\alpha }{1-\alpha }+\frac{\text{β}}{1-\text{β}}$                 $\left(\begin{array}{c}\because \alpha +{\alpha }^2+\mathrm{.....}\text{ to}\infty =\frac{\alpha }{1-\alpha }\\ \text{ β}+{\text{β}}^2+\mathrm{.......}\text{ to }\infty =\frac{\text{β}}{1-\text{β}}\end{array}\right)$  

                     $=\frac{\alpha -\alpha \text{β}+\text{β}-\alpha \text{β}}{\left(1-\alpha \right)\left(1-\text{β}\right)}=\frac{\left(\alpha +\text{β}\right)-2\alpha \text{β}}{1-\alpha -\text{β}+\alpha \text{β}}$

                     $=\frac{\frac{1}{15}+\frac{4}{375}}{1-\frac{1}{15}-\frac{2}{375}}=\frac{25+4}{375-25-2}=\frac{29}{348}=\frac{1}{12}$