Question

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

• A. $\frac{7}{116}$
• B. $\frac{1}{12}$
• C. $\frac{29}{358}$
• D. $\frac{21}{346}$

Answer 1

The correct option is B $\frac{1}{12}$

Given, $\alpha$ and $\beta$ are the roots of the equation $375x^2-25x-2=0$
Therefore $\alpha +\beta =-(-\frac{25}{375})=\left(\frac{25}{375}\right)$
and $\alpha \cdot \beta =\left(\frac{-2}{375}\right)$
$\underset{n\to \infty }{lim}\sum _{r=1}^n{\alpha }^r+\underset{n\to \infty }{lim}\sum _{r=1}^n{\beta }^r=\underset{n\to \infty }{lim}\sum _{r=1}^n({\alpha }^r+{\beta }^r)$
$=\underset{n\to \infty }{lim}({\alpha }^1+{\beta }^1)+({\alpha }^2+{\beta }^2)+({\alpha }^3+{\beta }^3)+\cdots +({\alpha }^n+{\beta }^n)$
$=(\alpha +{\alpha }^2+{\alpha }^3+\cdots \infty )+(\beta +{\beta }^2+{\beta }^3+\cdots \infty$
Here we have infinite G.P series therefore
$=\frac{\alpha }{1-\alpha }+\frac{\beta }{1-\beta }=\frac{\alpha +\beta -2\alpha \beta }{1-(\alpha +\beta )+\alpha \beta }$
$=\frac{\frac{25}{375}+\frac{4}{375}}{1-\frac{25}{375}-\frac{2}{375}}=\frac{29}{348}=\frac{1}{12}$