Question

If $\alpha$ and $\beta$ are the roots of the equation $6x^2-x-1=0$ and $S_r={\alpha }^r+{\beta }^r$ , then ${lim}_{n\to \infty }{\sum }_{r=1}^n{S}_r$ is

• A. $\frac{3}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $1$

Answer 1

The correct option is A $\frac{3}{4}$

$Solvingtheequationwehavetherootsas\alpha =\frac{1}{2},\beta =-\frac{1}{3}Theyarenumericallylessthan1.\therefore {lim}_{n\to \infty }{\sum }_{r=1}^n{S}_r=\alpha +{\alpha }^2+{\alpha }^3+.........\infty +\beta +{\beta }^2+{\beta }^3+.........\infty Theseriesin\alpha and\beta isaGPseries,$

$firsttermsbeing\alpha ,\beta andcommonratio\alpha and\beta respectively.\therefore Thesummationis\frac{\alpha }{1-\alpha }+\frac{\beta }{1-\beta }=\frac{\frac{1}{2}}{1-\frac{1}{2}}+\frac{(-\frac{1}{3})}{1+\frac{1}{3}}=\frac{3}{4}$