Multiple choice

If $\alpha $ and $\beta $ are the roots of the equation $6x^{2}-x-1=0$ and $S_{r}=\alpha ^{r}+\beta ^{r}$ , then $\lim_{n\rightarrow \infty }\sum_{r=1}^{n}S_{r}$ is

  1. $\displaystyle \frac{3}{4}$
  2. $\displaystyle \frac{1}{2}$
  3. $\displaystyle \frac{1}{4}$
  4. $1$
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A Correct answer
Explanation

The roots of 6x^2 - x - 1 = 0 are 1/2 and -1/3. S_r = (1/2)^r + (-1/3)^r. The sum is a geometric series: sum(1/2^r) + sum((-1/3)^r). Sum = (1/2)/(1-1/2) + (-1/3)/(1 - (-1/3)) = 1 + (-1/3)/(4/3) = 1 - 1/4 = 3/4.

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