JEE Main 2019 · 12 April · Shift I

JEE Main 2019 Mathematics question on Limits Continuity and Differentiability

If α\alpha and β\beta are the roots of the equation 375x2−25x−2=0375x^2 - 25x - 2 = 0, then lim⁡n→∞∑r=1nαr+lim⁡n→∞∑r=1nβr\lim_{n \to \infty} \sum_{r=1}^{n} \alpha^r + \lim_{n \to \infty} \sum_{r=1}^{n} \beta^r is equal to
  1. A.
    21346\dfrac{21}{346}
  2. B.
    29358\dfrac{29}{358}
  3. C.
    112\dfrac{1}{12}
  4. D.
    7116\dfrac{7}{116}
Show answer

Correct option: C

[JEE Main 2019 · 12 April · Shift I]

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