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JEE Main 2019 · 8 April · Shift II

JEE Main 2019 Mathematics question on Sequences and Series

The sum ∑k=120k⋅12k\sum_{k=1}^{20} k \cdot \frac{1}{2^k} is equal to
  1. A.
    2−112192 - \dfrac{11}{2^{19}}
  2. B.
    1−112201 - \dfrac{11}{2^{20}}
  3. C.
    2−32172 - \dfrac{3}{2^{17}}
  4. D.
    2−212202 - \dfrac{21}{2^{20}}
Show answer

Correct option: A

[JEE Main 2019 · 8 April · Shift II]

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