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JEE Mains · Maths · STD 11 - 8. sequence and series

Then sum \(\sum\limits_{k = 1}^{20} {k\frac{1}{{{2^k}}}} \) is equal to

  1. A \(2 - \frac{{11}}{{{2^{19}}}}\)
  2. B \(2 - \frac{{11}}{{{2^{20}}}}\)
  3. C \(2 - \frac{{21}}{{{2^{20}}}}\)
  4. D \(2 - \frac{{3}}{{{2^{17}}}}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2 - \frac{{11}}{{{2^{19}}}}\)

Step-by-step Solution

Detailed explanation

\(S = \sum\limits_{k = 1}^{20} {\frac{k}{{{2^k}}}} \) \(S = \frac{1}{2} + \frac{2}{{{2^2}}} + \frac{3}{{{2^3}}} + .... + \frac{{20}}{{{2^{20}}}}\) \(S\frac{1}{2} = \,\,\,\frac{1}{{{2^2}}} + \frac{2}{{{2^3}}} + .... + \frac{{20}}{{{2^{21}}}}\)…
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