JEE Mains · Maths · STD 11 - 8. sequence and series
Then sum \(\sum\limits_{k = 1}^{20} {k\frac{1}{{{2^k}}}} \) is equal to
- A \(2 - \frac{{11}}{{{2^{19}}}}\)
- B \(2 - \frac{{11}}{{{2^{20}}}}\)
- C \(2 - \frac{{21}}{{{2^{20}}}}\)
- D \(2 - \frac{{3}}{{{2^{17}}}}\)
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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