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STD 11 - 8. sequence and series — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 8. sequence and series1 Mark
MCQ
Then sum $\sum\limits_{k = 1}^{20} {k\frac{1}{{{2^k}}}} $ is equal to
  • ✓
    $2 - \frac{{11}}{{{2^{19}}}}$
  • B
    $2 - \frac{{11}}{{{2^{20}}}}$
  • C
    $2 - \frac{{21}}{{{2^{20}}}}$
  • D
    $2 - \frac{{3}}{{{2^{17}}}}$
✓

Answer

Correct option: A.
$2 - \frac{{11}}{{{2^{19}}}}$
a
$S = \sum\limits_{k = 1}^{20} {\frac{k}{{{2^k}}}} $

$S = \frac{1}{2} + \frac{2}{{{2^2}}} + \frac{3}{{{2^3}}} + .... + \frac{{20}}{{{2^{20}}}}$

$S\frac{1}{2} = \,\,\,\frac{1}{{{2^2}}} + \frac{2}{{{2^3}}} + .... + \frac{{20}}{{{2^{21}}}}$

$\frac{1}{2}S = \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + .... + \frac{1}{{{2^{20}}}} - \frac{{20}}{{{2^{21}}}}$

$ = \frac{{\frac{1}{2}\left( {1 - \frac{1}{{{2^{20}}}}} \right)}}{{\frac{1}{2}}} - \frac{{20}}{{{2^{21}}}}$

$ = 1 - \frac{{20}}{{{2^{21}}}}$

$S = 2 - \frac{{11}}{{{2^{19}}}}$

 

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