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 = \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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