Then sum _ k = 1 ^ 20 k1 2^k is equal to - Vidyadip

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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