The sum sumlimits_{k = 1}^{20} {frac{k}{{{2^k | vedclass

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Question

(A) Let $S = \sum\limits_{k = 1}^{20} {\frac{k}{{{2^k}}}} = \frac{1}{2} + \frac{2}{{{2^2}}} + \frac{3}{{{2^3}}} + \dots + \frac{{20}}{{{2^{20}}}}$Mult

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

Vedclass · Answer

(A) Let $S = \sum\limits_{k = 1}^{20} {\frac{k}{{{2^k}}}} = \frac{1}{2} + \frac{2}{{{2^2}}} + \frac{3}{{{2^3}}} + \dots + \frac{{20}}{{{2^{20}}}}$Multiply by $\frac{1}{2}$:$\frac{1}{2}S = \frac{1}{{{2^2}}} + \frac{2}{{{2^3}}} + \dots + \frac{{19}}{{{2^{20}}}} + \frac{{20}}{{{2^{21}}}}$Subtract the two equations:$S - \frac{1}{2}S = \frac{1}{2} + \left( \frac{2-1}{{{2^2}}} \right) + \left( \frac{3-2}{{{2^3}}} \right) + \dots + \left( \frac{20-19}{{{2^{20}}}} \right) - \frac{{20}}{{{2^{21}}}}$$\frac{1}{2}S = \left( \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \dots + \frac{1}{{{2^{20}}}} \right) - \frac{{20}}{{{2^{21}}}}$The term in the bracket is a Geometric Progression with $a = \frac{1}{2}$,$r = \frac{1}{2}$,and $n = 20$:Sum $= \frac{\frac{1}{2}(1 - (\frac{1}{2})^{20})}{1 - \frac{1}{2}} = 1 - \frac{1}{{{2^{20}}}}$So,$\frac{1}{2}S = 1 - \frac{1}{{{2^{20}}}} - \frac{20}{{{2^{21}}}} = 1 - \frac{2}{{{2^{21}}}} - \frac{20}{{{2^{21}}}} = 1 - \frac{22}{{{2^{21}}}} = 1 - \frac{11}{{{2^{20}}}}$$S = 2 - \frac{11}{{{2^{19}}}}$

The sum $\sum\limits_{k = 1}^{20} {\frac{k}{{{2^k}}}}$ is equal to

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