The sum \(\mathop \sum \limits_{k = 1}^{20} k\frac{1}{{{2^k}}}\)is equal to:

Let, \(S = \mathop \sum \limits_{k = 1}^{20} k\cdot\frac{1}{{{2^k}}}\) \(S = \frac{1}{2} + 2\cdot\frac{1}{{{2^2}}} + 3\cdot\frac{1}{{{2^3}}} + \ldots + 20\cdot\frac{1}{{{2^{20}}}}\) ----(1)Multiply 1/2 on both SIDES, we get\(\frac{1}{2}S = \frac{1}{{{2^2}}} + 2\cdot\frac{1}{{{2^3}}} + \ldots + 19\frac{1}{{{2^{20}}}} + 20\frac{1}{{{2^{21}}}}\) ----(2)On subtracting EQUATIONS (2) by (1)\(\frac{S}{2} = \left( {\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \ldots + \frac{1}{{{2^{20}}}}} \right) - 20\frac{1}{{{2^{31}}}}\) \(= \frac{{\frac{1}{2}\left( {1 - \;\frac{1}{{{2^{20}}}}} \right)}}{{1 - \;\frac{1}{2}}} - 20\cdot\frac{1}{{{2^{21}}}}\) \(= 1 - \frac{1}{{{2^{20}}}} - 10\cdot\frac{1}{{{2^{20}}}}\) \(\Rightarrow \frac{S}{2} = 1 - 11\cdot\frac{1}{{{2^{20}}}}\) \(\Rightarrow S = 2 - 11\cdot\frac{1}{{{2^{19}}}}\) \(\Rightarrow S = 2 - \frac{{11}}{{{2^{19}}}}\)