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If the solve the problem

Question:

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

  1. $2-\frac{3}{2^{17}}$

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

  3. $1-\frac{11}{2^{20}}$

  4. $2-\frac{21}{2^{20}}$


Correct Option: , 2

Solution:

$S=\sum_{k=1}^{20} \frac{1}{2^{k}}$

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

$S \times \frac{1}{2}=\frac{1}{2^{2}}+\frac{2}{2^{3}}+\ldots+\frac{19}{2^{20}}+\frac{20}{2^{21}}$

$\Rightarrow\left(1-\frac{1}{2}\right) \mathrm{S}=\frac{1}{2}+\frac{1}{2^{2}}+\ldots+\frac{1}{2^{20}}-\frac{20}{2^{21}}$

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

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