Question

The sum ${\displaystyle \sum _{k=1}^{20}k\frac{1}{2^k}}$ is equal to ______.

विकल्प

• ${\displaystyle 2-\frac{3}{2^{17}}}$

• ${\displaystyle 1-\frac{11}{2^{20}}}$

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

• ${\displaystyle 2-\frac{21}{2^{20}}}$

Answer 1

The sum ${\displaystyle \sum _{k=1}^{20}k\frac{1}{2^k}}$ is equal to ${\displaystyle \underline{2-\frac{11}{2^{19}}}}$.

Explanation:

Let, S = ${\displaystyle \sum _{k=1}^{20}k.\frac{1}{2^k}}$

S = ${\displaystyle \frac{1}{2}+2.\frac{1}{2^2}+3.\frac{1}{2^3}+....+20.\frac{1}{2^{20}}}$  ...(i)

${\displaystyle \frac{1}{2}S=\frac{1}{2^2}+2.\frac{1}{2^3}+....+19\frac{1}{2^{20}}+20\frac{1}{2^{21}}}$  ...(ii)

On subtracting equations (ii) by (i),

${\displaystyle \frac{S}{2}=(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{20}})-20\frac{1}{2^{21}}}$

= ${\displaystyle \frac{\frac{1}{2}(1-\frac{1}{2^{20}})}{1-\frac{1}{2}}-20.\frac{1}{2^{21}}}$

= ${\displaystyle 1-\frac{1}{2^{20}}-10.\frac{1}{2^{20}}}$

${\displaystyle \frac{S}{2}=1-11.\frac{1}{2^{20}}}$

⇒ S = ${\displaystyle 2-11.\frac{1}{2^{19}}}$

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