Question

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

• A. $2-\frac{11}{2^{19}}$
• B. $2-\frac{21}{2^{20}}$
• C. $1-\frac{11}{2^{20}}$
• D. $2-\frac{3}{2^{17}}$

Answer 1

The correct option is A $2-\frac{11}{2^{19}}$

Let $S=\sum _{k=1}^{20}k\frac{1}{2^k}$
$S=\frac{1}{2}+2\cdot \frac{1}{2^2}+3\cdot \frac{1}{2^3}+\cdots +20\cdot \frac{1}{2^{20}}\cdots \left(1\right)$
$\frac{1}{2}S=\frac{1}{2^2}+2\cdot \frac{1}{2^3}+\cdots +19\cdot \frac{1}{2^{20}}+20\cdot \frac{1}{2^{21}}\cdots \left(2\right)$

Subtracting $\left(1\right)$ and $\left(2\right)$, we get
$\frac{1}{2}S=(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots +\frac{1}{2^{20}})-20\cdot \frac{1}{2^{21}}$
$\Rightarrow \frac{1}{2}S=\frac{\frac{1}{2}(1-\frac{1}{2^{20}})}{\frac{1}{2}}-20\cdot \frac{1}{2^{21}}$
$\Rightarrow S=2-\frac{11}{2^{19}}$