Question:
The sum $\sum_{k=1}^{20} k \frac{1}{2^{k}}$ is equal to :
Correct Option: , 3
Solution:
Let, $S=\sum_{k=1}^{20} k \cdot \frac{1}{2^{k}}$
$S=\frac{1}{2}+2 \cdot \frac{1}{2^{2}}+3 \cdot \frac{1}{2^{3}}+\ldots .+20 \cdot \frac{1}{2^{20}}$......(i)
$\frac{1}{2} S=\frac{1}{2^{2}}+2 \cdot \frac{1}{2^{3}}+\ldots+19 \frac{1}{2^{20}}+20 \frac{1}{2^{21}}$...(ii)
On subtracting equations (ii) by (i),
$\frac{S}{2}=\left(\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots .+\frac{1}{2^{20}}\right)-20 \frac{1}{2^{21}}$
$=\frac{\frac{1}{2}\left(1-\frac{1}{2^{20}}\right)}{1-\frac{1}{2}}-20 \cdot \frac{1}{2^{21}}=1-\frac{1}{2^{20}}-10 \cdot \frac{1}{2^{20}}$
$\frac{S}{2}=1-11 \cdot \frac{1}{2^{20}} \Rightarrow S=2-11 \cdot \frac{1}{2^{19}}=2-\frac{11}{2^{19}}$