JEE PYQ: Sequence And Series Question 47
Question 47 - 2019 (08 Apr Shift 2)
The sum $\sum_{k=1}^{20} k \cdot \frac{1}{2^k}$ is equal to:
(1) $2 - \frac{3}{2^{17}}$
(2) $1 - \frac{11}{2^{20}}$
(3) $2 - \frac{11}{2^{19}}$
(4) $2 - \frac{21}{2^{20}}$
Show Answer
Answer: (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)
$\frac{1}{2}S = \frac{1}{2^2} + 2 \cdot \frac{1}{2^3} + \ldots + 19 \cdot \frac{1}{2^{20}} + 20 \cdot \frac{1}{2^{21}}$ …(ii)
On subtracting equations (ii) by (i),
$\frac{S}{2} = 1 - 1 \cdot \frac{1}{2^{20}} - 20 \cdot \frac{1}{2^{21}}$
$\Rightarrow S = 2 - 11 \cdot \frac{1}{2^{19}} \Rightarrow S = 2 - \frac{11}{2^{19}}$
BETA
