The sum sumlimits_{k = 1}^{20} {kfrac{1}{{{2^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $S = \sum\limits_{k = 1}^{20} {\frac{k}{{{2^k}}}} = \frac{1}{2} + \frac{2}{{{2^2}}} + \frac{3}{{{2^3}}} + \dots + \frac{20}{{{2^{20}}}}$Multip

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $S = \sum\limits_{k = 1}^{20} {\frac{k}{{{2^k}}}} = \frac{1}{2} + \frac{2}{{{2^2}}} + \frac{3}{{{2^3}}} + \dots + \frac{20}{{{2^{20}}}}$Multiply by $\frac{1}{2}$:$\frac{1}{2}S = \frac{1}{{{2^2}}} + \frac{2}{{{2^3}}} + \dots + \frac{19}{{{2^{20}}}} + \frac{20}{{{2^{21}}}}$Subtract the two equations:$S - \frac{1}{2}S = \frac{1}{2} + \left( \frac{2-1}{{{2^2}}} \right) + \left( \frac{3-2}{{{2^3}}} \right) + \dots + \left( \frac{20-19}{{{2^{20}}}} \right) - \frac{20}{{{2^{21}}}}$$\frac{1}{2}S = \left( \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \dots + \frac{1}{{{2^{20}}}} \right) - \frac{20}{{{2^{21}}}}$The sum inside the parenthesis is a geometric progression with $a = \frac{1}{2}$,$r = \frac{1}{2}$,and $n = 20$:Sum $= \frac{\frac{1}{2}(1 - (\frac{1}{2})^{20})}{1 - \frac{1}{2}} = 1 - \frac{1}{{{2^{20}}}}$So,$\frac{1}{2}S = 1 - \frac{1}{{{2^{20}}}} - \frac{20}{{{2^{21}}}} = 1 - \frac{2}{{{2^{21}}}} - \frac{20}{{{2^{21}}}} = 1 - \frac{22}{{{2^{21}}}} = 1 - \frac{11}{{{2^{20}}}}$$S = 2 - \frac{11}{{{2^{19}}}}$

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

Similar Questions

Find the sum $S_n = 1 + 2x + 3x^2 + 4x^3 + \dots$ up to $n$ terms.

Suppose $a_1, a_2, 2, a_3, a_4$ are in an arithmetico-geometric progression. If the common ratio of the corresponding geometric progression is $2$ and the sum of all $5$ terms of the arithmetico-geometric progression is $\frac{49}{2}$,then $a_4$ is equal to $...........$.

The sum of $1 + \frac{2}{5} + \frac{3}{5^2} + \frac{4}{5^3} + \dots$ up to $n$ terms is

If the $r^{th}$ term of a series is $(2r + 1)2^{-r}$,what is the sum of its infinite terms?

If $\tan \theta = \sqrt{\frac{3}{2}}$,the sum of the infinite series $1 + 2(1 - \cos \theta) + 3(1 - \cos \theta)^2 + 4(1 - \cos \theta)^3 + \dots \infty$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module