Sequences And Series Ques 78
The sum $\sum _{k=1}^{20} k \frac{1}{2^{k}}$ is equal to
(2019 Main, 8 April II)
(a) $2-\frac{11}{2^{19}}$
(b) $1-\frac{11}{2^{20}}$
(c) $2-\frac{3}{2^{17}}$
(d) $2-\frac{21}{2^{20}}$
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Answer:
Correct Answer: 78.(a)
Solution:
Formula:
Arithmetico-Geometrical Progression (A.G.P.):
- Let $S=\sum _{k=1}^{20} k (\frac{1}{2^{k}})$
$ S=\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\frac{4}{2^{4}}+\ldots+\frac{20}{2^{20}} $ $\quad$ …….(i)
On multiplying by $(\frac{1}{2})$ both sides, we get
$ \frac{S}{2}=\frac{1}{2^{2}}+\frac{2}{2^{3}}+\frac{3}{2^{4}}+\ldots+\frac{19}{2^{20}}+\frac{20}{2^{21}} $ $\quad$ …….(ii)
On subtracting Eq. (ii) from Eq. (i), we get
$ \begin{aligned} & S-\frac{S}{2}=\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots+\frac{1}{2^{20}}-\frac{20}{2^{21}} \\ & \Rightarrow \quad \frac{S}{2}=\frac{\frac{1}{2} (1-\frac{1}{2^{20}})}{1-\frac{1}{2}}-\frac{20}{2^{21}} \\ & \because \text { sum of GP }=\frac{a\left(1-r^{n}\right)}{1-r}, r<1 \end{aligned} $
$ \begin{gathered} \frac{S}{2}=1-\frac{1}{2^{20}}-\frac{20}{2^{21}}=1-\frac{1}{2^{20}}-\frac{10}{2^{20}}=1-\frac{11}{2^{20}} \\ \Rightarrow \quad S=2-\frac{11}{2^{19}} \end{gathered} $
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