Question

The sum of $\frac{3}{1\cdot 2}\left(\frac{1}{2}\right)+\frac{4}{2\cdot 3}{\left(\frac{1}{2}\right)}^2+\frac{5}{3\cdot 4}{\left(\frac{1}{2}\right)}^3+\cdots$ upto $20$ terms is

• A. $1-\frac{1}{21}{\left(\frac{1}{2}\right)}^{20}$
• B. $1+\frac{1}{21}{\left(\frac{1}{2}\right)}^{20}$
• C. $1-\frac{1}{20}{\left(\frac{1}{2}\right)}^{21}$
• D. $1+\frac{1}{20}{\left(\frac{1}{2}\right)}^{21}$

Answer 1

The correct option is A $1-\frac{1}{21}{\left(\frac{1}{2}\right)}^{20}$

Given : $S_n=\frac{3}{1\cdot 2}\left(\frac{1}{2}\right)+\frac{4}{2\cdot 3}{\left(\frac{1}{2}\right)}^2+\frac{5}{3\cdot 4}{\left(\frac{1}{2}\right)}^3+\cdots \Rightarrow S_{20}=\sum _{n=1}^{20}\frac{n+2}{n(n+1)}{\left(\frac{1}{2}\right)}^n\Rightarrow S_{20}=\sum _{n=1}^{20}\frac{2(n+1)-n}{n(n+1)}{\left(\frac{1}{2}\right)}^n\Rightarrow S_{20}=\sum _{n=1}^{20}[\frac{1}{n}{\left(\frac{1}{2}\right)}^{n-1}-\frac{1}{n+1}{\left(\frac{1}{2}\right)}^n]\Rightarrow S_{20}=1-\frac{1}{2}\left(\frac{1}{2}\right)+\frac{1}{2}\left(\frac{1}{2}\right)-\frac{1}{3}{\left(\frac{1}{2}\right)}^2+\frac{1}{3}{\left(\frac{1}{2}\right)}^2-\frac{1}{4}{\left(\frac{1}{2}\right)}^3⋮+\frac{1}{20}{\left(\frac{1}{2}\right)}^{19}-\frac{1}{21}{\left(\frac{1}{2}\right)}^{20}\therefore S_{20}=1-\frac{1}{21}{\left(\frac{1}{2}\right)}^{20}$