A) \[39+\frac{1}{{{2}^{20}}}\]
B) \[39+\frac{1}{{{2}^{19}}}\]
C) \[39+\frac{1}{{{2}^{20}}}\]
D) \[38+\frac{1}{{{2}^{19}}}\]
Correct Answer: D
Solution :
The general term of the series is \[=\frac{2*{{2}^{r}}-1}{{{2}^{r}}}\] we can write the given sum as \[1+\sum\limits_{r=1}^{19}{\frac{2*{{2}^{r}}-1}{{{2}^{r}}}}\] Now, \[\sum\limits_{r=1}^{19}{\frac{2*{{2}^{r}}-1}{{{2}^{r}}}=\sum\limits_{r=1}^{19}{2-\frac{1}{{{2}^{r}}}=2(19)-}}\] \[\frac{\frac{1}{2}(1-{{(\frac{1}{2})}^{19}})}{1-\frac{1}{2}}=38+\frac{{{(\frac{1}{2})}^{19}}-1}{1}=38+{{(\frac{1}{2})}^{19}}-1=37+{{(\frac{1}{2})}^{19}}\] As we can write the given sum as \[1+\sum\limits_{r=1}^{19}{\frac{2*{{2}^{r}}-1}{{{2}^{r}}}}\] \[\Rightarrow \]\[1+37+{{(\frac{1}{2})}^{19}}=38+{{(\frac{1}{2})}^{19}}\] Therefore correct answer is D \[{{T}_{n}}=\frac{{{2}^{n}}-1}{{{2}^{n-1}}}\] \[=2-\frac{1}{{{2}^{n-1}}}\] \[\sum{_{0}^{20}{{T}_{n}}=\sum{_{0}^{20}2-\sum{_{0}^{20}}}\frac{1}{{{2}^{n-1}}}}\] \[=2(20)-\frac{(1)\left( 1-\frac{1}{{{2}^{20}}} \right)}{1-\frac{1}{2}}\] \[=40-2\left( 1-\frac{1}{{{2}^{20}}} \right)\] \[=38+\frac{1}{{{2}^{19}}}\] Hence, the answer is option D Let \[{{S}_{n}}=1+\frac{3}{2}+\frac{7}{4}+\frac{15}{8}+\frac{31}{16}+....+{{T}_{n}}\] Divide each term by 2, \[\Rightarrow \]\[\frac{{{S}_{n}}}{2}=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\frac{31}{32}+....+{{T}_{n+1}}\] ?(i) \[\Rightarrow \]\[\frac{{{S}_{n}}}{2}=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+...+{{T}_{n}}+{{T}_{n+1}}\] \[\Rightarrow \]\[0=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{{{2}^{n}}}-{{T}_{n+1}}\] \[\Rightarrow \]\[{{T}_{n+1}}=\frac{1}{2}+\frac{1}{4}+....+\frac{1}{{{2}^{n}}}\] This is a sum of n terms of a G.P. with common ratio \[\frac{1}{2}<1\]and first term \[\frac{1}{2}\] \[\Rightarrow \]\[{{T}_{n+1}}=\frac{\frac{1}{2}(1-{{(\frac{1}{2})}^{n}})}{1-\frac{1}{2}}\] \[\Rightarrow \]\[{{T}_{n+1}}=\frac{{{2}^{n}}-1}{{{2}^{n}}}\] Now, \[\frac{{{S}_{n}}}{2}=\sum{\frac{{{2}^{n}}-1}{{{2}^{n}}}}\] \[\Rightarrow \]\[\frac{{{S}_{20}}}{2}=\sum\limits_{n=1}^{20}{\frac{{{2}^{n}}-1}{{{2}^{n}}}}\] \[\Rightarrow \]\[\frac{{{S}_{20}}}{2}=\sum\limits_{n=1}^{20}{1-\frac{1}{{{2}^{n}}}}\] \[\Rightarrow \]\[\frac{{{S}_{20}}}{2}=\sum\limits_{n=1}^{20}{1-\sum\limits_{n=1}^{20}{\frac{1}{{{2}^{n}}}}}\] \[\Rightarrow \]\[\frac{{{S}_{20}}}{2}=20-\left( \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{20} \right)\] \[\Rightarrow \]\[\frac{{{S}_{20}}}{2}=20-\left( \frac{\frac{1}{2}\left( 1-\left( \frac{1}{2} \right) \right)}{1-\frac{1}{2}} \right)\] \[\Rightarrow \]\[\frac{{{S}_{20}}}{2}=20-1+\frac{1}{{{2}^{20}}}\] \[\Rightarrow \]\[{{S}_{20}}=38+\frac{1}{{{2}^{19}}}\]
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