A) \[n-1+\frac{1}{{{2}^{n}}}\]
B) \[n+\frac{1}{{{2}^{n}}}\]
C) \[2n+\frac{1}{{{2}^{n}}}\]
D) \[n+1+\frac{1}{{{2}^{n}}}\]
Correct Answer: A
Solution :
\[\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+...\]upto n terms \[=\left( 1-\frac{1}{2} \right)+\left( 1-\frac{1}{4} \right)+\left( 1-\frac{1}{8} \right)+...\]upto n terms \[=(1+1+...n\,times)\] \[-\left[ \frac{1}{2}+{{\left( \frac{1}{2} \right)}^{2}}+{{\left( \frac{1}{2} \right)}^{3}}+...n\,\text{terms} \right]\] \[=n-\frac{\frac{1}{2}\left[ 1-{{\left( \frac{1}{2} \right)}^{n}} \right]}{\left( 1-\frac{1}{2} \right)}\] \[=n-1+\frac{1}{{{2}^{n}}}.\] Alternate Method: Let \[\frac{\begin{align} & S=1+3+7+15+...+{{t}_{n}} \\ & S=\,\,\,\,\,\,\,\,\,\,1+3+7+...\,\,\,\,\,+{{t}_{n}} \\ \end{align}}{\begin{align} & 0=1+2+4+8+...\,\,\,\,\,\,\,\,\,\,\,\,\,{{t}_{n}} \\ & {{t}_{n}}=1+2+{{2}^{2}}+{{2}^{3}}+...\,n\,\text{terms} \\ \end{align}}\] \[=\frac{1({{2}^{n}}-1)}{2-1}={{2}^{n}}-1\] \[\therefore \]\[nth\]terms of given series. \[{{T}_{n}}=\frac{{{2}^{n}}-1}{{{2}^{n}}}=1-\frac{1}{{{2}^{n}}}.\] Required sum \[=\sum{{{T}_{n}}=\sum{\left( 1-\frac{1}{{{2}^{n}}} \right)}}\] \[=\sum{1}-\sum{\left( \frac{1}{{{2}^{n}}} \right)}\] \[=n-\left( \frac{1}{2}+\frac{1}{{{2}^{2}}}+...+\frac{1}{{{2}^{n}}} \right)\] \[=n-\frac{\frac{1}{2}\left[ 1-{{\left( \frac{1}{2} \right)}^{n}} \right]}{\left( 1-\frac{1}{2} \right)}\] \[=n-1+\frac{1}{{{2}^{n}}}.\]
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