A) \[{{2}^{n}}-1\]
B) \[1-{{2}^{n}}\]
C) \[n+{{2}^{n}}-1\]
D) \[n-1+{{2}^{-n}}\]
E) \[n-{{2}^{n}}-1\]
Correct Answer: D
Solution :
\[\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+....\]up to n terms \[=\left( 1-\frac{1}{2} \right)+\left( 1-\frac{1}{4} \right)+\left( 1-\frac{1}{8} \right).....\]up to n terms \[=(1+1+1+....n\,terms)\] \[-\left( \frac{1}{2}+\frac{1}{{{2}^{2}}}+\frac{1}{{{2}^{3}}}+....\,n\,terms \right)\] \[=n-\frac{\frac{1}{2}\left( 1-\frac{1}{{{2}^{n}}} \right)}{1-\frac{1}{2}}=n-1+\frac{1}{{{2}^{n}}}\] \[=n-1+{{2}^{-n}}\]
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