A) \[{{2}^{n}}-n+1\]
B) \[1-{{2}^{n}}\]
C) \[n+{{2}^{-n}}-1\]
D) \[{{2}^{n}}-1\]
Correct Answer: C
Solution :
Let\[{{S}_{n}}\]be the sum of first\[n\]terms of the series \[\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+...\] \[\therefore \] \[{{S}_{n}}=\left( 1-\frac{1}{2} \right)+\left( 1-\frac{1}{4} \right)+\left( 1-\frac{1}{8} \right)+...\] \[+\left( 1-\frac{1}{{{2}^{n}}} \right)\] \[=n-\frac{1}{2}\frac{\left( 1-\frac{1}{{{2}^{n}}} \right)}{\left( 1-\frac{1}{2} \right)}\] \[=n-1+\frac{1}{{{2}^{n}}}=n-1+{{2}^{-n}}\]
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