A) \[{{2}^{n}}-n-1\]
B) \[1-{{2}^{-n}}\]
C) \[n+{{2}^{-n}}-1\]
D) \[{{2}^{n}}-1\]
Correct Answer: C
Solution :
The sum of the first \[n\] terms is \[{{S}_{n}}=\left( 1-\frac{1}{2} \right)+\left( 1-\frac{1}{{{2}^{2}}} \right)+\left( 1-\frac{1}{{{2}^{3}}} \right)+\left( 1-\frac{1}{{{2}^{4}}} \right)\]\[+......+\left( 1-\frac{1}{{{2}^{n}}} \right)\] \[=n-\left\{ \frac{1}{2}+\frac{1}{{{2}^{2}}}+.....+\frac{1}{{{2}^{n}}} \right\}\] =\[n-\frac{1}{2}\left( \frac{1-\frac{1}{{{2}^{n}}}}{1-\frac{1}{2}} \right)=n-\left( 1-\frac{1}{{{2}^{n}}} \right)=n-1+{{2}^{-n}}\]. Trick: Check for \[n=1,\ 2\ i.e.\] \[{{S}_{1}}=\frac{1}{2},\ {{S}_{2}}=\frac{5}{4}\] and (c) \[\Rightarrow {{S}_{1}}=\frac{1}{2}\] and \[{{S}_{2}}=2+{{2}^{-2}}-1=\frac{5}{4}\].
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