A) \[{{2}^{n+1}}-n\]
B) \[{{2}^{n+1}}-n-2\]
C) \[{{2}^{n}}-n-2\]
D) None of these
Correct Answer: B
Solution :
Let \[{{T}_{n}}\] be the \[{{n}^{th}}\] term and \[S\] the sum upto \[n\] terms. \[S=1+3+7+15+31+......+{{T}_{n}}\] Again \[S=1+3+7+15+...........\text{ }+{{T}_{n-1}}+{{T}_{n}}\] Subtracting, we get \[0=1+\left\{ 2+4+8+...({{T}_{n}}-{{T}_{n-1}}) \right\}-{{T}_{n}}\] \[\therefore \ \ {{T}_{n}}=1+2+{{2}^{2}}+{{2}^{3}}+.....\text{upto}\ n\ \text{terms}\] \[=\frac{1({{2}^{n}}-1)}{2-1}={{2}^{n}}-1\] Now \[S=\Sigma {{T}_{n}}=\Sigma {{2}^{n}}-\Sigma 1\] \[=(2+{{2}^{2}}+{{2}^{3}}+......+{{2}^{n}})-n\] \[=2\left( \frac{{{2}^{n}}-1}{2-1} \right)-n={{2}^{n+1}}-2-n\]. Aliter: \[1+3+7+......+{{T}_{n}}\] \[=2-1+{{2}^{2}}-1+{{2}^{3}}-1+..........+{{2}^{n}}-1\] \[=(2+{{2}^{2}}+......+{{2}^{n}})-n={{2}^{n+1}}-2-n\]. Trick: Check the options for\[n=1,\ 2\].
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