A) \[{{2}^{100}}-102\]
B) \[{{2}^{101}}-102\]
C) \[{{2}^{102}}-103\]
D) \[{{2}^{102}}\text{ }-104\]
Correct Answer: B
Solution :
Let \[S=1+3+7+15+31+....+{{T}_{n}}\] \[\frac{S=1+3+7+15+.....+{{T}_{n}}}{On\,\,subtracting\,\,0=(1+2+4+8+16+....}\] \[\overline{+n\,terms)-{{T}_{n}}}\] \[\Rightarrow \]\[{{T}_{n}}=1\{2+4+8+16+...+(n-1)\,terms\}\] \[\Rightarrow \] \[{{T}_{n}}=1+\{a=2,\,r=2,\,n=1\,terms\}\] \[\Rightarrow \]\[{{T}_{n}}=1\,+\frac{2({{2}^{n-1}}-1)}{2-1}\] \[\left[ \because \,\,{{S}_{n}}=\frac{a({{r}^{n}}-1)}{r-1} \right]\] \[\Rightarrow \] \[{{T}_{n}}=1+2({{2}^{n-1}}-1)\] \[\Rightarrow \] \[{{T}_{n}}=1+{{2}^{n}}-2\] \[\Rightarrow \] \[{{T}_{n}}={{2}^{n}}-1\] Now, \[S=\Sigma {{T}_{n}}=\Sigma {{2}^{n}}-\Sigma 1\] \[=({{2}^{1}}+{{2}^{2}}+{{2}^{3}}+....+{{2}^{n}})-n\] \[=\frac{2({{2}^{n}}-1)}{2-1}-n={{2}^{n+1}}-2-n\] If \[n=100.\] Then, \[S={{2}^{101}}-2-100\] \[={{2}^{101}}-102\]
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