A) \[{{2}^{8}}-\,1\]
B) \[{{2}^{16}}-\,1\]
C) \[{{2}^{32}}-\,1\]
D) \[{{2}^{64}}-\,1\]
Correct Answer: B
Solution :
[b] According to question, \[(2+1)\,\,({{2}^{2}}+1)\,\,({{2}^{4}}+1)\,\,({{2}^{8}}+1)=?\] \[\Rightarrow (2+1)\,\,({{2}^{8}}+1)\,\,({{2}^{2}}+1)\,\,({{2}^{4}}+1)\] \[\Rightarrow ({{2}^{9}}+2+{{2}^{8}}+1)\,\,({{2}^{6}}+{{2}^{2}}+{{2}^{4}}+1)\] \[\Rightarrow ({{2}^{15}}+{{2}^{14}}+{{2}^{13}}+{{2}^{12}}+{{2}^{11}}+{{2}^{10}}+{{2}^{9}}+{{2}^{8}}+{{2}^{7}}+{{2}^{6}}+{{2}^{5}}+\]\[{{2}^{8}}+{{2}^{7}}+{{2}^{6}}+{{2}^{5}}+{{2}^{4}}+{{2}^{3}}+{{2}^{2}}+2)+1\] Here, a=2, n=15 r=2 \[\therefore \,\,Sn=\frac{a\,\,({{r}^{n}}-1)}{r-1}\] \[Sn=\frac{2\,\,\left( {{2}^{15}}-1 \right)}{2-1}=2\,\,\left( {{2}^{15}}-1 \right)\] \[\Rightarrow \left( {{2}^{16}}-2 \right)\] \[\therefore \,\,\left( {{2}^{16}}-2 \right)+1=\left( {{2}^{16}}-1 \right)\]\[\therefore \,\,\left( 2+1 \right)\,\,\left( {{2}^{2}}+1 \right)\,\,\left( {{2}^{4}}+1 \right)\,\,\left( {{2}^{8}}+1 \right)=\left( {{2}^{16}}-1 \right)\]
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