A) \[{{2}^{n-1}}\]
B) \[{{2}^{n+1}}\]
C) \[{{2}^{n+2}}\]
D) not divisible by 2
Correct Answer: B
Solution :
\[R={{(3+\sqrt{5})}^{2n}},\,\,G\,\,=\,\,{{(3-\sqrt{5})}^{2n}}\] Let \[\left[ R \right] + 1 = I\] (\[\because \] [.] greatest integer function) \[\Rightarrow \,\,\,R+G=I\,\,\,\,\,\,\,\,\,(\because \,\,\,0<G<1)\] \[{{(3+\sqrt{5})}^{2n}}+{{(3-\sqrt{5})}^{2n}}=\,\,I\] seeing the option put \[\operatorname{n}= 1\] \[\operatorname{I} = 28 is divisible by 4 i.e. {{2}^{n+1}}\]
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