Evaluation of limn→∞{(2+√3)2n}, Where n∈N.
MyTry:: Let (2+√3)2n=I+f,
where $0
Now Let 0<(2−√3)2n<1,
So (2−√3)2n=f′
.
So (2+√3)2n+(2−√3)2n=I+f+f′=IntegerQuantity.
So f+f′∈Z.
Now Given 0<(f+f′)<2. So f+f′=1∈Z
So I+f+f′=IntegerQuantity⇒f=IntegerQuantity−f′
So limn→∞{(2+√3)2n}=Integerquantity−limn→∞{(2−√3)2n}
Now how can i solve after that, help me, Thanks
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