AP EAMCET Engineering 23 Sep 2020 Shift 1 Paper

It is given that.
(5√5+11)2n+1=R=[R]+f
and (5√5−11)2n+1=f′, where f,f′∈{0,1)
and On subtracting both, we get
⇒[R]+f−f′= even integer
∵[R] is an integer so, f−f′ must be integer and as f∈(0,1) and f′(0,1)
∴−1<f−f′<1, so possible integral value of f−f′ is zero.
∴‌‌f=f′
Now, (5√5+11)2n+1(5√5−11)2n+1=Rf′
⇒‌‌(125−121)2n+1=Rf {∵f′=f}
⇒‌‌Rf=42n+1