⇒[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