If(5+26)n=I+f;n,I∈N and 0≤f<1, then I is equal to
1f−f
11+f−f
11+f+f
11−f−f
∵I+f+f′=(5+26)n+(5−26)n=2k [even integer]
∴ f+f′=1
Now, (I+f)f′=(5+26)n(5−26)n=(1)n=1
⇒ (I+f)(1−f)=1 or I=1(1−f)−f