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Questions  

If(5+26)n=I+f;n,IN and 0f<1, then I is equal to 

a
1f−f
b
11+f−f
c
11+f+f
d
11−f−f

detailed solution

Correct option is D

∵I+f+f′=(5+26)n+(5−26)n=2k  [even integer]∴ f+f′=1Now, (I+f)f′=(5+26)n(5−26)n=(1)n=1⇒ (I+f)(1−f)=1 or I=1(1−f)−f

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