If (2+3)n=I+f where I and n are positive  integers and 0

If (2+3)n=I+f where I and n are positive  integers and 0<f<1, then (1f)(I+f)

  1. A

    -1

  2. B

    1

  3. C

    2

  4. D

    None of these 

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    Solution:

    I+f=(2+3)n=2n+nC12n13+nC22n2(3)2+nC32n3(3)3+  (i) 

     Now,     0<23<1        0<(23)n<1

     Let (23)n=f where 0<f<1

     f=2nnC12n13+nC22n2(3)2nC32n3(3)3+ (ii

    On adding Eqs. (i) and (ii), we get

    I+f+f=22n+nC22n23+-------------(iii)

     I+f+f= eveninteger 

    Now 0<f<1

    and 0<f<10<f+f<2

    Hence, from Eq. (iii) we conclude that f + f is an
    integer between 0 and 2.

     f+f=1f=1f   ----(iv)

     I+f=(2+3)n,f=1f=(23)n

     (I+f)(1f)=[(2+3)(23)]n=(43)n=1

     (I+f)(1f)=1

     

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