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Question

# If x=(√3+1)n such that n∈N and is an odd number then [x] is (where [x] denotes the greatest integers less than or equal to x)

A
2k, where kI
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B
2k+1, where kI
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C
4n
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D
8n
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Solution

## The correct option is A 2k, where k∈Ix=(√3+1)n= nC0(√3)n(1)0 + nC1(√3)n−1(1)1 + nC2(√3)n−2(1)2 + nC3(√3)n−3(1)3 + ...... + nCn(√3)0(1)nLet y=(√3−1)n = nC0(√3)n(−1)0 + nC1(√3)n−1(−1)1 + nC2(√3)n−2(−1)2 + nC3(√3)n−3(−1)3 + ...... + nCn(√3)0(−1)nClearly x−I=2k for some k∈I and also 0<y<1We can expand x as, x= [x]+ {x} where 0≤{x}<1 So , [x]+x −y=2k, since [x] is an integer, {x} - y must also be an integer as RHS is an integer .So x−y=0 since −1<x−y<1 So. [x]=2k for some k∈I

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