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Question

If x=(3+1)n, where n is odd positive integer, then [x] is
(where [x] denotes the greatest integer 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 kI
x=(3+1)n=[x]+f
 0<f<1
Let f=(31)n
0<f<11<ff<1

[x]+ff=(3+1)n(31)n=(1+3)n+(13)n(n odd positive integer)=2[1+ nC2(3)2+ nC4(3)4+]=2k, kIff=2k[x]ff=integer

As 1<ff<1, so
ff=0
So,
[x]=2k,kI

Mathematics

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