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Q.

If a term independent of x in the expansion of x+1x2n exists, then one of the possible value of n is

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a

2012

b

2011

c

2010

d

2009

answer is A.

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Detailed Solution

  x+1x2nG.T=Cr  n (x)n-r  1x2r =Cr  n  xn-3r

For Independent term  n-3r=0

                                                 n=3r

     n must be a multiple of 3

     n=2010

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