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JEE Main & Advanced JEE Main Solved Paper-2016

  • question_answer
    If the number of terms in the expansion of \[{{\left( 1-\frac{2}{x}+\frac{4}{{{x}^{2}}} \right)}^{n}},x\ne 0,\] , is 28, then the sum of the coefficients of all the terms in this expansion, is :- [JEE Main Solved Paper-2016 ]

    A) 729        

    B) 64

    C) 2187                      

    D) 243

    Correct Answer: A

    Solution :

                    Number of terms in the expansion of \[{{\left( 1-\frac{2}{x}+\frac{4}{{{x}^{2}}} \right)}^{n}}\] is \[^{n+2}{{C}_{2}}\](considering \[\frac{1}{x}\]and \[\frac{1}{{{x}^{2}}}\]distinct) \[\therefore \]\[^{n+2}{{C}_{2}}=28\Rightarrow n=6\] \[\therefore \]Sum of coefficients = (1 ? 2 + 4)6 = 729 But number of dissimilar terms actually will be 2n + 1 (as\[\frac{1}{x}\]and\[\frac{1}{{{x}^{2}}}\]are functions as same variable) Hence it contains error, so a bonus can be expected.                                


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