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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    Let\[{{(1+x)}^{n}}=1+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+.....+{{a}_{n}}{{x}^{n}}\].If\[{{a}_{1}},{{a}_{2}}\]and\[{{a}_{3}}\]are in AP, then the value of n is

    A)  4                                            

    B)  5

    C)  6                            

    D)         7

    E)  8

    Correct Answer: D

    Solution :

    It is given that,\[{{a}_{1}}{{=}^{n}}{{C}_{1}},{{a}_{2}}{{=}^{n}}{{C}_{2}}\]and\[{{a}_{3}}{{=}^{n}}{{C}_{3}}\]are in AP. \[\Rightarrow \]               \[{{2}^{n}}{{C}_{2}}{{=}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{3}}\] \[\Rightarrow \]               \[\frac{2n(n-1)}{2!}=n+\frac{n(n-1)(n-2)}{3!}\] \[\Rightarrow \]               \[n(n-1)=n+\frac{n(n-1)(n-2)}{6}\] \[\Rightarrow \]               \[6n-6=6+{{n}^{2}}-3n+2\] \[\Rightarrow \]               \[{{n}^{2}}-9n+14=0\] \[\Rightarrow \]               \[n=7,2\] If\[n=2,\]then there are only three terms in the expansion of\[{{(1+x)}^{n}}\]. Therefore;\[n=7\].


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