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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    If\[^{2n+1}{{p}_{n-1}}{{:}^{2n-1}}{{p}_{n}}=3:5,\]then the value of\[n\]is equal to

    A)  4                                            

    B)  3

    C)  2                            

    D)         1

    E)  5

    Correct Answer: A

    Solution :

     Given, \[^{2n+1}{{p}_{n-1}}{{:}^{2n-1}}{{p}_{n}}=3:5\] \[\Rightarrow \]               \[\frac{(2n+1)!}{(n+2)!}\times \frac{(n-1)!}{(2n-1)!}=\frac{3}{5}\] \[\Rightarrow \]               \[\frac{(2n+1)2n}{(n+2)(n+1)n}=\frac{3}{5}\] \[\Rightarrow \]               \[10(2n+1)=3({{n}^{2}}+3n+2)\] \[\Rightarrow \]               \[3{{n}^{2}}-11n-4=0\] \[\Rightarrow \]               \[(3n+1)(n-4)=0\] \[\Rightarrow \]               \[n=4\]                                 \[\left( n\ne -\frac{1}{3} \right)\]


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