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BCECE Engineering BCECE Engineering Solved Paper-2012

  • 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

    Correct Answer: A

    Solution :

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


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