CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    \[\left( 1+\frac{{{C}_{1}}}{{{C}_{0}}} \right)\left( 1+\frac{{{C}_{2}}}{{{C}_{1}}} \right)\left( 1+\frac{{{C}_{3}}}{{{C}_{2}}} \right).....\left( 1+\frac{{{C}_{n}}}{{{C}_{n-1}}} \right)\]is equal to:

    A)  \[\frac{n+1}{n!}\]                          

    B)  \[\frac{{{(n+1)}^{n}}}{(n-1)!}\]

    C)  \[\frac{{{(n-1)}^{n}}}{n!}\]         

    D)         \[\frac{{{(n+1)}^{n}}}{n!}\]

    E)  \[\frac{n-1}{n!}\]

    Correct Answer: D

    Solution :

    \[\left( 1+\frac{{{C}_{1}}}{{{C}_{0}}} \right)\left( 1+\frac{{{C}_{2}}}{{{C}_{1}}} \right)\left( 1+\frac{{{C}_{3}}}{{{C}_{2}}} \right)....\left( 1+\frac{{{C}_{n}}}{{{C}_{n-1}}} \right)\] \[=\left( 1+\frac{n}{1} \right)\left( 1+\frac{n(n-1)}{2n} \right).....\left( 1+\frac{1}{n} \right)\] \[=\left( \frac{1+n}{1} \right)\left( \frac{1+n}{2} \right).....\left( \frac{1+n}{n} \right)\] \[=\frac{{{(1+n)}^{n}}}{n!}\]


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