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JEE Main & Advanced Sample Paper JEE Main Sample Paper-12

  • question_answer
    The value of the expression\[\left( 1+\frac{1}{\omega } \right)\left( 1+\frac{1}{{{\omega }^{2}}} \right)+\left( 2+\frac{1}{\omega } \right)\left( 2+\frac{1}{{{\omega }^{2}}} \right)\]\[+\left( 3+\frac{1}{\omega } \right)\left( 3+\frac{1}{{{\omega }^{2}}} \right)+.....+\left( n+\frac{1}{\omega } \right)\left( n+\frac{1}{{{\omega }^{2}}} \right)\]where \[\omega \] is an imaginary cube root of unity, is

    A)  \[\frac{n({{n}^{2}}-2)}{3}\]                        

    B)  \[\frac{n({{n}^{2}}+2)}{3}\]

    C)  \[\frac{n({{n}^{2}}-1)}{3}\]                        

    D)  none of these

    Correct Answer: B

    Solution :

     \[\left( 1+\frac{1}{\omega } \right)\left( 1+\frac{1}{{{\omega }^{2}}} \right)+\left( 2+\frac{1}{\omega } \right)\left( 2+\frac{1}{{{\omega }^{2}}} \right)\] \[+\left( 3+\frac{1}{\omega } \right)\left( 3+\frac{1}{{{\omega }^{2}}} \right)+......+\left( n+\frac{1}{\omega } \right)\left( n+\frac{1}{{{\omega }^{2}}} \right)\] Consider, \[\left( r+\frac{1}{\omega } \right)\left( r+\frac{1}{{{\omega }^{2}}} \right)\] \[={{r}^{2}}+(\omega +{{\omega }^{2}})r+1=({{r}^{2}}-r+1)\] \[=\sum\limits_{r=1}^{n}{({{r}^{2}}-r+1)}=\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+n\] \[=\frac{n}{6}[2{{n}^{2}}+3n+1-3n-3+6]\] \[=\frac{n}{6}(2{{n}^{2}}+4)=\frac{n({{n}^{2}}+2)}{3}\]


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