2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-37

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


You need to login to perform this action.
You will be redirected in 3 sec spinner