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11th Class Mathematics Complex Numbers and Quadratic Equations Question Bank Critical Thinking

  • question_answer
    Let \[\omega \] is an imaginary cube roots of unity then the value of\[2(\omega +1)({{\omega }^{2}}+1)+3(2\omega +1)(2{{\omega }^{2}}+1)+.....\]\[+(n+1)(n\omega +1)(n{{\omega }^{2}}+1)\]  is [Orissa JEE 2002]

    A) \[{{\left[ \frac{n(n+1)}{2} \right]}^{2}}+n\]

    B) \[{{\left[ \frac{n(n+1)}{2} \right]}^{2}}\]

    C) \[{{\left[ \frac{n(n+1)}{2} \right]}^{2}}-n\]

    D) None of these

    Correct Answer: A

    Solution :

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


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