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JEE Main & Advanced Mathematics Binomial Theorem and Mathematical Induction Question Bank Binomial theorem for any index

  • question_answer
    \[{{\sum\limits_{k=1}^{n}{k\left( 1+\frac{1}{n} \right)}}^{k-1}}=\] [EAMCET 2002; Pb. CET 2002]

    A) \[n(n-1)\]

    B) \[n(n+1)\]

    C) \[{{n}^{2}}\]

    D) \[{{(n+1)}^{2}}\]

    Correct Answer: C

    Solution :

    \[\sum\limits_{k\,=1}^{n}{k\,{{\left( 1+\frac{1}{n} \right)}^{k-1}}}\] \[=1+2{{\left( 1+\frac{1}{n} \right)}^{1}}+3{{\left( 1+\frac{1}{n} \right)}^{2}}+....\]upto n terms = \[1+2t+3{{t}^{2}}+...\]upto n terms = \[{{(1-t)}^{-2}}={{\left[ 1-\left( 1+\frac{1}{n} \right) \right]}^{-2}}\] = \[{{\left( \frac{1}{n} \right)}^{-2}}={{(n)}^{2}}={{n}^{2}}\].


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