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JEE Main & Advanced Mathematics Binomial Theorem and Mathematical Induction Question Bank Self Evaluation Test - Principle of Mathematical Induction

  • question_answer
    If \[P=n({{n}^{2}}-{{1}^{2}})({{n}^{2}}-{{2}^{2}})({{n}^{2}}-{{3}^{2}})...({{n}^{2}}-{{r}^{2}}),\] \[n>r,n\in N\] then P is necessarily divisible by

    A) \[(2r+2)!\]

    B) \[(2r+4)!\]

    C) \[(2r+1)!\]

    D) None of these

    Correct Answer: C

    Solution :

    [c] \[P=n(n+1)(n-1)(n+2)(n-2)...(n+r)(n-r).\] \[=\{n(n+1)(n+2)...(n+r)\}\{(n-1)(n-2)...(n-r)\}\] \[=(n+r)(n+r-1)...(n+1)(n)(n-1)...(n-r)\] Clearly P is product of \[(2r+1)\] consecutive integers, so divisible by \[(2r+1)!\]


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