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JEE Main & Advanced Mathematics Sequence & Series Question Bank nth term of special series, Sum to n terms and Infinite number of terms

  • question_answer
    The sum 1(1!) + 2(2!) + 3(3!) + ....+n (n!) equals  [AMU 1999; DCE 2005]

    A) \[3\,(n\,!)\,+\,n-3\]

    B) \[(n+1)!\,-\,(n-1)!\]

    C) \[(n+1)\,!\,-1\]

    D) \[2\,(n\,!)-2n-1\]

    Correct Answer: C

    Solution :

    \[{{S}_{n}}=1(1!)+2(2!)+3(3!)+.....+n(n!)\] =\[(2-1)(1!)+(3-1)(2!)+(4-1)(3!)+.....\]\[+[(n+1)-1](n!)\] = \[(2.1!-1!)+(3.2!-2!)+(4.3!-3!)+....\]\[+[(n+1)(n!)-(n!)]\] =\[(2!-1!)+(3!-2!)+(4!-3!)+....+[(n+1)!-(n)!]\] = \[(n+1)!-1!\].


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