JEE Main & Advanced Mathematics Other Series Question Bank Critical Thinking

  • question_answer 5) \[1+\frac{1+2}{1\,!}+\frac{1+2+3}{2\,!}+\frac{1+2+3+4}{3\,!}+....\infty =\]

    A) 0

    B) 1

    C) \[\frac{7e}{2}\]

    D) \[2\,e\]

    Correct Answer: C

    Solution :

    \[\frac{1}{0!}+\frac{1+2}{1!}+\frac{1+2+3}{2!}+....\infty \] \[{{n}^{th}}\]term \[{{T}_{n}}=\frac{1+2+3+4+.....+n}{(n-1)!}=\frac{n(n+1)}{2(n-1)!}\] \[{{T}_{n}}=\frac{1}{2}\left[ \frac{1}{(n-3)!}+\frac{4}{(n-2)!}+\frac{2}{(n-1)!} \right]\] Therefore sum\[{{S}_{\infty }}=\frac{7e}{2}\].

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