2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Sequence & Series

JEE Main & Advanced Mathematics Sequence & Series Question Bank Exponential series

  • question_answer
    \[\frac{{{1}^{2}}.2}{1\,!}+\frac{{{2}^{2}}.3}{2\,!}+\frac{{{3}^{2}}.4}{3\,!}+.....\infty =\] [UPSEAT 1999]

    A) \[6\,e\]

    B) \[7\,e\]

    C) \[8\,e\]

    D) \[9\,e\]

    Correct Answer: B

    Solution :

    \[S=\frac{{{1}^{2}}.2}{1!}+\frac{{{2}^{2}}.3}{2!}+\frac{{{3}^{2}}.4}{3!}+....\] Here \[{{T}_{n}}=\frac{{{n}^{2}}.(n+1)}{n!}=\frac{n(n+1)}{(n-1)!}\]             \[=\frac{(n-1)(n-2)+4n-2}{(n-1)!}=\frac{1}{(n-3)!}+\frac{4(n-1)+2}{(n-1)!}\] \[\frac{1}{(n-3)!}\] \[\Rightarrow \,\,\,S=\sum{{{T}_{n}}=e+4e+2e=7e}\].


You need to login to perform this action.
You will be redirected in 3 sec spinner