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

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

    A) \[2\,e\]

    B) \[3\,e\]

    C) \[(0.5)-\frac{{{(0.5)}^{2}}}{2}+\frac{{{(0.5)}^{3}}}{3}-\frac{{{(0.5)}^{4}}}{4}+....\]

    D) \[5\,e\]

    Correct Answer: D

    Solution :

    \[S=1+\frac{{{2}^{2}}}{1\ !}+\frac{{{3}^{2}}}{2\ !}+\frac{{{4}^{2}}}{3\ !}+.......+\frac{{{n}^{2}}}{(n-1)\ !}+.......\infty \] Here \[{{T}_{n}}=\frac{{{n}^{2}}}{(n-1)\ !}=\frac{(n-1)(n-2)}{(n-1)\ !}+\frac{3(n-1)+1}{(n-1)\ !}\]                     \[=\frac{1}{(n-3)\ !}+\frac{3}{(n-2)\ !}+\frac{1}{(n-1)\ !}\] \[\Rightarrow S=\Sigma {{T}_{n}}=e+3e+e=5e\].


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