JEE Main & Advanced Mathematics Other Series Question Bank Critical Thinking

  • question_answer 6) \[1.5+\frac{2.6}{1\,!}+\frac{3.7}{2\,!}+\frac{4.8}{3\,!}+.....\] is equal to

    A) \[13\,e\]

    B) \[15\,e\]

    C) \[9\,e+1\]

    D) \[5\,e\]

    Correct Answer: A

    Solution :

    \[1.5+\frac{2.6}{1!}+\frac{3.7}{2!}+\frac{4.8}{3!}+....\] \[{{T}_{n}}=\frac{n(n+4)}{(n-1)!}=\frac{(n-1)(n+4)}{(n-1)!}+\frac{(n+4)}{(n-1)!}\]   \[=\frac{n+4}{(n-2)!}+\frac{1}{(n-2)!}+\frac{5}{(n-1)!}\] \[=\frac{1}{(n-3)!}+\frac{7}{(n-2)!}+\frac{5}{(n-1)!}\] \[{{S}_{\infty }}=\sum\limits_{n=1}^{\infty }{\frac{1}{(n-3)!}+7\sum\limits_{n=1}^{\infty }{\frac{1}{(n-2)!}+5\sum\limits_{n=1}^{\infty }{\frac{1}{(n-1)!}}}}\]              \[=e+7e+5e=13e\].

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