MCQMediumJEE Main 2025 · 29 January, Shift 1Sum of Series

Mathematics Question from JEE Main 2025 · 29 January, Shift 1

The value of limn(k=1nk3+6k2+11k+5(k+3)!)\lim_{n \to \infty} \left( \sum_{k=1}^{n} \frac{k^3 + 6k^2 + 11k + 5}{(k + 3)!} \right) is:

  • A

    43\frac{4}{3}

  • B

    53\frac{5}{3}

  • C

    22

  • D

    73\frac{7}{3}

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