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Jee Main 2025 — Medium JEE math MCQ

The value of \( \lim_{n \to \infty} \left( \sum_{k=1}^{n} \frac{k^4 + 4k^2 + 11k + 5}{(k+3)!} \right) \) is:
  1. A. \quad 4/3 \\
  2. B. \quad 2 \\
  3. C. \quad 7/3 \\
  4. D. \quad 5/3 \\

Solution

The correct option is **D**. (D. \quad 5/3 \\ \end{align*} \])

MATH

medium • PYQ Reworded
Question
Read carefully, then pick the best option.
The value of lim⁡n→∞(∑k=1nk4+4k2+11k+5(k+3)!) \lim_{n \to \infty} \left( \sum_{k=1}^{n} \frac{k^4 + 4k^2 + 11k + 5}{(k+3)!} \right) is:
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Jee Main 2025 — Medium JEE Mathematics MCQ | MyGoalPrep