JEE Main 2025January, Unspecified (session-level dataset)MathematicsMediumSequence and Series179× since 2002The value of limn→∞(∑k=1nk3+6k2+11k+6(k+3)!)\lim_{n \to \infty} \left( \sum_{k=1}^n \frac{k^3 + 6k^2 + 11k + 6}{(k+3)!} \right)limn→∞(∑k=1n(k+3)!k3+6k2+11k+6) is:AOption A. 43\frac{4}{3}34BOption B. 2COption C. 73\frac{7}{3}37DOption D. 53\frac{5}{3}35CheckStep-by-step hintSkip · show solution
