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Question #40636

Difficulty: 8/10 | 🟢 Free

The Question

Let $M_n$ be the greatest binomial coefficient in the expansion of $(1+x)^{2n}$ , and let $A_n = \sum_{k=0}^{2n} \binom{2n}{k} e^{k/n}$ . Given the asymptotic expansion $n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \exp\left(\frac{1}{12n}\right)$ for large $n$ , evaluate the limit: $\lim_{n \to \infty} \left( \frac{A_n}{M_n \sqrt{\pi n} \, e} \right)^n$

Options

A.
$e^{1/4}$
B.
$e^{3/8}$
C.
$e^{1/2}$
D.
$e^{5/8}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Greatest Binomial Coefficient|Sigma Notation Properties|Indeterminate Forms

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous evaluation of $1^\infty$ indeterminate forms involving combinatorial sums and asymptotic approximations.

The primary trap lies in premature truncation of Taylor series and Stirling's approximation.

Students often use the first-order approximations $M_n \sim \frac{4^n}{\sqrt{\pi n}}$ and $1+e^{1/n} \sim 2+1/n$ , which completely misses the $O(1/n)$ correction terms in the exponent.

When raised to the power of $n$ , these $O(1/n)$ terms become $O(1)$ constants ( $e^{-1/8}$ and $e^{1/4}$ respectively), dominating the final limit.

Failing to account for the second-order derivative in the Maclaurin expansion of $\ln(1+e^t)$ leads to the incorrect option A.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Binomial Theorem - Class 11
Concept	Greatest Binomial Coefficient
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free