Question

Let \( a_n \) denote the term independent of \( x \) in the expansion of \[ \left[x + \frac{\sin(1/n)}{x^2}\right]^{3n}, \] then \[ \lim_{n\to\infty} \frac{(a_n)n!}{\,{}^{3n}P_n} \] equals:

• A. \( 0 \)
• B. \( 1 \)
• C. \( e \)
• D. \( \frac{e}{\sqrt{3}} \)

Hint:

For constant term in binomial: \begin{itemize} \item Match powers carefully. \item Use asymptotics like \( (1+1/n)^n \to e \). \end{itemize}

Answer 1

The Correct Option is C

Concept: General term: \[ T_{k+1} = \binom{3n}{k} x^{3n-k} \left(\frac{\sin(1/n)}{x^2}\right)^k \] Power of \( x \): \[ 3n - k - 2k = 3n - 3k \] For constant term: \[ 3n - 3k = 0 \Rightarrow k = n \] Step 1: {\color{red}Find constant term.} \[ a_n = \binom{3n}{n} \sin^n\!\left(\frac{1}{n}\right) \] Step 2: {\color{red}Substitute into expression.} \[ \frac{(a_n)n!}{\,{}^{3n}P_n} = \frac{\binom{3n}{n} \sin^n(1/n)\, n!}{\frac{(3n)!}{(2n)!}} \] \[ = \frac{\frac{(3n)!}{n!(2n)!} \sin^n(1/n)\, n!}{\frac{(3n)!}{(2n)!}} \] \[ = \sin^n(1/n) \] Step 3: {\color{red}Evaluate limit.} \[ \sin(1/n) \sim \frac{1}{n} \] \[ \sin^n(1/n) \sim \left(\frac{1}{n}\right)^n \] Using: \[ \lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n = e \] Hence limit → \( e \).