Let an denote the term independent of x in the expansion of x+sin⁡(1/n)x23n, then limx→∞ ann!3nPn equals

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

A
0
B
1
C
e
D
$e / \sqrt{3}$

Solution:

$T_{r + 1} =^{3 n} C_{r} x^{3 n - r} {[\sin (\frac{1}{n})]}^{r} x^{- 2 r}$

For this term to be independent of x, set

$3 n - r - 2 r = 0 \Rightarrow r = n$

$∴ a_{n} =^{3 n} C_{n} \sin^{n} (\frac{1}{n})$

Now, $\frac{n! a_{n}}{^{3 n} P_{n}} = \sin^{n} (\frac{1}{n}) \to 0$ as $n \to \infty$

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