If the coefficients of $$x^{-2}$$ and $$x^{-4}$$, in the expansion of $$\left(x^{1/3} + \frac{1}{2x^{1/3}}\right)^{18}$$, $$(x \gt 0)$$, are $$m$$ and $$n$$ respectively, then $$\frac{m}{n}$$ is equal to

We begin with the binomial expression $$\left(x^{1/3} + \frac{1}{2x^{1/3}}\right)^{18}.$$ To expand it we use the binomial theorem, which states that for any real numbers $$u$$ and $$v$$ and a non-negative integer $$n$$, we have $$ (u+v)^n=\sum_{r=0}^{n}\binom{n}{r}u^{\,n-r}v^{\,r}. $$

Here $$u= x^{1/3},\qquad v=\frac{1}{2x^{1/3}},\qquad n=18.$$ So the general term of the expansion is

$$ T_{r+1}= \binom{18}{r}\left(x^{1/3}\right)^{18-r}\left(\frac{1}{2x^{1/3}}\right)^{r}, \qquad 0\le r\le 18. $$

First we simplify the power of $$x$$ inside this term. We have

$$ \left(x^{1/3}\right)^{18-r}=x^{(18-r)/3},\qquad \left(\frac{1}{2x^{1/3}}\right)^{r}= \frac{1}{2^{r}}\,x^{-r/3}. $$

Multiplying these two powers of $$x$$, we obtain

$$ x^{(18-r)/3}\,x^{-r/3}=x^{\frac{18-r-r}{3}}=x^{\frac{18-2r}{3}}. $$

Thus the complete general term is

$$ T_{r+1}= \binom{18}{r}\frac{1}{2^{r}}\,x^{\frac{18-2r}{3}}. $$

We need the coefficients of the terms $$x^{-2}$$ and $$x^{-4}$$. So we equate the exponent $$\dfrac{18-2r}{3}$$ with $$-2$$ and $$-4$$ successively.

For $$x^{-2}$$:

$$ \frac{18-2r}{3}=-2 \;\;\Longrightarrow\;\; 18-2r=-6 \;\;\Longrightarrow\;\; 2r=24 \;\;\Longrightarrow\;\; r=12. $$

For $$x^{-4}$$:

$$ \frac{18-2r}{3}=-4 \;\;\Longrightarrow\;\; 18-2r=-12 \;\;\Longrightarrow\;\; 2r=30 \;\;\Longrightarrow\;\; r=15. $$

Hence

• The coefficient $$m$$ of $$x^{-2}$$ is obtained from $$r=12$$: $$ m=\binom{18}{12}\frac{1}{2^{12}}. $$
• The coefficient $$n$$ of $$x^{-4}$$ is obtained from $$r=15$$: $$ n=\binom{18}{15}\frac{1}{2^{15}}. $$

We now evaluate the required binomial coefficients.

Because $$\binom{18}{12}=\binom{18}{6}$$ (symmetry property), we calculate

$$ \binom{18}{6}= \frac{18!}{6!\,12!} =\frac{18\cdot17\cdot16\cdot15\cdot14\cdot13}{6\cdot5\cdot4\cdot3\cdot2\cdot1} =18\,564. $$

Similarly

$$ \binom{18}{15}=\binom{18}{3} =\frac{18\cdot17\cdot16}{3\cdot2\cdot1} =816. $$

Therefore

$$ m=18\,564\left(\frac{1}{2}\right)^{12},\qquad n=816\left(\frac{1}{2}\right)^{15}. $$

We now find their ratio:

$$$ \frac{m}{n}= \frac{18\,564\left(\dfrac{1}{2}\right)^{12}} {816\left(\dfrac{1}{2}\right)^{15}} \;=\; \frac{18\,564}{816}\times\left(\frac{1}{2}\right)^{12-15} \;=\; \frac{18\,564}{816}\times 2^{3}. $$$

Calculating step by step,

$$ \frac{18\,564}{816}=22.75=\frac{1\,547}{68}, $$

$$ \left(\frac{18\,564}{816}\right)\times 2^{3} =\frac{1\,547}{68}\times 8 =\frac{12\,376}{68} =182. $$

So we obtain

$$ \frac{m}{n}=182. $$

Hence, the correct answer is Option B.