Here's a clear, educational, and well-structured solution to the problem.

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Key Concept: The General Term of a Binomial Expansion

For a binomial expansion of the form $(A+B)^N$, the general term (or $(r+1)^{\text{th}}$ term) is given by the formula: $T_{r+1} = \binom{N}{r} A^{N-r} B^r$ where $\binom{N}{r} = \frac{N!}{r!(N-r)!}$ is the binomial coefficient. This formula allows us to find any specific term in the expansion without writing out all the terms.

Problem Analysis: We are given the binomial expansion $\left(\frac{1}{3} x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{2}{3}}}\right)^{18}$. Let's identify the components for our general term formula:

• $A = \frac{1}{3} x^{\frac{1}{3}}$
• $B = \frac{1}{2 x^{\frac{2}{3}}} = \frac{1}{2} x^{-\frac{2}{3}}$ (It's often helpful to rewrite terms with variables in the denominator using negative exponents).
• $N = 18$

Our objective is to find the coefficients of the seventh term ($T_7$) and the thirteenth term ($T_{13}$), denoted as $m$ and $n$ respectively. Then, we need to calculate the value of $\left(\frac{n}{m}\right)^{\frac{1}{3}}$.

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Step 1: Determine the coefficient of the seventh term ($T_7$)

For the seventh term, $T_7$, we need to find the value of $r$. Since $T_{r+1}$ is the $(r+1)^{\text{th}}$ term, for $T_7$, we have $r+1=7$, which implies $r=6$.

Now, substitute $N=18$, $r=6$, $A=\frac{1}{3} x^{\frac{1}{3}}$, and $B=\frac{1}{2} x^{-\frac{2}{3}}$ into the general term formula: $T_7 = T_{6+1} = \binom{18}{6} \left(\frac{1}{3} x^{\frac{1}{3}}\right)^{18-6} \left(\frac{1}{2} x^{-\frac{2}{3}}\right)^6$ $T_7 = \binom{18}{6} \left(\frac{1}{3} x^{\frac{1}{3}}\right)^{12} \left(\frac{1}{2} x^{-\frac{2}{3}}\right)^6$

To find the coefficient, we separate the numerical factors from the variable factors ($x$ terms): $T_7 = \binom{18}{6} \left(\frac{1}{3}\right)^{12} (x^{\frac{1}{3}})^{12} \left(\frac{1}{2}\right)^6 (x^{-\frac{2}{3}})^6$ Next, we simplify the powers of $x$ using the exponent rule $(a^p)^q = a^{pq}$: $T_7 = \binom{18}{6} \left(\frac{1}{3}\right)^{12} x^{\frac{1}{3} \cdot 12} \left(\frac{1}{2}\right)^6 x^{-\frac{2}{3} \cdot 6}$ $T_7 = \binom{18}{6} \left(\frac{1}{3}\right)^{12} x^4 \left(\frac{1}{2}\right)^6 x^{-4}$ Combine the $x$ terms using $x^p \cdot x^q = x^{p+q}$: $T_7 = \binom{18}{6} \left(\frac{1}{3}\right)^{12} \left(\frac{1}{2}\right)^6 x^{4-4}$ $T_7 = \binom{18}{6} \left(\frac{1}{3}\right)^{12} \left(\frac{1}{2}\right)^6 x^0$ Since $x^0 = 1$, the seventh term is a constant term (independent of $x$). The coefficient of the seventh term, $m$, is the numerical part: $m = \binom{18}{6} \left(\frac{1}{3}\right)^{12} \left(\frac{1}{2}\right)^6$

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Step 2: Determine the coefficient of the thirteenth term ($T_{13}$)

For the thirteenth term, $T_{13}$, we have $r+1=13$, which means $r=12$.

Substitute $N=18$, $r=12$, $A=\frac{1}{3} x^{\frac{1}{3}}$, and $B=\frac{1}{2} x^{-\frac{2}{3}}$ into the general term formula: $T_{13} = T_{12+1} = \binom{18}{12} \left(\frac{1}{3} x^{\frac{1}{3}}\right)^{18-12} \left(\frac{1}{2} x^{-\frac{2}{3}}\right)^{12}$ $T_{13} = \binom{18}{12} \left(\frac{1}{3} x^{\frac{1}{3}}\right)^6 \left(\frac{1}{2} x^{-\frac{2}{3}}\right)^{12}$

Separate numerical and variable parts: $T_{13} = \binom{18}{12} \left(\frac{1}{3}\right)^6 (x^{\frac{1}{3}})^6 \left(\frac{1}{2}\right)^{12} (x^{-\frac{2}{3}})^{12}$ Simplify the powers of $x$: $T_{13} = \binom{18}{12} \left(\frac{1}{3}\right)^6 x^{\frac{1}{3} \cdot 6} \left(\frac{1}{2}\right)^{12} x^{-\frac{2}{3} \cdot 12}$ $T_{13} = \binom{18}{12} \left(\frac{1}{3}\right)^6 x^2 \left(\frac{1}{2}\right)^{12} x^{-8}$ Combine the $x$ terms: $T_{13} = \binom{18}{12} \left(\frac{1}{3}\right)^6 \left(\frac{1}{2}\right)^{12} x^{2-8}$ $T_{13} = \binom{18}{12} \left(\frac{1}{3}\right)^6 \left(\frac{1}{2}\right)^{12} x^{-6}$ In problems asking for "the coefficient of the term", it generally refers to the numerical part, even if the variable part is not $x^0$. The coefficient of the thirteenth term, $n$, is: $n = \binom{18}{12} \left(\frac{1}{3}\right)^6 \left(\frac{1}{2}\right)^{12}$

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Step 3: Calculate the ratio $\frac{n}{m}$

Now we have expressions for $m$ and $n$: $m = \binom{18}{6} \left(\frac{1}{3}\right)^{12} \left(\frac{1}{2}\right)^6$ $n = \binom{18}{12} \left(\frac{1}{3}\right)^6 \left(\frac{1}{2}\right)^{12}$

Let's form the ratio $\frac{n}{m}$: $\frac{n}{m} = \frac{\binom{18}{12} \left(\frac{1}{3}\right)^6 \left(\frac{1}{2}\right)^{12}}{\binom{18}{6} \left(\frac{1}{3}\right)^{12} \left(\frac{1}{2}\right)^6}$ A crucial property of binomial coefficients is symmetry: $\binom{N}{r} = \binom{N}{N-r}$. Applying this, we can simplify $\binom{18}{12}$: $\binom{18}{12} = \binom{18}{18-12} = \binom{18}{6}$. Substitute this into the ratio: $\frac{n}{m} = \frac{\binom{18}{6} \left(\frac{1}{3}\right)^6 \left(\frac{1}{2}\right)^{12}}{\binom{18}{6} \left(\frac{1}{3}\right)^{12} \left(\frac{1}{2}\right)^6}$ The binomial coefficients $\binom{18}{6}$ cancel out: $\frac{n}{m} = \frac{\left(\frac{1}{3}\right)^6 \left(\frac{1}{2}\right)^{12}}{\left(\frac{1}{3}\right)^{12} \left(\frac{1}{2}\right)^6}$ Now, simplify the powers using the exponent rule $\frac{a^p}{a^q} = a^{p-q}$: $\frac{n}{m} = \left(\frac{1}{3}\right)^{6-12} \cdot \left(\frac{1}{2}\right)^{12-6}$ $\frac{n}{m} = \left(\frac{1}{3}\right)^{-6} \cdot \left(\frac{1}{2}\right)^6$ Recall that $a^{-p} = \frac{1}{a^p}$. So, $\left(\frac{1}{3}\right)^{-6} = 3^6$. $\frac{n}{m} = 3^6 \cdot \left(\frac{1}{2}\right)^6$ Using the exponent rule $a^p b^p = (ab)^p$: $\frac{n}{m} = \left(3 \cdot \frac{1}{2}\right)^6$ $\frac{n}{m} = \left(\frac{3}{2}\right)^6$

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Step 4: Calculate $\left(\frac{n}{m}\right)^{\frac{1}{3}}$

We have found $\frac{n}{m} = \left(\frac{3}{2}\right)^6$. Now we need to calculate its cube root: $\left(\frac{n}{m}\right)^{\frac{1}{3}} = \left(\left(\frac{3}{2}\right)^6\right)^{\frac{1}{3}}$ Apply the exponent rule $(a^p)^q = a^{pq}$: $\left(\frac{n}{m}\right)^{\frac{1}{3}} = \left(\frac{3}{2}\right)^{6 \cdot \frac{1}{3}}$ $\left(\frac{n}{m}\right)^{\frac{1}{3}} = \left(\frac{3}{2}\right)^2$ Finally, calculate the square: $\left(\frac{n}{m}\right)^{\frac{1}{3}} = \frac{3^2}{2^2}$ $\left(\frac{n}{m}\right)^{\frac{1}{3}} = \frac{9}{4}$

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Tips and Common Mistakes:

• Correct 'r' value: Remember that for the $k^{\text{th}}$ term, $r = k-1$. A frequent error is to use $r=k$.
• Exponent Laws: Be meticulous when applying exponent rules, especially with fractional and negative exponents. Any misstep in simplifying $x$ terms or combining powers can lead to an incorrect numerical coefficient.
• Binomial Coefficient Symmetry: The property $\binom{N}{r} = \binom{N}{N-r}$ is a powerful tool for simplifying ratios of binomial coefficients, as demonstrated in this problem. Always look for opportunities to use it.
• Interpretation of "Coefficient": When asked for "the coefficient of the $k^{\text{th}}$ term," it usually refers to the numerical part only. If the problem intended to include specific powers of $x$, it would typically ask for the "coefficient of $x^p$."

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Summary: We began by identifying the components of the binomial expansion and applied the general term formula. We calculated the numerical coefficients $m$ and $n$ for the seventh and thirteenth terms, respectively, carefully handling the $x$ terms. The key simplification involved using the symmetry property of binomial coefficients, $\binom{18}{12} = \binom{18}{6}$, which allowed for cancellation. After simplifying the powers of constants, we found the ratio $\frac{n}{m} = \left(\frac{3}{2}\right)^6$. Finally, taking the cube root of this ratio yielded $\left(\frac{n}{m}\right)^{\frac{1}{3}} = \frac{9}{4}$.

The final answer is $\boxed{\frac{9}{4}}$.