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

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Key Concepts and Formulas

The binomial theorem states that for any non-negative integer $n$, the expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{r=0}^{n} {}^nC_r a^{n-r} b^r$ The general term, or the $(r+1)^{\text{th}}$ term, in this expansion is denoted by $T_{r+1}$ and is given by: $T_{r+1} = {}^nC_r a^{n-r} b^r$

To find the numerically greatest term in the expansion of $(A+B)^n$, we analyze the ratio of consecutive terms. The $(r+1)^{\text{th}}$ term, $T_{r+1}$, is the numerically greatest term if it is greater than or equal to both its preceding term ($T_r$) and its succeeding term ($T_{r+2}$). That is, $|T_{r+1}| \ge |T_r|$ and $|T_{r+1}| \ge |T_{r+2}|$.

The ratio of the $(r+1)^{\text{th}}$ term to the $r^{\text{th}}$ term is: $\frac{T_{r+1}}{T_r} = \frac{{}^nC_r a^{n-r} b^r}{ {}^{n}C_{r-1} a^{n-(r-1)} b^{r-1} } = \frac{n-r+1}{r} \cdot \frac{b}{a}$

Part 1: Finding the Least Value of $n$ ($n_0$)

We are given the binomial expansion $(3+6x)^n$. We need to find the least integer value of $n$, denoted as $n_0$, such that the $9^{\text{th}}$ term is the greatest term when $x=\frac{3}{2}$.

1. Identify $a$, $b$, and the ratio of consecutive terms: In the expansion of $(3+6x)^n$, we have $a=3$ and $b=6x$. The ratio of the $(r+1)^{\text{th}}$ term to the $r^{\text{th}}$ term is: $\frac{T_{r+1}}{T_r} = \frac{n-r+1}{r} \cdot \frac{6x}{3} = \frac{n-r+1}{r} \cdot 2x$ Now, substitute the given value $x = \frac{3}{2}$: $\frac{T_{r+1}}{T_r} = \frac{n-r+1}{r} \cdot 2 \left(\frac{3}{2}\right) = \frac{3(n-r+1)}{r}$ Since $3$, $n$, and $r$ (for $r \ge 1$) are positive, and $x = \frac{3}{2}$ is positive, all terms in the expansion will be positive. Therefore, we can remove the absolute value signs and just use $T_{r+1} \ge T_r$ and $T_{r+1} \ge T_{r+2}$.

2. Apply the condition for $T_9$ to be the greatest term: For $T_9$ to be the greatest term, it must satisfy two conditions: a. $T_9 \ge T_8$ (The $9^{\text{th}}$ term is greater than or equal to the $8^{\text{th}}$ term) b. $T_9 \ge T_{10}$ (The $9^{\text{th}}$ term is greater than or equal to the $10^{\text{th}}$ term)

Condition a: $T_9 \ge T_8$ This means the ratio $\frac{T_9}{T_8} \ge 1$. Here, $T_9$ is the $(r+1)^{\text{th}}$ term, so we set $r=8$. Substitute $r=8$ into the ratio formula: $\frac{3(n-8+1)}{8} \ge 1$ $\frac{3(n-7)}{8} \ge 1$ Multiply both sides by 8: $3(n-7) \ge 8$ $3n - 21 \ge 8$ Add 21 to both sides: $3n \ge 29$ Divide by 3: $n \ge \frac{29}{3} \implies n \ge 9.66...$

Condition b: $T_9 \ge T_{10}$ This means the ratio $\frac{T_{10}}{T_9} \le 1$. Here, $T_{10}$ is the $(r+1)^{\text{th}}$ term, so we set $r=9$. Substitute $r=9$ into the ratio formula: $\frac{3(n-9+1)}{9} \le 1$ $\frac{3(n-8)}{9} \le 1$ Simplify the fraction: $\frac{n-8}{3} \le 1$ Multiply both sides by 3: $n-8 \le 3$ Add 8 to both sides: $n \le 11$

3. Determine the least value of $n$ ($n_0$): Combining the two inequalities, we have: $9.66... \le n \le 11$ Since $n$ must be an integer (as it's the power in a binomial expansion), the possible integer values for $n$ are 10 and 11. The problem asks for the least value of $n$, which is $n_0$. Therefore, $n_0 = 10$.

Tip for Greatest Term Problems: Remember that $n$ must always be an integer. When you get a range for $n$ from inequalities, carefully select the integer values.

Part 2: Calculating the Ratio $k$ of Coefficients

We need to find the ratio $k$ of the coefficient of $x^6$ to the coefficient of $x^3$ in the expansion of $(3+6x)^{n_0}$. Since $n_0 = 10$, we are considering the expansion of $(3+6x)^{10}$.

1. Write the general term and identify the coefficient of $x^r$: The general term in the expansion of $(3+6x)^{10}$ is: $T_{r+1} = {}^{10}C_r 3^{10-r} (6x)^r = {}^{10}C_r 3^{10-r} 6^r x^r$ The coefficient of $x^r$ is $C_r = {}^{10}C_r 3^{10-r} 6^r$.

2. Find the coefficient of $x^6$: To find the coefficient of $x^6$, we set $r=6$: $C_6 = {}^{10}C_6 3^{10-6} 6^6 = {}^{10}C_6 3^4 6^6$

3. Find the coefficient of $x^3$: To find the coefficient of $x^3$, we set $r=3$: $C_3 = {}^{10}C_3 3^{10-3} 6^3 = {}^{10}C_3 3^7 6^3$

4. Calculate the ratio $k$: $k = \frac{\text{Coefficient of } x^6}{\text{Coefficient of } x^3} = \frac{{}^{10}C_6 3^4 6^6}{{}^{10}C_3 3^7 6^3}$ To simplify, we group the combination terms and the exponential terms: $k = \left( \frac{{}^{10}C_6}{{}^{10}C_3} \right) \cdot \left( \frac{3^4}{3^7} \right) \cdot \left( \frac{6^6}{6^3} \right)$ Using exponent rules ($\frac{a^m}{a^n} = a^{m-n}$): $k = \left( \frac{{}^{10}C_6}{{}^{10}C_3} \right) \cdot 3^{4-7} \cdot 6^{6-3}$ $k = \left( \frac{{}^{10}C_6}{{}^{10}C_3} \right) \cdot 3^{-3} \cdot 6^3$ Now, let's simplify the combination term. Recall that ${}^nC_r = \frac{n!}{r!(n-r)!}$ and the property ${}^nC_r = {}^nC_{n-r}$. So, ${}^{10}C_6 = {}^{10}C_{10-6} = {}^{10}C_4$. $\frac{{}^{10}C_6}{{}^{10}C_3} = \frac{{}^{10}C_4}{{}^{10}C_3} = \frac{\frac{10!}{4!6!}}{\frac{10!}{3!7!}}$ Invert and multiply: $\frac{10!}{4!6!} \cdot \frac{3!7!}{10!} = \frac{3!7!}{4!6!}$ Expand the factorials to simplify: $\frac{3! \cdot (7 \cdot 6!)}{(4 \cdot 3!) \cdot 6!} = \frac{7}{4}$ Substitute this back into the expression for $k$: $k = \frac{7}{4} \cdot \frac{1}{3^3} \cdot 6^3$ $k = \frac{7}{4} \cdot \frac{1}{27} \cdot 216$ Since $216 = 8 \times 27$: $k = \frac{7}{4} \cdot \frac{216}{27} = \frac{7}{4} \cdot 8$ $k = 7 \times 2$ $k = 14$

Common Mistake: Be very careful with simplifying factorials and powers. A common error is to cancel terms incorrectly or make arithmetic mistakes during the final calculation. Always double-check your steps.

Part 3: Final Calculation: $k + n_0$

We found $n_0 = 10$ and $k = 14$. Therefore, $k + n_0 = 14 + 10 = 24$

The final answer is $\boxed{24}$.

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Summary and Key Takeaway

This problem effectively tests a student's comprehensive understanding of the Binomial Theorem. It requires:

1. Correctly applying the condition for the numerically greatest term: This involves setting up and solving inequalities using the ratio of consecutive terms. It is crucial to remember that the power 'n' must be an integer.
2. Accurately calculating ratios of coefficients: This involves identifying the general term, substituting the correct 'r' values, and simplifying complex fractional expressions involving combinations and powers. Mastery of factorial and exponent rules is essential.

Solving such problems reinforces the fundamental principles and practical applications of the binomial theorem.