To find the ratio of coefficients $\frac{a_7}{a_{13}}$ in the expansion of $(2x^2 + 3x + 4)^{10}$, we will use the Multinomial Theorem and exploit a symmetry property of the coefficients.

1. Understanding the Multinomial Theorem

For an expansion of the form $(x_1 + x_2 + \dots + x_k)^n$, the general term is given by: $T = \frac{n!}{n_1! n_2! \dots n_k!} x_1^{n_1} x_2^{n_2} \dots x_k^{n_k}$ where $n_1, n_2, \dots, n_k$ are non-negative integers such that $n_1 + n_2 + \dots + n_k = n$.

In our problem, we have $(2x^2 + 3x + 4)^{10}$. Here, $n=10$, and the terms are $x_1 = 2x^2$, $x_2 = 3x$, $x_3 = 4$.

The general term in this expansion is: $T(n_1, n_2, n_3) = \frac{10!}{n_1! n_2! n_3!} (2x^2)^{n_1} (3x)^{n_2} (4)^{n_3}$ where $n_1 + n_2 + n_3 = 10$.

Let's simplify this term to identify the coefficient of $x^r$: $T(n_1, n_2, n_3) = \frac{10!}{n_1! n_2! n_3!} 2^{n_1} (x^2)^{n_1} 3^{n_2} x^{n_2} 4^{n_3}$ $T(n_1, n_2, n_3) = \frac{10!}{n_1! n_2! n_3!} 2^{n_1} 3^{n_2} 4^{n_3} x^{2n_1 + n_2}$ The coefficient $a_r$ of $x^r$ is the sum of all such terms where the power of $x$ is $r$. So, for $a_r$, we must have: $2n_1 + n_2 = r$ And the coefficient $a_r$ is: $a_r = \sum_{\substack{n_1+n_2+n_3=10 \\ 2n_1+n_2=r}} \frac{10!}{n_1! n_2! n_3!} 2^{n_1} 3^{n_2} 4^{n_3}$

2. Establishing a Relationship Between $n_1, n_3$ and $r$

We have two conditions on $n_1, n_2, n_3$:

1. $n_1 + n_2 + n_3 = 10$
2. $2n_1 + n_2 = r$

We can eliminate $n_2$ from these equations. From (2), $n_2 = r - 2n_1$. Substitute this into (1): $n_1 + (r - 2n_1) + n_3 = 10$ $-n_1 + r + n_3 = 10$ Rearranging this, we get a crucial relationship: $n_3 - n_1 = 10 - r$ This equation shows that for any term contributing to $a_r$, the difference $n_3 - n_1$ is constant and depends only on $r$.

3. Exploiting Coefficient Symmetry

We are asked to find $\frac{a_7}{a_{13}}$. Notice that $7 + 13 = 20$, which is the maximum power of $x$ in the expansion (since $(x^2)^{10} = x^{20}$). This suggests a symmetry property.

Consider a general polynomial $P(x) = (C_k x^k + \dots + C_1 x + C_0)^n = \sum_{i=0}^{nk} a_i x^i$. If we consider the polynomial $Q(x) = (C_0 x^k + C_1 x^{k-1} + \dots + C_k)^n = \sum_{i=0}^{nk} b_i x^i$, then its coefficients $b_i$ are related to $a_i$ by $b_i = a_{nk-i}$. This property arises from the identity: $x^{nk} P\left(\frac{1}{x}\right) = Q(x)$.

In our problem: $P(x) = (2x^2 + 3x + 4)^{10}$. Here $C_2=2, C_1=3, C_0=4$. The maximum power $nk = 2 \times 10 = 20$. The coefficients of $P(x)$ are $a_r$. The "reversed" polynomial is $Q(x) = (4x^2 + 3x + 2)^{10}$. Let its coefficients be $b_r$. According to the property, $b_r = a_{20-r}$.

We need to find $\frac{a_7}{a_{13}}$. Since $13 = 20 - 7$, we can write $a_{13} = a_{20-7} = b_7$. So, the problem reduces to finding $\frac{a_7}{b_7}$.

4. Comparing Coefficients $a_7$ and $b_7$

The coefficient $a_7$ is the sum of terms $T(n_1, n_2, n_3)$ where $2n_1 + n_2 = 7$: $a_7 = \sum_{\substack{n_1+n_2+n_3=10 \\ 2n_1+n_2=7}} \frac{10!}{n_1! n_2! n_3!} 2^{n_1} 3^{n_2} 4^{n_3}$ The coefficient $b_7$ is the sum of terms $T'(n_1, n_2, n_3)$ for the polynomial $(4x^2 + 3x + 2)^{10}$, where $2n_1 + n_2 = 7$: $b_7 = \sum_{\substack{n_1+n_2+n_3=10 \\ 2n_1+n_2=7}} \frac{10!}{n_1! n_2! n_3!} 4^{n_1} 3^{n_2} 2^{n_3}$ Notice that both sums are over the exact same set of $(n_1, n_2, n_3)$ triplets.

Let's consider the ratio of a generic term from $a_7$ to the corresponding term from $b_7$ (for the same $n_1, n_2, n_3$ values): $\frac{\text{Term for } a_7}{\text{Term for } b_7} = \frac{\frac{10!}{n_1! n_2! n_3!} 2^{n_1} 3^{n_2} 4^{n_3}}{\frac{10!}{n_1! n_2! n_3!} 4^{n_1} 3^{n_2} 2^{n_3}}$ The factorial and $3^{n_2}$ parts cancel out: $= \frac{2^{n_1} 4^{n_3}}{4^{n_1} 2^{n_3}}$ We know $4 = 2^2$, so substitute this: $= \frac{2^{n_1} (2^2)^{n_3}}{(2^2)^{n_1} 2^{n_3}} = \frac{2^{n_1} 2^{2n_3}}{2^{2n_1} 2^{n_3}} = \frac{2^{n_1 + 2n_3}}{2^{2n_1 + n_3}} = 2^{(n_1 + 2n_3) - (2n_1 + n_3)} = 2^{n_3 - n_1}$

5. Final Calculation

From Step 2, we found $n_3 - n_1 = 10 - r$. For $r=7$, this means $n_3 - n_1 = 10 - 7 = 3$. Since $n_3 - n_1 = 3$ for all triplets $(n_1, n_2, n_3)$ that satisfy $2n_1+n_2=7$ and $n_1+n_2+n_3=10$, the ratio of corresponding terms is constant: $\frac{\text{Term for } a_7}{\text{Term for } b_7} = 2^3 = 8$ Since every individual term contributing to $a_7$ is 8 times the corresponding term contributing to $b_7$, the sum of these terms will also maintain this ratio: $a_7 = 8 \cdot b_7$ Now, substitute $b_7 = a_{13}$ back into the equation: $a_7 = 8 \cdot a_{13}$ Finally, we can find the required ratio: $\frac{a_7}{a_{13}} = 8$

The final answer is $\boxed{8}$.