Let `(2x^(2)+3x+4)^(10)=sum_(r=0)^(20)a_(r )x^(r )`, then the value of `(a_(7))/(a_(13))` is (a) 6 (b) 8 (c) 12 (d) 16

Options

A. `6`
B. `8`
C. `12`
D. `16`

Explanation

To solve the problem, we need to find the ratio \( \frac{a_7}{a_{13}} \) where \( a_r \) represents the coefficients of \( x^r \) in the expansion of \( (2x^2 + 3x + 4)^{10} \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression:
\[ (2x^2 + 3x + 4)^{10} = \sum_{r=0}^{20} a_r x^r \]
Here, \( a_r \) are the coefficients of \( x^r \) in the expansion.

2. **Substituting \( x \) with \( \frac{2}{x} \)**:
To find \( a_7 \) and \( a_{13} \), we substitute \( x \) with \( \frac{2}{x} \):
\[ \left(2\left(\frac{2}{x}\right)^2 + 3\left(\frac{2}{x}\right) + 4\right)^{10} \]
Simplifying this gives:
\[ \left(\frac{8}{x^2} + \frac{6}{x} + 4\right)^{10} \]

3. **Combining Terms**:
We can express this as:
\[ \left(\frac{8 + 6x + 4x^2}{x^2}\right)^{10} = \frac{(8 + 6x + 4x^2)^{10}}{x^{20}} \]
This means:
\[ (2x^2 + 3x + 4)^{10} = \sum_{r=0}^{20} a_r x^r = \sum_{r=0}^{20} a_r 2^{r-10} x^{20-r} \]

4. **Finding Coefficients**:
From the expansion, the coefficient of \( x^7 \) corresponds to \( a_7 \) and the coefficient of \( x^{13} \) corresponds to \( a_{13} \).
We can relate these coefficients:
\[ a_7 = a_{20-13} \cdot 2^{13-10} = a_{13} \cdot 2^3 \]

5. **Calculating the Ratio**:
Therefore, we have:
\[ \frac{a_7}{a_{13}} = 2^3 = 8 \]

6. **Final Answer**:
The value of \( \frac{a_7}{a_{13}} \) is:
\[ \boxed{8} \]