In the expansion of `(ax+b)^(2020)`, if the coefficient of `x^(2) and x^(3)` are equal, then the value of `(9)/(100)((b)/(a))` is equal to

Answer

To solve the problem, we need to find the value of \(\frac{9}{100} \cdot \frac{b}{a}\) given that the coefficients of \(x^2\) and \(x^3\) in the expansion of \((ax + b)^{2020}\) are equal.

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The general term in the binomial expansion of \((ax + b)^{2020}\) is given by:
\[ T_k = \binom{n}{k} (ax)^k b^{n-k} \]
where \(n = 2020\).

2. **Coefficient of \(x^2\)**:
For \(x^2\), we need the term where \(k = 2\):
\[ T_2 = \binom{2020}{2} (ax)^2 b^{2020-2} = \binom{2020}{2} a^2 x^2 b^{2018} \]
The coefficient of \(x^2\) is:
\[ C_2 = \binom{2020}{2} a^2 b^{2018} \]

3. **Coefficient of \(x^3\)**:
For \(x^3\), we need the term where \(k = 3\):
\[ T_3 = \binom{2020}{3} (ax)^3 b^{2020-3} = \binom{2020}{3} a^3 x^3 b^{2017} \]
The coefficient of \(x^3\) is:
\[ C_3 = \binom{2020}{3} a^3 b^{2017} \]

4. **Set the Coefficients Equal**:
According to the problem, the coefficients of \(x^2\) and \(x^3\) are equal:
\[ C_2 = C_3 \]
This gives us:
\[ \binom{2020}{2} a^2 b^{2018} = \binom{2020}{3} a^3 b^{2017} \]

5. **Simplify the Equation**:
We can simplify this equation:
\[ \frac{\binom{2020}{2}}{\binom{2020}{3}} = \frac{a^3 b^{2017}}{a^2 b^{2018}} \]
The left-hand side simplifies to:
\[ \frac{2020 \cdot 2019 / 2}{2020 \cdot 2019 \cdot 2018 / 6} = \frac{6}{3 \cdot 2018} = \frac{2}{2018} \]
The right-hand side simplifies to:
\[ \frac{a}{b} \]
Thus, we have:
\[ \frac{2}{2018} = \frac{a}{b} \]

6. **Express \(\frac{b}{a}\)**:
Rearranging gives:
\[ \frac{b}{a} = \frac{2018}{2} = 1009 \]

7. **Calculate \(\frac{9}{100} \cdot \frac{b}{a}\)**:
Now we can find:
\[ \frac{9}{100} \cdot \frac{b}{a} = \frac{9}{100} \cdot 1009 = \frac{9 \cdot 1009}{100} = 90.81 \]

### Final Answer:
The value of \(\frac{9}{100} \cdot \frac{b}{a}\) is \(90.81\).