If the coefficients of `x^3` and `x^4` in the expansion of `(1""+a x+b x^2)""(1-2x)^(18)` in powers of x are both zero, then (a, b) is equal to (1) `(16 ,(251)/3)` (3) `(14 ,(251)/3)` (2) `(14 ,(272)/3)` (4) `(16 ,(272)/3)`

Options

A. `(16,251/3)`
B. `(14,251/3)`
C. `(14, 272/3)`
D. `(16,272/3)`

Explanation

To solve the problem, we need to find the coefficients of \( x^3 \) and \( x^4 \) in the expansion of \( (1 + ax + bx^2)(1 - 2x)^{18} \) and set them both to zero.

### Step 1: Coefficient of \( x^3 \)

The coefficient of \( x^3 \) in the expansion can be calculated using the binomial theorem.

The expansion of \( (1 - 2x)^{18} \) gives us:
\[ \text{Coefficient of } x^3 = \binom{18}{3} (-2)^3 = \binom{18}{3} (-8) \]

Now, we also need to consider the contributions from \( (1 + ax + bx^2) \):
- From \( 1 \): contributes \( \binom{18}{3} (-8) \)
- From \( ax \): contributes \( a \cdot \binom{18}{2} (-2)^2 = a \cdot \binom{18}{2} \cdot 4 \)
- From \( bx^2 \): contributes \( b \cdot \binom{18}{1} (-2) = b \cdot 18 \cdot (-2) \)

Thus, the equation for the coefficient of \( x^3 \) becomes:
\[ -8 \binom{18}{3} + 4a \binom{18}{2} - 36b = 0 \]

Calculating the binomial coefficients:
\[ \binom{18}{3} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816 \]
\[ \binom{18}{2} = \frac{18 \times 17}{2} = 153 \]

Substituting these values into the equation:
\[ -8 \cdot 816 + 4a \cdot 153 - 36b = 0 \]
\[ -6528 + 612a - 36b = 0 \]
This simplifies to:
\[ 612a - 36b = 6528 \quad \text{(Equation 1)} \]

### Step 2: Coefficient of \( x^4 \)

Now, we find the coefficient of \( x^4 \):
- From \( 1 \): contributes \( \binom{18}{4} (-2)^4 = \binom{18}{4} \cdot 16 \)
- From \( ax \): contributes \( a \cdot \binom{18}{3} (-2)^3 = a \cdot \binom{18}{3} \cdot (-8) \)
- From \( bx^2 \): contributes \( b \cdot \binom{18}{2} (-2)^2 = b \cdot \binom{18}{2} \cdot 4 \)

Thus, the equation for the coefficient of \( x^4 \) becomes:
\[ 16 \binom{18}{4} - 8a \binom{18}{3} + 4b \binom{18}{2} = 0 \]

Calculating \( \binom{18}{4} \):
\[ \binom{18}{4} = \frac{18 \times 17 \times 16 \times 15}{4 \times 3 \times 2 \times 1} = 3060 \]

Substituting this into the equation:
\[ 16 \cdot 3060 - 8a \cdot 816 + 4b \cdot 153 = 0 \]
\[ 48960 - 6528a + 612b = 0 \quad \text{(Equation 2)} \]

### Step 3: Solving the System of Equations

Now we have two equations:
1. \( 612a - 36b = 6528 \)
2. \( -6528a + 612b = -48960 \)

We can solve these equations simultaneously.

From Equation 1:
\[ b = \frac{612a - 6528}{36} \]

Substituting \( b \) into Equation 2:
\[ -6528a + 612\left(\frac{612a - 6528}{36}\right) = -48960 \]

Multiplying through by 36 to eliminate the fraction:
\[ -6528 \cdot 36 a + 612(612a - 6528) = -48960 \cdot 36 \]

This will yield a linear equation in terms of \( a \). Solving for \( a \) and substituting back will give us \( b \).

After solving, we find:
\[ a = 16, \quad b = \frac{272}{3} \]

### Final Answer

Thus, the values of \( (a, b) \) are:
\[ (a, b) = \left(16, \frac{272}{3}\right) \]