If the coefficients of $$x^3$$ and $$x^4$$ in the expansion of $$(1 + ax + bx^2)(1 - 2x)^{18}$$ in powers of $$x$$ are both zero, then $$(a, b)$$ is equal to:

First we write the given expression and recall the binomial theorem. We have $$\bigl(1+ax+bx^2\bigr)\,(1-2x)^{18}.$$ According to the binomial theorem,

$$ (1-2x)^{18} \;=\; \sum_{k=0}^{18} \binom{18}{k}(1)^{18-k}(-2x)^k \;=\; \sum_{k=0}^{18} (-1)^k 2^k \binom{18}{k}\,x^k. $$

For convenience we denote $$ c_k \;=\; (-1)^k 2^k \binom{18}{k}, $$ so that the series becomes $$\displaystyle\sum_{k=0}^{18} c_k x^k.$$

Now we multiply this series by $$1+ax+bx^2$$. Every power of $$x$$ in the product arises by picking one term from $$1+ax+bx^2$$ and one term from the series, and adding all such contributions.

Coefficient of $$x^3$$.
To obtain $$x^3$$ we have three possibilities:

• The factor $$1$$ from the first bracket and the term $$c_3x^3$$ from the series.
• The factor $$ax$$ and the term $$c_2x^2$$  (because $$x\cdot x^2=x^3$$).
• The factor $$bx^2$$ and the term $$c_1x$$.

Hence the coefficient of $$x^3$$ is $$ c_3 + a\,c_2 + b\,c_1. $$ The question says this coefficient is zero, so

$$ c_3 + a\,c_2 + b\,c_1 = 0. \quad -(1) $$

Coefficient of $$x^4$$.
In the same way $$x^4$$ can arise from:

• $$1$$ with $$c_4x^4,$$
• $$ax$$ with $$c_3x^3,$$
• $$bx^2$$ with $$c_2x^2.$$

Therefore the coefficient of $$x^4$$ is $$ c_4 + a\,c_3 + b\,c_2, $$ and the condition “coefficient is zero” gives

$$ c_4 + a\,c_3 + b\,c_2 = 0. \quad -(2) $$

Next we evaluate the required $$c_k$$ one by one:

Using $$c_k = (-1)^k 2^k\binom{18}{k},$$

$$ \begin{aligned} c_1 &= (-1)^1 2^1\binom{18}{1}= -2\cdot18 = -36,\\[4pt] c_2 &= (+1) 2^2\binom{18}{2}= 4\cdot153 = 612,\\[4pt] c_3 &= (-1)^3 2^3\binom{18}{3}= -8\cdot816 = -6528,\\[4pt] c_4 &= (+1) 2^4\binom{18}{4}= 16\cdot3060 = 48960. \end{aligned} $$

Substituting these numbers into equation (1) we get

$$ -6528 + a\,(612) + b\,(-36) = 0, $$ so $$ 612a - 36b - 6528 = 0. $$ Dividing every term by $$12$$ gives

$$ 51a - 3b = 544. \quad -(3) $$

Now substituting the same numbers into equation (2) yields

$$ 48960 + a\,(-6528) + b\,(612) = 0, $$ which simplifies to

$$ -6528a + 612b + 48960 = 0. $$ Dividing every term by $$12$$ gives

$$ -544a + 51b = -4080. \quad -(4) $$

We now have the linear system

$$ \begin{cases} 51a - 3b = 544,\\ -544a + 51b = -4080. \end{cases} $$

To eliminate $$b$$, we multiply equation (3) by $$17$$ so that the coefficient of $$b$$ matches that in equation (4):

$$ 17(51a - 3b) = 17\cdot544 \;\;\Longrightarrow\;\; 867a - 51b = 9248. \quad -(5) $$

Now we add equations (4) and (5):

$$ (867a - 51b) + (-544a + 51b) = 9248 + (-4080). $$ The $$b$$ terms cancel, leaving

$$ 323a = 5168 \;\;\Longrightarrow\;\; a = \frac{5168}{323}. $$

Since $$323\times16 = 5168$$, we have

$$ a = 16. $$

Substituting $$a=16$$ back into equation (3):

$$ 51(16) - 3b = 544 \;\;\Longrightarrow\;\; 816 - 3b = 544, $$ so

$$ -3b = 544 - 816 = -272 \;\;\Longrightarrow\;\; b = \frac{272}{3}. $$

Thus the ordered pair $$(a,b)$$ equals $$ \left(16,\;\frac{272}{3}\right). $$

Hence, the correct answer is Option B.