用 $a$ 和 $n$ 表示 $(2+x)^n$ 展开式中的第三项。 写出 $(1-3x)^5$ 展开式中的前三项。 在 $(1-3x)^5(2+ax)^6$ 的展开式中，$x$ 的系数为 $-384$。求 $a$ 的值。

1. Apply the binomial theorem to expand the first three terms

$(1-3x)^5 = 1^5 + \binom{5}{1}(1)^4(-3x) + \binom{5}{2}(1)^3(-3x)^2 + \dots$

2. Simplify the terms

$= 1 + 5(-3x) + 10(9x^2) + \dots$
$= 1 - 15x + 90x^2$

1. Find the first two terms of the expansion of $(2+ax)^6$.

The first two terms of $(2+ax)^6$ are:
$2^6 + \binom{6}{1}(2)^5(ax) = 64 + 6(32)(ax) = 64 + 192ax$

2. Identify the terms in the product $(1-3x)^5(2+ax)^6$ that result in an $x$ term.

From part (ii), the first two terms of $(1-3x)^5$ are $1 - 15x$.
Multiplying the expansions:
$(1 - 15x + \dots)(64 + 192ax + \dots)$
The terms containing $x$ are:
$(1)(192ax) + (-15x)(64) = 192ax - 960x$

3. Equate the coefficient of $x$ to $-384$ and solve for $a$.

The coefficient of $x$ is $192a - 960$.
$192a - 960 = -384$
$192a = 576$
$a = 3$