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

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

2. Simplify the terms

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

1. Find the first two terms of the expansion of $(1-3x)^5$.

$(1-3x)^5 = 1 + 5(-3x) + \dots = 1 - 15x + \dots$

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

$(2+ax)^6 = 2^6 + \binom{6}{1}(2^5)(ax) + \dots = 64 + 6(32)ax + \dots = 64 + 192ax + \dots$

3. Find the coefficient of $x$ in the product of the two expansions.

The terms in $x$ from the product $(1 - 15x + \dots)(64 + 192ax + \dots)$ are:
$1(192ax) + (-15x)(64) = 192ax - 960x$
So, the coefficient of $x$ is $192a - 960$.

4. Equate the coefficient to $-384$ and solve for $a$.

$192a - 960 = -384$
$192a = 576$
$a = 3$