2021-M2-03

$\begin{array}{cl}
&(1-4x)^n \\
= & 1+C^n_1(-4x)+C^n_2(-4x)^2+\ldots+C^n_{n-1}(-4x)^{n-1}+(-4x)^n \\
= & 1-4nx+\dfrac{n(n-1)}{2}(16x^2)+\ldots+n(-4x)^{n-1}+(-4x)^n \\
= & 1-4nx+8n(n-1)x^2+\ldots+n(-4x)^{n-1}+(-4x)^n
\end{array}$

Since the coefficient of $x^2$ is $240$, we have

$\begin{array}{rcl}
8n(n-1) & = & 240 \\
n^2-n & = & 30 \\
n^2-n-30 & = & 0 \\
(n-6)(n+5) & = & 0
\end{array}$

$\therefore n=6$ or $n=-5$ (rejected).

$\begin{array}{cl}
& (1-4x)^6\left(1+\dfrac{2}{x}\right)^5 \\
= & \left[(-4x)^6+6(-4x)^5+15(-4x)^4+\ldots\right]\left[1+5\times\dfrac{2}{x}+10\times \left(\dfrac{2}{x}\right)^2+\ldots\right] \\
= & (4096x^6-6144x^5+3840x^4+\ldots)\left(1+\dfrac{10}{x}+\dfrac{40}{x^2}+\ldots\right)
\end{array}$

Therefore, the coefficient of $x^4$

$\begin{array}{cl}
= & 4096\times40+(-6144)\times 10+3840\times 1 \\
= & 106\ 240
\end{array}$