Let’s build up the Riemann sum first. The interval is $[2,5]$, so its length is $3$, and when you divide it into $n$ equal subintervals, each will be of length $\frac3n$, so $\Delta x$ (not $dx$) is indeed $\frac3n$. The ends of the subintervals $-$ the $x_k$’s $-$ are $2+\frac3n,2+2\left(\frac3n\right)$, and so on, with $x_k=2+\frac{3k}n$. Thus, your $n$-th Riemann sum, $R_n$, is
$\begin{align*} R_n=\sum_{k=1}^n(4-2x_k)\frac3n&=\sum_{k=1}^n\left(4-2\left(2+\frac{3k}n\right)\right)\frac3n\\ &=\sum_{k=1}^n\left(4-4-\frac{6k}n\right)\frac3n\\ &=\sum_{k=1}^n\left(-\frac{6k}n\right)\frac3n\\ &=\sum_{k=1}^n\left(-\frac{18k}{n^2}\right)\;. \end{align*}$
Now that’s just $-\frac{18}{n^2}-\frac{18\cdot 2}{n^2}-\frac{18\cdot 3}{n^2}-\ldots-\frac{18n}{n^2}\;,$
with a factor of $-\dfrac{18}{n^2}$ in every term that we can factor out to get $-\frac{18}{n^2}(1+2+3+\ldots+n)=-\frac{18}{n^2}\sum_{k=1}^nk\;.$ As you said in the question, you know that $\sum_{k=1}^nk$, the sum of the first $n$ positive integers, is $\frac{n(n+1)}2$, so $\begin{align*}R_n&=-\frac{18}{n^2}\sum_{k=1}^nk=-\frac{18}{n^2}\cdot\frac{n(n+1)}2\\&=-\frac{9(n+1)}n=-9\cdot\frac{n+1}n\\&=-9\left(1+\frac1n\right)\;.\end{align*}$
Finally, $\begin{align*}\int_2^5(4-2x)dx&=\lim_{n\to\infty}R_n\\&=\lim_{n\to\infty}-9\left(1+\frac1n\right)\\&=-9\lim_{n\to\infty}\left(1+\frac1n\right)\;.\end{align*}$
Now, what’s $\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)$?