How to find the RHS of $\sum_{1 \leq a \leq \lfloor{\frac{n}{2}}\rfloor} (n-2a) $

Question

How does

$$\sum_{1 \leq a \leq \lfloor{\frac{n}{2}}\rfloor} (n-2a) = \lfloor{\frac{n}{2}}\rfloor (n - \lfloor{\frac{n}{2}}\rfloor - 1)$$

I cannot see how you get to the RHS. This looks close to number of elements less than n/2 times by the remaining. However, I cannot see the logic for this right now.

Answer 1

$\large\textbf{Hint :}$ \begin{aligned}\sum_{1\leq a\leq\left\lfloor\frac{n}{2}\right\rfloor}{\left(n-2a\right)}=n\left\lfloor\frac{n}{2}\right\rfloor -2\sum_{a=1}^{\left\lfloor\frac{n}{2}\right\rfloor}{a}\end{aligned}

What does the sum $ \sum\limits_{k=1}^{p}{k} $, for $ p\in\mathbb{N}$, simplify to ?