I'm pretty sure $$\sum\limits\_{i=1}^{a-1} i = \frac{a(a-1)}{2}$$ using the relationship $\sum\limits\_{i=1}^{n} i = \frac{n(n+1)}{2}$.

It looks similar to $$\sum\limits\_{i=a}^{n} i = \frac{(n+a)(n-a+1)}{2}$$ but I'm not sure how or why does$$\sum\limits\_{i=a}^{n} i = \frac{(n+a)(n-a+1)}{2}?$$