Evaluating $\sum\limits_{i=1}^{a-1} i = \frac{a(a-1)}{2}$

Question

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$$\sum\limits_{i=a}^{n} i = \frac{(n+a)(n-a+1)}{2}?$$

Answer 1

Hint: $$\sum_A^B=\sum_1^B - \sum_1^{A-1}$$

Answer 2

$$\sum\limits_{i=a}^{n} i = \sum\limits_{i=1}^{n} i- \sum\limits_{i=1}^{a-1} i$$

Now substitute and subtract.