Why does $\sum_{i=1}^n a = (n+1)a$

Question

I'm currently working my way through Eccles's "An Introduction to Mathematical Reasoning" and an alternative proof of $\sum_{i=0}^n (a+ib) = \frac 12 (n+1)(2a+bn)$ states that $$\sum_{i=0}^n (a+ib) = \sum_{i=1}^n a + b\sum_{i=1}^n i = (n+1)a + \frac 12n(n+1)b$$ After some careful thought I understand everything except why $\sum_{i=1}^n a = (n+1)a$. Isn't this just the series "1+2+3+4..." which is represented by $$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$ or am I misunderstanding?

Answer 1

Whe have$$\sum_{i=0}^na=\overbrace{a+a+\cdots+a}^{n+1\text{ times}}=(n+1)a.$$