Closed form summation with $n-1$ on top.

Question

Why does the close form of the summation $$\sum_{i=0}^{n-1} 1$$ equals $n$ instead of $n-1$?

Answer 1

We have that

$$\sum_{i=0}^{n-1}= \overbrace{{1+1+1+\cdots+1}}^\color{red}{\text{n terms from 0 to n-1}}=n$$

Answer 2

$$\sum_{i=1}^{n-1}1=n-1$$ When $i$ starts at $0$ you get an extra term $$\sum_{i=0}^{n-1} 1=n$$

Answer 3

Suppose $n = 2$. Then $n - 1 = 1$, and the sum becomes $$\sum_{i=0}^1 1.$$ Clearly, there are two terms in this sum, corresponding to $i = 0$ and $i = 1$. So the sum evaluates to $2$.

In the general case, when the lower index of summation begins at $1$, then the counting of the terms is in a sense "natural" and we have $$\sum_{i=1}^{n-1} 1 = n-1.$$ But when the lower index of summation starts at $i = 0$, then there is an extra term.