I know this is a basic question, but I don't follow why this is true. Nor do I understand it by intuition.
$$ \sum_{i=1}^{n}1=n \tag 1 $$
Why is the sum not $1+n$? I mean, $\sum_{i=1}^{n}1=1+0+0+ \cdots +n=1+n$?
And also this sum
$$ \sum_{i=0}^{n-1}1=n \tag 2 $$
Why not $n-1$? I tried $\sum_{i=0}^{n-1}1=0 + 0+ \cdots + n-1=n-1$.
The symbol $\sum_{i=a}^b$ denotes not a number but an operation. You need to put some function of $i$ after it to make sense of it. Then it means $$ \sum_{i=a}^b f(i) = f(a)+f(a+1)+f(a+2)+\dots+f(b)$$ (as a side note, notice that it only has sense if $b-a\in\mathbb N$).
If there's nothing after the symbol, it's probably because you factored something out before the symbol, but that does not leave nothing behind the symbol it leaves factor $1$. That means that writing, for example, $$ \sum_{i=a}^b c = c \sum_{i=a}^b $$ is incorrect, you can only write $$ \sum_{i=a}^b c = \sum_{i=a}^b (c\cdot 1) = c \cdot \sum_{i=a}^b 1 $$
In such case, when you have $f(i) = 1= const.$, the symbol denotes the sum of an appropriate number of equal addends. Assuming that $a$ and $b$ are integer, the number of addends is the same as the number of integer numbers between $a$ and $b$, including $a$ and $b$ themselves. If you calculate it, it's equal to $b-a+1$ That is $$ \sum_{i=a}^b c = c+c+c+\dots+c = (b-a+1)c$$ in particular $$ \sum_{i=1}^n 1 = \sum_{i=0}^{n-1} 1 = n$$ or $$ \sum_{i=1}^n 0 = \sum_{i=0}^{n-1} 0 = 0$$