Why $\sum_{x=y}^n 1 = n-y+1 ?$ Why $\sum_{x=y}^n 1 = n-y+1 ?$ We know that $1 \leq y \leq x \leq n$.
$$\sum_{x=y}^n 1$$
Is there a formula for solving this? I would have said that it should equal n-y, but why the +1?
 A: Consider the following sequence of numbers:
$$
13, 14, 15, \ldots, 41, 42
$$
How many numbers are in this sequence? Well, one way to do this is assign each number a label, starting with $s_1$ and ending at $s_N$. Then our answer would be $N$. Indeed, we have:
$$
\begin{array}{c|c|c|c|c|c}
s_1 & s_2 & s_3 & \cdots & s_{N-1} & s_N \\ \hline
13 & 14 & 15 & \cdots & 41 & 42
\end{array}
$$
Let's look at the pattern. Given that $s_x = 15$, how could we have deduced that $x = 3$? It's easy: just subtract $12$. So if $s_N = 42$, it follows that $N = 42 - 12 = 30$.

Likewise, consider the following sequence of numbers:
$$
y, y+1, y+2, \ldots, n-1, n
$$
To count how many numbers are in this sequence, we assign each number a label, starting with $s_1$ and ending at $s_N$. Then our answer would be $N$. Indeed, we have:
$$
\begin{array}{c|c|c|c|c|c}
s_1 & s_2 & s_3 & \cdots & s_{N-1} & s_N \\ \hline
y & y+1 & y+2 & \cdots & n-1 & n
\end{array}
$$
Let's look at the pattern. Given that $s_x = y+2$, how could we have deduced that $x = 3$? It's easy: just subtract $y$ and add $1$. So if $s_N = n$, it follows that $N = n - y + 1$.
A: It can't be $n-y$ because $\sum_{x=n}^n1$ isn't an empty sum, but rather a sum containing one $1$. You can continue for $n\ge y$ by induction on $n$. Maybe you thought $x=n$ isn't included in the sum; it is.
A: You could think about it as pages you still need to read on a book that you started.
Suppose the book has $200$ pages and you're on page $100$.
If you're at the very beginning of page $100$, you have to read the pages $101$ to $200$, which are $100$ (just like the number of numbers in $\{1,2,3,4,5,6,7,8,9,10\}$ is $10$) plus page $100$ itself, so $(200-100) + 1$ in total.
In this summation, you're including $y$, and that's where the $+1$ comes from, so the $\le$ from $y \leq x$ is crucial here.
A: $$\sum_{x=y}^n 1=\underbrace{\sum_{x=1}^n 1}_{=n}- \underbrace{\sum_{x=1}^{y-1} 1}_{=y-1}= n-(y-1)$$
