This is perhaps very basic but I am currently very lost on how to think in order to end up with the answer: $9-n$ in the following summation:
$$\sum_{i=n-4}^4 1 = 9-n $$
My first idea was to rewrite the summation with something like: $5+ \sum_{i=1}^4 1$, but that is clearly not a correct way of thinking. I would be very greatful for any kind of help or enlightment regarding this question!
How many numbers are there from $n-4$ to $4$ inclusive?
$$4-(n-4)+1$$
define a new index $j=(n-4)-i$, then as $i$ runs from $n-4$ to $4$, $j$ will run from $j=0$ to $j=n-8$. So $$\sum_{i=n-4}^{i=4}1=\sum_{j=0}^{n-8}1=n-9.$$ In the last sum, ask yourself, how many times are you are adding $1$?
It's quite trivial. In any summation, say $$\sum_{i=a}^{b} f(i),$$ the number of entries are $b-a+1$. In your case $f(i)$ is $1$. So the answer is $b-a+1$.
You can increase both top and bottom indices as follows: $$\sum_{i=n-4}^4 1=\sum_{i=n-4\color{red}{+4}}^{4\color{red}{+4}}1=\sum_{i=n}^8 1=\underbrace{1}_{n}+\underbrace{1}_{n+1}+\cdots+\underbrace{1}_{8}=1\cdot(8-n+1)=9-n.$$
