How to prove that $\sum_{i=j}^nn-i = \sum_{i=1}^{n-j}i$? Trying to solve question 2-3 from Skiena's Algorithm Design Manual which asks to find the runtime of the following loop:
for i := 1 to n do
   for j := 1 to i do
      for k := j to i + j do
         for l := 1 to i + j - k do
            r := r + 1

The actual sum used to solve this is $\sum_{i=1}^n\sum_{j=1}^i\sum_{k=j}^{i+j}\sum_{l=1}^{i+j-k}1 = \sum_{i=1}^n\sum_{j=1}^i\sum_{k=j}^{i+j}i+j-k$, but I've tried to focus the question on the step that is confusing me (used i instead of k, n instead of i+j):
$$\sum_{i=j}^nn-i = \sum_{i=1}^{n-j}i$$
Edit: I didn't really ask the right question... I'm wondering how you find the sum on the right (without knowing it) from the sum on the left. It seems that writing out or visualizing the terms like Pedro and Michael did would help seeing the sum on the right from 0 to n-j, I was just hoping for some kind of rule that helps convert from one sum to the other. 

Set $n-i=k,i=n\iff k=0,i=j\iff k=?$  $$\sum_{i=j}^n(n-i)=\sum_{k=0}^{n-j}k=\sum_{k=1}^{n-j}k$$
