How to compute the time complexity of a triple nested loop represented by $\sum_{i=1}^{2n} \sum_{j=1}^{n} \sum_{k=j}^{n} i-j$ var r = 0;
for(var i=1; i<=2*n; i++) {
  for(var j=1; j<=n; j++) {
    for(var k=j; k<=n; k++) {
      r = r + (i - j);
    }
  }
}

I'm trying to use summation to solve for it, but I'm having a bit of trouble.
I got this summation:
$$
\sum_{i=1}^{2n} \sum_{j=1}^{n} \sum_{k=j}^{n} i-j
$$
But after trying to get a closed form for this, I got $-n^4 + n^2$, which is surely not the answer.
Could you guys help me?
 A: $$\sum_{i=1}^{2n} \sum_{j=1}^{n} \sum_{k=j}^{n} i-j$$
$$\sum_{i=1}^{2n} \sum_{j=1}^{n}  (i-j)\cdot(n-j+1)$$
$$\sum_{i=1}^{2n} \sum_{j=1}^{n}  (ni-ij+i-nj+j^2-j)$$
$$\sum_{i=1}^{2n}   (ni\cdot n-i\frac{n(n+1)}{2}+in-n\frac{n(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2})$$
$$\sum_{i=1}^{2n}   \left(i( n^2-\frac{n(n+1)}{2}+n)-\frac{n(n^2+3n+2)}{6}\right)$$
$$\sum_{i=1}^{2n}   \left(i\frac{n(n+1)}{2}-\frac{n(n^2+3n+2)}{6}\right)$$
$$  \frac{2n(2n+1)}{2}\frac{n(n+1)}{2}-2n\frac{n(n^2+3n+2)}{6}=\frac23 n^4 + \frac12 n^3 - \frac16 n^2.$$
But this isn't the time complexity of the loop. It is the value of the triple summation in closed form.
A: According to Wolfram Alpha,
$$
\sum_{i=1}^{2n} \sum_{j=1}^{n} \sum_{k=j}^{n} i-j
= \frac23 n^4 + \frac12 n^3 - \frac16 n^2.
$$
But that is not the time complexity of the loop.
Instead, it is the value of the variable $r$ at the end of the loop.
To get the time complexity, ignore the values that are getting added or subtracted to $r$ inside the loop. You just want the big-O class of the total number of operations performed.
A simple trick to count the operations is to replace $i - j$ with $1.$ This will undercount the operations, but not by more than some constant factor, and big-O means you don't have to worry about the constant factor.
A: $$\sum_{i=1}^{2n} \sum_{j=1}^{n} \sum_{k=j}^{n} i-j = \sum\limits_{i=1}^{2n} \sum\limits_{j=1}^n (i-j) \cdot (n-j+1) = \sum\limits_{i=1}^{2n} in^2 -\frac{in(n+1)}{2} + in - \frac{n^2(n+1)}{2} + \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} = \frac{2n^3(2n+1)}{2} - \frac{2n^2(n+1)(2n+1)}{4} + \frac{2n^2(2n+1)}{2} - \frac{2n^3(n+1)}{2} + \frac{2n^2(n+1)(2n+1)}{6} - \frac{2n^2(n+1)}{2} = \frac{2}{3}n^4 + \frac{n^3}{2} - \frac{n^2}{6} $$
