How was the closed form of $\sum_{j=i+1}^{n}(n-i-j+2)$ calculated?

Question

I am struggling to understand the closed form calculation of the following triple summation:

$$ \sum_{i=1}^{n} \sum_{j=i+1}^{n} \sum_{k=i+j-1}^{n} 1 $$

The first step I understand:

$$ \sum_{i=1}^{n} \sum_{j=i+1}^{n} \sum_{k=i+j-1}^{n} 1=\sum_{i=1}^{n} \sum_{j=i+1}^{n}(-i-j+n+2) $$

But the next steps are what confuses me.

$$ =\sum_{i=1}^{n} \sum_{j=i+1}^{n}(-i-j+n+2)=\sum_{i=1}^{n}\left(\frac{1}{2}(i-n)(3 i-n-3)\right) $$

$$ =\frac{1}{2}(n-1) n $$

I would like to know, in excruciating detail, how $$ \sum_{j=i+1}^{n}(-i-j+n+2) = \frac{1}{2}(i-n)(3 i-n-3) $$

Answer 1

Let $r=j-i$, then $j=r+i$ and $$\begin{align}\sum_{j=i+1}^{n}(-i-j+n+2)&=\sum_{r=1}^{n-i}(-2i-r+n+2)\\ &=-\sum_{r=1}^{n-i}r+(n+2-2i)\sum_{r=1}^{n-i}1\\ &=-\frac{(n-i)(n-i+1)}{2}+(n+2-2i)(n-i)\\ &=\frac{(n-i)(-n+i-1+2n+4-4i)}{2}\\ &=\frac{(n-i)(n+3-3i)}{2}. \end{align}$$ which is the equal to $\frac{1}{2}(i-n)(3 i-n-3)$. As regards the last sum, let $j=n-i$, then $i=n-j$ and $$\begin{align} \sum_{i=1}^{n}\frac{(n-i)(n+3-3i)}{2}& =\frac{1}{2}\sum_{j=1}^{n-1}j(3j+3-2n)\\ &=\frac{3}{2}\sum_{j=1}^{n-1}j^2+\frac{(3-2n)}{2}\sum_{j=1}^{n-1}j\\ \end{align}$$ Can you finish the job?

Answer 2

Another way.

$\begin{array}\\ \sum_{j=i+1}^{n}(-i-j+n+2) &=\sum_{j=i+1}^{n}(-i+n+2)-\sum_{j=i+1}^{n}j\\ &=(n-i)(n-i+2)-\sum_{j=1}^{n-i}(j+i)\\ &=(n-i)(n-i+2)-\sum_{j=1}^{n-i}j-\sum_{j=1}^{n-i}i\\ &=(n-i)(n-i+2)-\frac12(n-i)(n-i+1)-i(n-i)\\ &=(n-i)\left((n-i+2)-\frac12(n-i+1)-i\right)\\ &=(n-i)\dfrac{2(n-i+2)-(n-i+1)-2i}{2}\\ &=(n-i)\dfrac{n-3i+3}{2}\\ \end{array} $