Why is this nested sum formula true I've been trying to get this sum:
$\sum_{i}^{n} \sum_{j=0}^{n-i}j$ into a closed formula but couldn't really understand how to "unpack" that nested sum.
It occured to me that the answer is:
$$\sum_{i=1}^n \left(\sum_{j=0}^{n-i} j\right) = \frac16 n (n^2-1)$$

But I can't figure out why.
 A: Use the identity 
$$\sum_{r=a}^{n}{r\choose a}={n+1\choose a+1}$$
Hence 
$$\begin{align}
\sum_{i=1}^n\sum_{j=0}^{n-i}j&=\sum_{i=1}^n\sum_{j=0}^{n-i}\binom j{\color{red}{1}}
=\sum_{i=1}^n\binom{n-i+1}2\\
&=\sum_{r=1}^n\binom r{\color{red}2}\qquad \text{putting $r=n-i+1$}\\
&=\binom{n+1}{\color{red}3}
=\frac{(n+1)n(n-1)}{1\cdot 2\cdot 3}\\
&=\frac16n(n^2-1)\qquad \blacksquare \end{align}$$
A: From sum of first n natural numbers, we get
$$\sum_{i = 1}^{n}\sum_{j = 0}^{n - i}j = \sum_{i = 1}^{n}\frac{(n - i)(n - i + 1)}{2}$$
$$=\frac{1}{2}\sum_{i = 1}^{n}n^2 - (2n + 1)i +i^2+n$$
$$=\frac{n^3}{2} + \frac{n^2}{2} - (2n + 1)\frac{1}{2}\sum_{i = 1}^{n}i + \frac{1}{2}\sum_{i = 1}^{n}i^2$$
Using sum of first n natural numbers and sum of first n squares, we get
$$=\frac{n^3}{2} + \frac{n^2}{2} - \frac{n(n+1)(2n + 1)}{4} + \frac{n(n+1)(2n+1)}{12}$$
$$=\frac{n^3}{2} + \frac{n^2}{2} - \frac{n(n+1)(2n + 1)}{6}$$
$$=\frac{n^3}{2} + \frac{n^2}{2} - \frac{2n^3 + 3n^2 + n}{6}$$
$$= \frac{3n^3 + 3n^2 - 2n^3 - 3n^2 - n}{6}$$
$$= \frac{n^3 - n}{6}$$
$$= \frac{n(n^2 - 1)}{6}$$
A: You know that:
$$\sum_{j=0}^{n-i} j = \frac{(n-i)(n-i+1)}{2}$$
Then:
$$\sum_{i=0}^{n}(n-i)(n-i+1) = \sum_{i=0}^{n}i(i+1) = \sum_{i=0}^{n}i^2 +\sum_{i=0}^{n}i = \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} = \frac{n(n+1)(2n+4)}{6}$$
The remaining part you can find easily as difference between the sum above and its value on i=0:
$$\sum_{i=1}^{n} = \frac{n(n+1)(n+2)}{6} - \frac{n(n+1)}{2} = \frac{n(n+1)(n-1)}{6}$$
A: A little figure immediately shows that the sum ($=:Q$) can be written as
$$Q=\sum_{(i,j)\in T} j\ ,$$ where $T$ is following triangle of lattice points:
$$T:=\bigl\{(i,j)\in{\mathbb Z}\>\bigm|\>i\geq1, \ j\geq 0,\ i+j\leq n\bigr\}\ .$$
Interchanging the order of summation then gives
$$Q=\sum_{j=0}^{n-1} j\left(\sum_{i=1}^{n-j} 1\right)=n \sum_{j=0}^{n-1} j-\sum_{j=0}^{n-1} j^2={(n-1)n^2\over2}-{(n-1)n(2n-1)\over6}={n(n^2-1)\over6}\ .$$
