<s>[INST] 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.
 [/INST] 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
$$=\frac{n^3}{2} + \frac{n^2}{2} - \frac{n(n+1)(2n + 1)}{4} + \frac{n(n+1)(2n+1)}{12}$$
$$=\frac{n^3}{2} + \frac{n^2}{2} - \frac{n(n+1)(2n + 1)}{6}$$
$$=\frac{n^3}{2} + \frac{n^2}{2} - \frac{2n^3 + 3n^2 + n}{6}$$
$$= \frac{3n^3 + 3n^2 - 2n^3 - 3n^2 - n}{6}$$
$$= \frac{n^3 - n}{6}$$
$$= \frac{n(n^2 - 1)}{6}$$
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