# Don't Understand Double Summation with Gauss's formula

So as we know, if we have a summation from $$1$$ to $$n$$, the simple formula is

$$\sum_{i=1}^n i=\frac{n(n+1)}2.$$

But if we have two summations, one from $$i=1$$ to $$n$$ and another one $$j>i$$ to $$n$$, the formula we get is

$$\sum_{i=1}^n\sum_{j>i}^nj=\frac{n(n-1)}2.$$

I'm not understanding this, how does it go from $$n+1$$ to $$n-1$$, especially since it's a double summation?

• I have edited your question. You can re-edit it if my edits are not correct. – Tianlalu Nov 19 '18 at 3:11

Fact 1: Define $$S_n := \sum_{k=1}^n k$$. Then $$S_n = n(n+1)/2$$ for every nonnegative integer $$n$$.
(You can see this by induction. Certainly the formula is correct for $$n = 0$$. With $$S_{n-1} = (n-1)n/2$$, we have $$S_{n} = n + S_{n-1} = (n+1)n/2$$, as required.)
Fact 2: Define $$T_n = \sum_{1 \leq i < j \leq n} 1$$. Then $$T_n = n(n-1)/2$$.
This is just the number of tuples $$(k, l)$$ with $$1 \leq k \leq n$$ and $$1 \leq l \leq n$$ there are with $$k < l$$. That's just $$\binom{n}{2}$$, or if you like it's also $$(n^2 - n)/2 = n(n-1)/2$$.