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


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

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

I think you might be confusing two different things.

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.)

Then there is the following fact

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 $(n^2 - n)/2 = n(n-1)/2$.


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