The nth triangle number can be found by adding the first n natural numbers together. The formula for the nth triangle number, $T_n$, is given by:

$$T_n = 1+2+3+...+n = \frac{n(n+1)}{2}$$

We can prove this formula by considering the sum of the first n natural numbers, which is:

$$T_n = 1+2+3+...+n$$

If we add this sum to itself, but invert the order, we get:

$$2T_n = (1+2+3+...+n) + (n+(n-1)+(n-2)+...+1)$$

Hence

$$2T_n = (1+n)+(2+(n-1))+(3+(n-2))+...+(n+1)$$

We can see that we have $n$ terms and they each add up to $n+1$, hence

$$2T_n = n*(n+1)$$

From which directly follows that

$$T_n = \frac{n(n+1)}{2}$$

which is what we wanted to proof.
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