<s>[INST] explain to me in a simple to understand way what the equation for finding the nth triangle number is and how it can be proved by using only high school level math. please give each step of a proof using LaTeX. [/INST] Sure! The $n$th triangle number is given by the formula:
$T(n) = (n) + (n-1) + (n-2) + ... + 1$
The name "triangle number" comes from the fact that if you build a triangle of height $n$, the number of elements in the triangle will be the $n$th triangle number. Here are the first few examples:
```
 * (n=2, T(n) = 3)
**
```

```
   * (n=3, T(n)=6
  **
 ***
```

If we continue this pattern we will see that the base of the triangle is length $n$ and the height of the triangle is length $n$ as well. The formula for the area of the triangle is $(b \times h) / 2$ or in this case $\frac{n^2}{2}$. That is close to the formula for the $n$th triangle number which is $\frac{n(n+1)}{2}$. The reason we increase the second $n$ by 1 is because we aren't getting the real area, but rather counting dots.</s>\ <s>[INST] Could you explain it in a simpler way? [/INST] 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.</s>