Sure, I'll try to simplify it further.

The $n$th triangle number is the sum of the first $n$ positive integers. For example, the 3rd triangle number is 1+2+3=6. We can write the formula as:

$T(n) = 1 + 2 + 3 + ... + n$

To prove this formula, we can use a method called mathematical induction.

First, we need to show that the formula is true for the smallest possible value of $n$, which is 1:

$T(1) = 1$

This is true, because the sum of the first positive integer is simply 1.

Next, we assume that the formula is true for some value of $n=k$, and use this assumption to show that it is true for the next value of $n=k+1$.

Assuming that the formula is true for $n=k$, we have:

$T(k) = 1 + 2 + 3 + ... + k = \frac{k(k+1)}{2}$

To prove that the formula is also true for $n=k+1$, we add the next term, which is $k+1$, to both sides:

$T(k+1) = 1 + 2 + 3 + ... + k + (k+1)$

Using the formula for $T(k)$, we can substitute in the expression for $T(k)$ and simplify:

$T(k+1) = \frac{k(k+1)}{2} + (k+1) = \frac{(k+1)(k+2)}{2}$

This is the formula for $T(k+1)$, so we have shown that if the formula is true for $n=k$, then it is also true for $n=k+1$.

By using the principle of mathematical induction, we have proven that the formula for the $n$th triangle number is given by:

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