Why can't the nth triangular number be expressed as the area of an equilateral triangle? It should be self-intuitive that the $nth$ triangular number is an equilateral triangle with base $n$, and thus its area should equal the value of the triangular number. So, I was wondering: why doesn't this work?
(Note: When $n=2$, then it has the value of $n=3$, iff we use PT on $n=2$)
 A: By what you mean as area of an triangle, that is the area of an triangle in terms of squares of unit length. But balls arranged in equilateral triangle do not pack in square cells; rather, they are more closely related to hexagonal tiling.
There are two directions: either arrange triangle numbers in the form of right-angled isosceles triangles, or use hexagons as base unit.
First, consider the $n$th triangle number arranged as a right-angled triangle, using $1\times 1$ square cells. Overlay a right-angled triangle of base and height $n$.
The $n$th triangle number is $\frac{n(n+1)}{2} = \frac{n^2 + n}{2}$, but the area of the triangle is only $\frac {n^2}2$. The difference comes from the series of $n$ little triangles lying outside the hypotenuse, each with base and height $1$.

Alternatively, consider the $n$th triangle number arranged as an equilateral triangle, using regular hexagons. The hexagons has sides $\frac1{\sqrt3}$, which makes the distance between any two parallel edges $1$. The area of such hexagon is 
$$A_h = 6\cdot\frac12\cdot\frac12\cdot\frac1{\sqrt3} = \frac{\sqrt3}{2}$$
Overlay an equilateral triangle of sides $n$ on the hexagonal arrangement. The area of this triangle is $\frac12\cdot n^2\cdot \frac{\sqrt3}{2}$, which is $\frac{n^2}2$ times of the basic hexagon. But this time, outside each of the $3$ side of the triangle, there are $n$ isosceles triangles in excess, each little triangle of area $\frac16$ of the basic hexagon. These add to the area of the triangle to obtain the area of the hexagonal arrangement,
$$\frac{n^2}{2}A_h + 3n\frac{A_h}6 = \frac{n(n+1)}{2}A_h$$
