Triangular numbers are of the form
T1 = 1
T2 = 1+2
T3 = 1+2+3
these can be represented as figurates.
for example, a pool game has 15 balls = 1+2+3+4+5 = T5
Beginning from T1, every 3rd triangle figurate has a centroid ball.
So for T4 the centroid is:
*
* *
* * *
* * * *
in this case, the centroid ball is the 2nd ball on the 3rd row as it is the centermost ball. Thus, the centroid is 5, i.e. the 5th ball counting from top to bottom.
What formula can be used to calculate the centroid ball for these special triangular numbers? (i.e. every 3rd triangular number starting from T1)
So for our example Centroid(T4) = 5
By geometrical considerations, any odd row contains the centroid of a triangle. The $n^{th}$ centroid is then on the $(2n-1)^{th}$ line. On the other hand, passing from one triangle to the successive one, three new additional rows are included, so that the $n^{th}$ triangle contains $(3n-2)^{th}$ lines.
Now let us consider the triangular number $Tk$, where $k=3n-2$. To express in terms of $k$ the row number containing the centroid, we can note that $n=(k+2)/3$, so that the $(2n-1)^{th}$ line corresponds to the $[(2k+1)/3]^{th}$ line.
The total number of points contained in the first $[(2k+1)/3]-1$ lines is given by
$$\frac 12 \left(\frac{2k+1}{3}-1\right) \frac{2k+1}{3}\\ =\frac 19 (k-1)(2k+1)$$
The row containing the centroid has $(2k+1)/3$ points, so if we start from the first point of the row and stop at the centroid the number of counted points is
$$\frac{(2k+1)/3+1}{2}=\frac{k+2}{3}$$
We conclude that the position of the centroid for the triangular number $Tk$, when counting all points from the first row, is
$$\frac 19 (k-1)(2k+1)+\frac{k+2}{3}\\ = \frac 19 (2 k^2 + 2 k + 5) $$
For $T(1)$, $T(4)$, $T(7)$, $T(10)$... this gives $1,5,13,25...$.