Let

Recall that all polygons can be "triangulated."
That is, all polygons can be divided into triangles by choosing
a vertex and drawing a segment to every non-adjacent vertex. A
square can be divided into 2 triangles; a pentagon into 3 triangles;
a hexagon into 4 triangles. In general, a **K-gon** can be
divided into **(K-2)** triangles. Consider the following pictures
of polygonal numbers that have been triangulated with red segments.
:

Notice the blue dot located at the bottom left corner of each figure.
Let this represent the first number of that **K-gon** sequence.
The second number is represented by the number of dots used to
make the smallest version of the figure. The next "level"
of the figure (including all the dots within it) represents the
next **K-gonal **number. In other words, to represent the **nth**
**K-gonal** number

begin at the blue dot (Count the blue dot as the 1stnumber)

follow an outer edge to the nthdot

trace out the figure at that level

The total number of dots around the edges and
inside that level is the **nth k-gonal **number.

Observe that the triangles within each polygon
follow that pattern as well. For example, the ** fourth**
pentagonal number (

Since every **K-gon** can be triangulated
into **(K-2)** triangles, every **K-gonal** number can be
expressed in terms of a triangular number. recall that the **nth**
triangular number is given by

For example,

54contains three sets ofT4.Does it follow that54 = 3*T4 ???If we count the dots, we find that54 =22.SinceT4 = 10, then3*T4 = 3*(10) = 30.What happened???

We must be careful to realize that the inner
triangles share some dots. We **CANNOT** simply state that
the number of dots in the **nth K-gon **is equal to** (# of
triangles)*****Tn = (K-2)*Tn. **We must consider the
dots that are "overcounted."

The dots that are overcounted lie along the
red
diagonals. There are **(K-3)** diagonals in a **K-gon. **There
are** n** dots on each diagonal. Two triangles share a diagonal
as a common side. For every diagonal, each dot on the diagonal
is counted twice. That's like having twice as many diagonals!!!
In other words, we must subtract the number of dots on the diagonals
that are counted twice:

Kn = (# of triangles)*Tn - (# of diagonals)*(number of dots on each diagonal)

Kn = (K-2)*Tn - (K-3)*n

.

Whew, wasn't that a task? Let's put that it
all together. So, the general form for **Kn** (the **nth** **k-gonal **number) is:

So,

.

And,

.

You can use the general form to get the formulas for each polygonal number.