Welcome to the forum.

I had not learnt of polygonal numbers so I had to look it up. There is a good article here:

http://en.wikipedia.org/wiki/Polygonal_number

I cannot follow exactly what you are describing but you can build a formula like this:

Diagram below (from Wiki and modified). Here n = 6.

The table shows a column for differences where the differences are n - 2 (= 4)

The second column is an arithmetic progression with first term, a = 1, and common difference, d = n-2

Terms in arithmetic progressions are given by t = a + (m-1)d

The third column shows the polygonal numbers. Each number = previous number + next term in the ap.

So you can build the formula from this.

Hope this helps. Post again,

Bob

]]>Draw a polygon haing number of sides required n this number is constant for the whole series n equal to 2 + diff of arith progress

Then choose on of its angles n draw diagonals n the sides of he angle n the diagonal are to be indefinetly produced

After that I take these 2 sides n diagonals o he first polygon as I often as I choose n draw from corresponding points marked by compass lines parallel to first polygon n divide them in as many equal parts or as many points as there are actualy in the diagonals in the 2 sides produced.

Please give geometrical represantation of the proof if you can

God bless n thank you]]>