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#1 2014-05-02 15:44:07

Registered: 2014-05-02
Posts: 18

Polygonal numbers

I had this from Euler's Algebra proof to represent polygonal numbers of any kind but I don't understand what it means it goes like this:
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


#2 2014-05-02 20:24:36

bob bundy
Registered: 2010-06-20
Posts: 7,270

Re: Polygonal numbers

hi Sven,

Welcome to the forum.

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

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, smile


View Image: polygonal numbers.gif



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