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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
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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:
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
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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