Question

How to prove that the number of diagonals of a polygon is n(n+3)/2

Answer 1

Answer:

Let dn be the number of diagonals of a polygon.

d3=0 (polygon is triangle) Assume that the number of diagonals of a n-polygon is n(n−3)2.

Сonsider n+1−poligon A1A2…An+1. Under the assumption the number of diagonals of n−poligon A1A2…An is n(n−3)2. Besides there are n−1 diagonals An+1A2, An+1A3,…,An+1An−1, and A1An for n+1−poligon A1A2…An+1. Therefore the number of diagonals of a n+1-polygon A1A2…An+1 equal to dn+1=n(n−3)2+n−1=n2−3n+2n−22=(n+1)((n+1)−3)2. Sorry for my English.

Answer 2

Answer:

Let dn be the number of diagonals of a polygon.

d3=0 (polygon is triangle) Assume that the number of diagonals of a n-polygon is n(n−3)2.

Сonsider n+1−poligon A1A2…An+1. Under the assumption the number of diagonals of n−poligon A1A2…An is n(n−3)2. Besides there are n−1 diagonals An+1A2, An+1A3,…,An+1An−1, and A1An for n+1−poligon A1A2…An+1. Therefore the number of diagonals of a n+1-polygon A1A2…An+1 equal to dn+1=n(n−3)2+n−1=n2−3n+2n−22=(n+1)((n+1)−3)2. Sorry for my English.

hope it helps you ☺☺❤

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