aritra barua wrote:If we mark our total diagonals as M,we find M=C(2016,2)-2016.Then our required diagonals will be less than the total number of diagonals.By cutting out a pattern.....we follow that for an N sided polygon,the required diagonals as per the question is 2n+(n-1)+(n-2).........+2+1...So,the required number of diagonals=2*2016+2015+2014....+2+1...which is less than C(2016,2)-2016
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