Group number 1 1

Group 2 No.3

Numbers 3: 5

......

The number of the n group 2n- 1

Suppose there are x numbers before n groups. According to the first n terms of arithmetic progression and the formula Sn = [(A 1+An) × N]/2 = N2.

There's arithmetic progression.

x= 1+3+5+...+2(n- 1)- 1 =(n- 1)？ =n？ -2n﹢ 1

Group n is n individual

So: the first number of n groups is n? -2n﹢2, and the last number is n? a；

According to the question, each group is a arithmetic progression with a tolerance of 1.

So an = [(a 1+an) × n]/2 = [(n? -2n+2)+n？ ](2n- 1)﹜/2=(n？ -n+ 1)(2n- 1)=2n？ -3n？ +3n- 1