2[ 1+2+3+' ' ' ' ' ' '+(n- 1)]=[ 1+2+3+' ' ' ' '+(n-2)+(n-3)+(n- 1)]+[(n- 1)+(n-2)+(n-3)+' ' ' '+3+2+ 1]

1 and n- 1 add to get n, 2 and n-2 add to get n, and so on, that is, there is n- 1 N, so the result is (n-1) n.

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