数列1 3+4 5+6+7 7+8+9+10 的前n项和

a(n)＝(2n-1)＋…＋(3n-2)＝n[(2n-1)+(3n-2)]/2＝(5/2)n²－(3/2)n
S(n)＝∑a(n)＝(5/2)∑n²－(3/2)∑n＝(5/2)[n(n+1)(2n+1)/6]－(3/2)[n(n+1)/2]＝n(n+1)(5n-2)/6