数列{an}的前n项和Sn=[3n(41-n)]/2 求数列{an的绝对值}的前30项的和

an = Sn - S(n-1) = [3n(41-n)]/2 - [3(n-1)(41-n+1)]/2 = 63 -3n
an 为等差数列.
63-3n ≥ 0 ,n≤21
前21项都为正,和为,a1= 63-3 =60,a21 = 0,S21 =(60+0)*21/2 = 630
从第22 项开始,an为负,an的绝对值变为 3n-63
首项a22 = 66-63=3,a30=3*30-63= 27,S22_30 = (3+27)*8/2=120
前30项和=630+120=750