Question

已知等差数列{an}的前n项和Sn=2n^2-25n

1）求通项公式an
2)求Sn的最大值
3）求数列{|an|}的前n项和

Answer 1

a(1)=s(1)=2-25=-23,
s(n) = 2n^2 - 25n
s(n+1) = 2(n+1)^2 - 25(n+1),
a(n+1) = s(n+1)-s(n) = 2(2n+1) - 25 = 4n - 23,
a(n) = 4(n-1) - 23 = 4n - 27

a(n+1) = 4n - 23 = s(n+1)-s(n)
1<=n<=5时，s(n+1)-s(n) = 4n-23 < 0, s(n)单调递减，-23=s(1)>=s(n)>=s(6)=4*6-27=-3.
n>=6时，s(n+1)-s(n)=4n-23>0,s(n)单调递增，-3=s(6)<=s(n).
s(n) = 2n^2 - 25n，当n->无穷大时，s(n)->无穷大。。
因此，s(n)没有最大值。
s(n)的最小值为s(6)=-3.

a(n) = 4n-27,

1<=n<=6时，a(n)<0, |a(n)| = -a(n) = 27-4n,
t(n) = |a(1)|+|a(2)|+...+|a(n)| = -a(1)-a(2)-...-a(n) = -s(n) = 25n-2n^2.

n>=7时，a(n)=4n-27>0, |a(n)| = a(n).
t(n) = |a(1)|+|a(2)|+...+|a(6)|+|a(7)|+...+|a(n)|
= -a(1)-a(2)-...-a(6)+a(7)+...+a(n)
= a(1)+a(2)+...+a(6)+a(7)+...+a(n) - 2[a(1)+a(2)+...+a(6)]
= s(n) - 2s(6)
= 2n^2 - 25n- 2[2*6^2 - 25*6]
= 2n^2 - 25n - 2[72 - 150]
= 2n^2 - 25n + 2*78
= 2n^2 - 25n + 156