Question

已知数列{an}的通项公式an=-2n+11，如果bn=│an│ (n∈N)，求数列{bn}的前n项和

Answer 1

-2n+11 >0
2n<11
n<11/2
n=5
ie
bn= -2n+11 ; n=1,2,..,5
= 2n-11 ; n=6,7,8,....
for n<=5
Sn = (-2n+11+9)n/2 = (11-n)n
for n>=6
Sn = b1+b2+..+bn
= (b1+b2+..+b5) + (b6+b7+..+bn)
= 5(9+1)/2 + (2n-10).(n-5)/2
= 25+(n-5)^2
=n^2-10n+50

Answer 2

Sn=-n^2+10n(n<=5)
Sn=n^2-10n+50(n>=6)
(^-^)