数列{an}中a1=8,a4=2,且满足a(n+2)-2a(n+1)+an=0求通项公式(2)设Sn=‖a1‖+‖a2‖+```‖an‖求Sn
问题描述:
数列{an}中a1=8,a4=2,且满足a(n+2)-2a(n+1)+an=0求通项公式(2)设Sn=‖a1‖+‖a2‖+```‖an‖求Sn
答
a(n+2)-2a(n+1)+an=0
a(n+2)+an=2a(n+1)
所以该数列为等差数列
a4=a1+3d
2=8+3d
d=-2
an=a1+(n-1)d
=8-2(n-1)
=10-2n
an=10-2n>=0
n