已知数列an满足a(n+1)=an-a(n-1) (n≥2) ,a1=1 ,a2=3 ,记Sb=a1+a2+...+an
1个回答
(1)
n≥2时,
a(n+1)=an-a(n-1)
a(n+2)=a(n+1)-an
a(n+2)+a(n+1)=a(n+1)-a(n-1)
a(n+2)=-a(n-1)
a(n+3)=-an
a(n+6)=-a(n+3)=an
数列{an}是以6为周期的周期数列
a100=a(6×16+4)=a4=-a(4-3)=-a1=-1
(2)
a3=a2-a1=3-1=2
a(n+1)=an-a(n-1)
an=a(n-1)-a(n-2)
a(n-1)=a(n-2)-a(n-3)
…………
a3=a2-a1
累加
a3+a4+...+an=a(n-1)-a1
Sn=a1+a2+a3+...+an=a1+a2+a(n-1)-a1=a2+a(n-1)
S100=a2+a(100-1)=a2+a99
a99=a(6×16+3)=a3=2
S100=a2+a99=3+2=5
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