数列an满足an+1 - an + an-1=0 (n≥2),且a1=1,a2=-1,则a2011= a.1 b.-1 c.2 d.-2
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数列an满足an+1-an+an-1=0(n≥2),且a1=1,a2=-1,则a2011=a.1b.-1c.2d.-2数列an满足an+1-an+an-1=0(n≥2),且a1=1,a2=-1,则a2
数列an满足an+1 - an + an-1=0 (n≥2),且a1=1,a2=-1,则a2011= a.1 b.-1 c.2 d.-2
数列an满足an+1 - an + an-1=0 (n≥2),且a1=1,a2=-1,则a2011= a.1 b.-1 c.2 d.-2
数列an满足an+1 - an + an-1=0 (n≥2),且a1=1,a2=-1,则a2011= a.1 b.-1 c.2 d.-2
an+1 - an + an-1=0 (n≥2)且a1=1,a2=-1,
an-an-1+an-2=0
相加,得
an+1+an-2=0
an+1=-an-2=an-5
所以
以6为周期
2011÷6=335...1
所以
a2011=a1=1
选a
A
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