数列{an}满足a1=3,a2=6,an+2=an+1-an,求a2008
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数列{an}满足a1=3,a2=6,an+2=an+1-an,求a2008数列{an}满足a1=3,a2=6,an+2=an+1-an,求a2008数列{an}满足a1=3,a2=6,an+2=an+
数列{an}满足a1=3,a2=6,an+2=an+1-an,求a2008
数列{an}满足a1=3,a2=6,an+2=an+1-an,求a2008
数列{an}满足a1=3,a2=6,an+2=an+1-an,求a2008
移向得到,
An=An+1-An+2
=An+2-An+3-An+2=-An+3
即An=-An+3
∴有An+3=-An+6
∴An=An+6
由于2008=6×334+4
∴A2008=A4=-A1=-3
楼上的是正确的,通常按以下方法处理容易理解些
根据前2项,结合递推式依次算出,后面的项,确定周期
a1=3
a2=6
a3=a2-a1=3
a4=a3-a2=-3
a5=a4-a3=-6
a6=a5-a4=-3
-----------------
以下开始重复
a7=a6-a5=3
a8=a7-a6=6
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楼上的是正确的,通常按以下方法处理容易理解些
根据前2项,结合递推式依次算出,后面的项,确定周期
a1=3
a2=6
a3=a2-a1=3
a4=a3-a2=-3
a5=a4-a3=-6
a6=a5-a4=-3
-----------------
以下开始重复
a7=a6-a5=3
a8=a7-a6=6
故周期是6
a2008=a(6*334+4)=a4=-3
收起
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