数列an满足a1=3 a2=6,a(n-2)=a(n-1)-an 则a2004?
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数列an满足a1=3a2=6,a(n-2)=a(n-1)-an则a2004?数列an满足a1=3a2=6,a(n-2)=a(n-1)-an则a2004?数列an满足a1=3a2=6,a(n-2)=a(
数列an满足a1=3 a2=6,a(n-2)=a(n-1)-an 则a2004?
数列an满足a1=3 a2=6,a(n-2)=a(n-1)-an 则a2004?
数列an满足a1=3 a2=6,a(n-2)=a(n-1)-an 则a2004?
a(n-2)=a(n-1)-an
an =a(n-1)-a(n-2)
a1=3
a2=6
a3= a2- a1=3
a4=a3- a2=-3
a5= a4-a3= -6
a6= a5-a4= -3
a7= a6-a5= 3
a8= a7-a6= 6
a9= a8-a7=3
a10= a9- a8=-3
a11= a10-a9= -6
..
可以看出规律:当n≥3时,每6项一个循环.
2004/3=668
即a2004= a3=3
解该数列应是周期数列由a(n-2)=a(n-1)-an
得an=a(n-1)-a(n-2)
a1=3,
a2=6,
a3=a2-a1=3
a4=a3-a2=-3
a5=a4-a3=-6
a6=a5-a4=-3
a7=a6-a5=3
a8=a7-a6=6
a9=a8-a7=3
...............
即该数列是以6为周期的周期数列
a2004=a(333*6+6)=a6=-3
由题可知,an=a(n-1)-a(n-2).即:a3=a2-a1=3,a4=a3-a2=a2-a1-a2=-a1=-3,a5=a4-a3=-6,a6=a5-a4=-6-(-3)=-3,a7=a6-a5=3,a8=a7-a6=6,a9=a8-a7=3.观察可知:该数列每6个循环一次,故a2004=a6=-3
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