a1=3,a(n+1)=an^2求通项公式a1=1,a2=-3,a(n+1)=an+a(n+2)求a2004
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a1=3,a(n+1)=an^2求通项公式a1=1,a2=-3,a(n+1)=an+a(n+2)求a2004a1=3,a(n+1)=an^2求通项公式a1=1,a2=-3,a(n+1)=an+a(n+
a1=3,a(n+1)=an^2求通项公式a1=1,a2=-3,a(n+1)=an+a(n+2)求a2004
a1=3,a(n+1)=an^2求通项公式
a1=1,a2=-3,a(n+1)=an+a(n+2)求a2004
a1=3,a(n+1)=an^2求通项公式a1=1,a2=-3,a(n+1)=an+a(n+2)求a2004
a(n+1)=an^2
二边取对数得到lga(n+1)=2lgan
即{lgan}是一个首项是lg3,公比是2的等比数列,则有lgan=lg3*2^(n-1)=lg3^[2^(n-1)]
那么有an=3^[2^(n-1)]
a(n+2)=a(n+1)-an
a3=a2-a1=-4
a4=a3-a2=-4+3=-1
a5=a4-a3=-1+4=3
a6=a5-a4=4
a7=a6-a5=1
a8=a7-a6=-3
.
可见,每隔着6个为一循环,2004/6=334
故有a2004=a6=4
①∵a(n+1)=an^2
∴an=a(n-1)^2=a(n-2)^4=a(n-3)^8=······=a1^2(n-1)
∴an=3^(2n-2)
②由题意:
a1=1
a2=-3
a3=-4
a4=-1
a5=3
a6=4
a7=1······
∴2004/6=334
∴a2004=4
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(1)数列{an}中,a1=1,a2=-3,a(n+1)=an+a(n+2),则a2005=____(2)已知数列{an}满足a1=1,a1×a2×a3…an=n^2,求an.