a+b=a(n+1)/an,ab=1/an6a-2ab+6b=36(a+b)-2ab=36a(n+1)∴6a(n+1)/an-2/an=3∴6a(n+1)=3an+2∴a(n+1)=1/2an+1/3∴[a(n+1)-2/3]/(an-2/3)=[1/2an+1/3-2/3]/(an-2/3)=1/2∴{an-2/3}是等比数列公比为1/2∴an-2/3=(1-2/3)...
已知 关于x的2次方程(an)x2-(an+1)x+1=0的两根a,b满足6a-2ab+6b=3,且a1=1求证数列an
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