数列An满足A1=0,A(n+1)=(An-√3)/(√3An+1)则A2009=
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数列An满足A1=0,A(n+1)=(An-√3)/(√3An+1)则A2009=
A2=(A1-3^0.5)/(3^0.5×A1+1)=(0-3^0.5)/(3^0.5×0+1)=-3^0.5
A3=(A2-3^0.5)/(3^0.5×A2+1)=(-3^0.5-3^0.5)/(3^0.5×(-3^0.5)+1)=3^0.5
A4=(A3-3^0.5)/(3^0.5×A3+1)=(3^0.5-3^0.5)/(3^0.5×3^0.5+1)=0
由此可猜出
A(3n-2)=0
A(3n-1)=-3^0.5
A(3n)=3^0.5
A2009=A(3×670-1)=-3^0.5
进一步证明
3^0.5×A(n+1)=(3^0.5×A(n)-3)/(3^0.5×A(n)+1)
3^0.5×A(n+2)=(3^0.5×A(n+1)-3)/(3^0.5×A(n+1)+1)
3^0.5×A(n+3)=(3^0.5×A(n+2)-3)/(3^0.5×A(n+2)+1)
一层一层带入
3^0.5×A(n+2)
=((3^0.5×A(n)-3)/(3^0.5×A(n)+1)-3)/((3^0.5×A(n)-3)/(3^0.5×A(n)+1)+1)
=((3^0.5×A(n)-3-3(3^0.5×A(n)+1))/((3^0.5×A(n)-3)+(3^0.5×A(n)+1))
=(-3^0.5×A(n)-3)/(3^0.5×A(n)-1)
3^0.5×A(n+3)
=((-3^0.5×A(n)-3)/(3^0.5×A(n)-1)-3)/((-3^0.5×A(n)-3)/(3^0.5×A(n)-1)+1)
=((-3^0.5×A(n)-3-3(3^0.5×A(n)-1))/((-3^0.5×A(n)-3+(3^0.5×A(n)-1))
=3^0.5×An
A(n+3)=An
A2009=A2=-3^0.5
A3=(A2-3^0.5)/(3^0.5×A2+1)=(-3^0.5-3^0.5)/(3^0.5×(-3^0.5)+1)=3^0.5
A4=(A3-3^0.5)/(3^0.5×A3+1)=(3^0.5-3^0.5)/(3^0.5×3^0.5+1)=0
由此可猜出
A(3n-2)=0
A(3n-1)=-3^0.5
A(3n)=3^0.5
A2009=A(3×670-1)=-3^0.5
进一步证明
3^0.5×A(n+1)=(3^0.5×A(n)-3)/(3^0.5×A(n)+1)
3^0.5×A(n+2)=(3^0.5×A(n+1)-3)/(3^0.5×A(n+1)+1)
3^0.5×A(n+3)=(3^0.5×A(n+2)-3)/(3^0.5×A(n+2)+1)
一层一层带入
3^0.5×A(n+2)
=((3^0.5×A(n)-3)/(3^0.5×A(n)+1)-3)/((3^0.5×A(n)-3)/(3^0.5×A(n)+1)+1)
=((3^0.5×A(n)-3-3(3^0.5×A(n)+1))/((3^0.5×A(n)-3)+(3^0.5×A(n)+1))
=(-3^0.5×A(n)-3)/(3^0.5×A(n)-1)
3^0.5×A(n+3)
=((-3^0.5×A(n)-3)/(3^0.5×A(n)-1)-3)/((-3^0.5×A(n)-3)/(3^0.5×A(n)-1)+1)
=((-3^0.5×A(n)-3-3(3^0.5×A(n)-1))/((-3^0.5×A(n)-3+(3^0.5×A(n)-1))
=3^0.5×An
A(n+3)=An
A2009=A2=-3^0.5
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