已知数列an满足:a1=-1⼀2,an^2+(a(n+1) +2)*an+2a(n+1) +1=0

2025-01-05 01:37:57
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(1)化简得a(n+1)=-(an+1)/(1+1/(an+1))=-1/(1/(an+1)^2+1/(an+1))
下面用数学归纳法:假设-1(2)题干不清楚
(3)若数列{ a2n-1}为递增数列,则a2n-1<a(2n-1)-1即要证{ a2n}递减∴可证{ a2n}递减∵a(n+1)=-1/(1/(an+1)^2+1/(an+1))若an>a(n+1)则an*(1/(an+1)^2+1/(an+1))>-1 化简得2an^2+3an+2>0∵△<0∴2an^2+3an+2>0成立,即证