已知an=1/n+1+1/n+2+1/n+3+...+1/2n,n属于N,那么a小n+1=a小n+?若实数x满足对任意正数a>0,均有x^20,均有x^2
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已知an=1/n+1+1/n+2+1/n+3+...+1/2n,n属于N,那么a小n+1=a小n+?若实数x满足对任意正数a>0,均有x^20,均有x^2
已知an=1/n+1+1/n+2+1/n+3+...+1/2n,n属于N,那么a小n+1=a小n+?若实数x满足对任意正数a>0,均有x^20,均有x^2
已知an=1/n+1+1/n+2+1/n+3+...+1/2n,n属于N,那么a小n+1=a小n+?若实数x满足对任意正数a>0,均有x^20,均有x^2
a小(n+1)=1/(n+2)+1/(n+3)+1/(n+4)+.1/(2n)+1/(2n+1)+1/(2n+2)
a小(n+1)=a(小n)+1/(2n+1)+1/(2n+2)-1/(n+1)=a小n +1/[2n+1)/(2n+2)]
对任意正数a>0,均有x^2
a(n+1)=1/(n+2)+1/(n+3)+....+1/(2n+2)
∴a(n+1)-a(n)=1/(2n+1)+1/(2n+2)-1/(n+1)
(2)x^2<1+a(a>0)恒成立则x^2≤1恒成立,
∴-1≤x≤1
(2n+1)*(2n+2)分之一 x为【-1,1】
(1) 由题意知:
a(n+1)=1/(n+2)+1/(n+3)+1/(n+4)+......1/(2n)+1/(2n+1)+1/(2n+2)
a(n+1)=an+1/(2n+1)+1/(2n+2)-1/(n+1)=an +1/[2n+1)/(2n+2)]
(2)对任意正数a>0,均有x²<1+a 即x²≤1 解得:-1≤ x ≤1
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