已知数列{a(n)}中,a(1)=2,a(n)-a(n-1)-2n=0(n≥2,n∈N),设Bn=1/a(n+1)+1/
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已知数列{a(n)}中,a(1)=2,a(n)-a(n-1)-2n=0(n≥2,n∈N),设Bn=1/a(n+1)+1/a(n+2)+1/a(n+3
已知数列{a(n)}中,a(1)=2,a(n)-a(n-1)-2n=0(n≥2,n∈N).设Bn=1/a(n+1)+1/a(n+2)+1/a(n+3)+……+1/a(2n),若对任意的正整数n,当m∈【-1,1】时,不等式t^2-2mt+1/6>Bn恒成立,求实数t的取值范围
已知数列{a(n)}中,a(1)=2,a(n)-a(n-1)-2n=0(n≥2,n∈N).设Bn=1/a(n+1)+1/a(n+2)+1/a(n+3)+……+1/a(2n),若对任意的正整数n,当m∈【-1,1】时,不等式t^2-2mt+1/6>Bn恒成立,求实数t的取值范围
![已知数列{a(n)}中,a(1)=2,a(n)-a(n-1)-2n=0(n≥2,n∈N),设Bn=1/a(n+1)+1/](/uploads/image/z/19985967-63-7.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ba%EF%BC%88n%EF%BC%89%7D%E4%B8%AD%2Ca%EF%BC%881%EF%BC%89%3D2%2Ca%EF%BC%88n%EF%BC%89-a%EF%BC%88n-1%EF%BC%89-2n%3D0%28n%E2%89%A52%2Cn%E2%88%88N%29%2C%E8%AE%BEBn%3D1%2Fa%EF%BC%88n%2B1%EF%BC%89%2B1%2F)
a(n)-a(n-1)-2n=0
=>
an=a(n-1)+2n
=a(n-2)+2(n-1)+2n)
=a1+2*(2+3+..+n)
=2+(n-1)(n+2)
=n(n+1)
Bn=1/a(n+1)+1/a(n+2)+...+1/a(2n)
=1/(n+1)(n+2)+1/(n+2)(n+3)+...+1/2n*(2n+1)
=1/(n+1)-1/(n+2)+...+1/(2n)-1/(2n+1)
=1/(n+1)-1/(2n+1)
=n/((n+1)(2n+1))
=n/(2n^2+3n+1)
=1/(2n+1/n+3)
又
2n+1/n+3>=2+1+3=6 (n=1时取等号)
=》BnBn恒成立,则需要:
t^2-2mt+1/6>1/6
=>t(t-2m)>0
又-2
=>
an=a(n-1)+2n
=a(n-2)+2(n-1)+2n)
=a1+2*(2+3+..+n)
=2+(n-1)(n+2)
=n(n+1)
Bn=1/a(n+1)+1/a(n+2)+...+1/a(2n)
=1/(n+1)(n+2)+1/(n+2)(n+3)+...+1/2n*(2n+1)
=1/(n+1)-1/(n+2)+...+1/(2n)-1/(2n+1)
=1/(n+1)-1/(2n+1)
=n/((n+1)(2n+1))
=n/(2n^2+3n+1)
=1/(2n+1/n+3)
又
2n+1/n+3>=2+1+3=6 (n=1时取等号)
=》BnBn恒成立,则需要:
t^2-2mt+1/6>1/6
=>t(t-2m)>0
又-2
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