已知二次函数,f(x)=x平方+ax(a∈R)
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已知二次函数,f(x)=x平方+ax(a∈R)
当a=2时,设n∈N*,S=n/f(n)+(n+1)/f(n+1)+...+(3n-1)/f(3n-1)+3n/f(3n)
求证3/4<S<2
当a=2时,设n∈N*,S=n/f(n)+(n+1)/f(n+1)+...+(3n-1)/f(3n-1)+3n/f(3n)
求证3/4<S<2
设an=n/f(n)=n/(n²+2n)=1/(n+2)
记S=h(n)
h(n+1)-h(n)=[(n+1)/f(n+1)+(n+2)/f(n+2)+...+(3n+2)/f(3n+2)+(3n+3)/f(3n+3)]-[n/f(n)+(n+1)/f(n+1)+...+(3n-1)/f(3n-1)+3n/f(3n)]
=(3n+1)/f(3n+1)+(3n+2)/f(3n+2)+(3n+3)/f(3n+3)-n/f(n)
=1/(3n+3)+1/(3n+4)+1/(3n+5)-1/3
=[(3n+4)(3n+5)+(3n+3)(3n+5)+(3n+3)(3n+4)-(n+1)(3n+4)(3n+5)]/(3n+3)(3n+4)(3n+5)
=(-9n³-9n²-47n+27)/(3n+3)(3n+4)(3n+5)
令v(x)=-9n³-9n²-47n+27,v'(x)=-27n²-18n-47
记S=h(n)
h(n+1)-h(n)=[(n+1)/f(n+1)+(n+2)/f(n+2)+...+(3n+2)/f(3n+2)+(3n+3)/f(3n+3)]-[n/f(n)+(n+1)/f(n+1)+...+(3n-1)/f(3n-1)+3n/f(3n)]
=(3n+1)/f(3n+1)+(3n+2)/f(3n+2)+(3n+3)/f(3n+3)-n/f(n)
=1/(3n+3)+1/(3n+4)+1/(3n+5)-1/3
=[(3n+4)(3n+5)+(3n+3)(3n+5)+(3n+3)(3n+4)-(n+1)(3n+4)(3n+5)]/(3n+3)(3n+4)(3n+5)
=(-9n³-9n²-47n+27)/(3n+3)(3n+4)(3n+5)
令v(x)=-9n³-9n²-47n+27,v'(x)=-27n²-18n-47
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