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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
人气:440 ℃ 时间:2020-03-26 13:02:21
解答
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+...
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