已知二次函数,f(x)=x平方+ax(a∈R)

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

f(n)=n(n+2),n∈N*，n/f(n)=1/(n+2)
S=n/f(n)+(n+1)/f(n+1)+...+(3n-1)/f(3n-1)+3n/f(3n)
=1/(n+2) +1/(n+3)+……+1/(3n+2)
S1=1/f(1) +2/f(2) +3/f(3)=1/3 +1/4 +1/5=47/60>3/4
而n/f(n)>0
∴S>S1...

S=n/f(n)+(n+1)/f(n+1)+...+(3n-1)/f(3n-1)+3n/f(3n)=1/(N+A)+1/(N+A+1)+1/(N+A+2)+...+1/(3N+A) A=2 s=1/(N+2)+1/(N+3)+1/(N+4)+1/(N+5)...1/(3N+2) 1/(3n+2)*2N

1年前

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