Question

已知二次函数,f(x)=x平方+ax(a∈R)

Answer 1

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