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+...