若f(n)=[1/(n+1)]+[1/(n+2)]+[1/(n+3)]+``` ```+(1/2n),则f(n+1)-f(n)=
问题描述:
若f(n)=[1/(n+1)]+[1/(n+2)]+[1/(n+3)]+``` ```+(1/2n),则f(n+1)-f(n)=
答
[1/(2n+1)]-[1/(2n+2)]
若f(n)=[1/(n+1)]+[1/(n+2)]+[1/(n+3)]+``` ```+(1/2n),则f(n+1)-f(n)=
[1/(2n+1)]-[1/(2n+2)]