f(n)=1/(n+1)+1/(n+2)+...+1/2n是什么意思，要求f(n+1)-f(n)=?

f(n)表示关于n的函数
1/(n+1)+1/(n+2)+...+1/2n则表示函数关系

f(n+1)就是用n+1代替f(n)=1/(n+1)+1/(n+2)+...+1/2n中的n
所以f(n+1)=1/[(n+1)+1]+1/[(n+1)+2]+...+1/[2(n+1)]
=1/(n+2)+1/(n+3)+...+1/2n+1/(2n+1)+1/(2n+2)
所以f(n+1)-f(n)
=[1/(n+2)+1/(n+3)+...+1/2n+1/(2n+1)+1/(2n+2)]-[1/(n+1)+1/(n+2)+...+1/2n]
=1/(2n+1)+1/(2n+2)-1/(n+1)
=1/(4n^2+6n+2)

f(n+1)-f(n)=1/2(n+1)+1/(2n+1)-1/(n+1)=1/(2n+1)-1/2(n+1)

f(n)=1/(n+1)+1/(n+2)+...+1/2n.
f(n+1)-f(n)
=[1/(n+2)+1/(n+3)+...+1/2(n+1)]-
[1/(n+1)+1/(n+2)+...+1/2n]
=1/[2(N+1)]-1/(N+1)
=-1/[2(N+1)]

f(n+1)=1/(n+2)+...+1/2n+1/(2n+1)+1/(2n+2)
f(n)=1/(n+1)+1/(n+2)+...+1/2n
两式相减得f(n+1)-f(n)=1/(2n+2)+1/(2n+1)-1/(n+1)

f(n+1)-f(n)
={（1/(n+2)+1/(n+3)+...+1/[2（n+1）]}-（1/(n+1)+1/(n+2)+...+1/2n）
=1/[2（n+1）]-1/(n+1)
=-1/[2（n+1）]