6. (10th Hope Cup - High School Freshman Math Test) Given that when $n$ is a natural number, $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots$ $+\frac{1}{2 n+1}$, then $f(n+1)-f(n)$ equals
A. $\frac{1}{2 n+2}$
B. $\frac{1}{2 n+3}$
C. $\frac{1}{2 n+2}+\frac{1}{2 n+3}$
D. $\frac{1}{2 n+3}-\frac{1}{2 n+2}$
6. Choose B.

Brief solution: $f(n+1)=\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2 n+3}, f(n+1)-f(n)=\frac{1}{2 n+3}-\frac{1}{2 n+2}$.