Prove that $\frac{1}{1*3}+\frac{1}{3*5}+\frac{1}{5*7}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$ Trying to prove that above stated question for $n \geq 1$. A hint given is that you should use $\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})$. Using this, I think I reduced it to $\frac{1}{2}(\frac{1}{n^2}-(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}))$. Just not sure if it's correct, and what to do with the second half. 

Hint: Multiply and divide by $2$.
What you have done is correct. All you need to observe is terms cancel out in pairs starting from $2nd$ term.
$$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...)$$
So we are left with $1st$ and the last terms.
$$\frac{1}{2}(1-\frac{1}{2n+1})$$
$$\frac{n}{2n+1}$$
