Proving sum of series $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\dots+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$
Solution 1:
We have$$\frac{2k^2 + 3k + 1}{4k^2 + 8k + 3} = \frac{(2k+1)(k+1)}{(2k+1)(2k+3)}.$$ The rest should be straightforward.