I wasnt looking more a direct answer, more like, a guide.
I have the formula: n/(2n+1) since I noticed that the Denominator was twice the numerator plus 1.
However, I have reached a halt here because I am unsure how to proceed.
You have the right formula
First you prove the formula works for your starting integer, 1 in this case.
\(\displaystyle \dfrac{1}{(2 * 1 - 1)(2 * 1 + 1)} = \dfrac{1}{(2 - 1)(2 * 1 + 1)} = \dfrac{1}{1(2 * 1 + 1)} = \dfrac{1}{2 * 1 + 1}.\)
To do your second step, you need to specify your proposition generally in terms of k
\(\displaystyle \displaystyle \left(\sum_{i=1}^k\dfrac{1}{(2i - 1)(2i + 1)}\right) = \dfrac{k}{2k + 1}.\)
Now, given your first step it is certainly warranted to say:
\(\displaystyle \exists\ integer\ k \ge 1\ such\ that\ \displaystyle \left(\sum_{i=1}^k\dfrac{1}{(2i - 1)(2i + 1)}\right) = \dfrac{k}{2k + 1}.\)
So you have a true equation about some integer k above. Now take either the left hand or right hand side of that equation and create the corresponding expression for k + 1. Technically it makes no difference whether you choose the right or left hand side, but in practice it usually works best to take the side that is a sum or product ending at k (if there is one) and turn that into a sum or product ending at k + 1. So let's do that
\(\displaystyle \displaystyle \left(\sum_{i=1}^{k+1}\dfrac{1}{(2i - 1)(2i + 1)}\right).\)
Now try to equate that to an expression that involves only the corresponding expression about k and terms in k + 1. In this case, that is really easy because we chose the summation to work with.
\(\displaystyle \displaystyle \left(\sum_{i=1}^{k+1}\dfrac{1}{(2i - 1)(2i + 1)}\right) = \left(\sum_{i=1}^k\dfrac{1}{(2i - 1)(2i + 1)}\right) + \dfrac{1}{\{2(k + 1) - 1\}\{2(k + 1) + 1\}}.\) Good so far?
Now we substitute from our equation involving k.
\(\displaystyle \displaystyle \left(\sum_{i=1}^k\dfrac{1}{(2i - 1)(2i + 1)}\right) + \dfrac{1}{\{2(k + 1) - 1\}\{2(k + 1) + 1\}} = \dfrac{k}{2k + 1} + \dfrac{1}{\{2(k + 1) - 1\}\{2(k + 1) + 1\}}.\)
Can you finish it now?
