T_n = \frac{1}{(2n-1)(2n+1)} = \frac{1}{2}\left(\frac{1}{2n-1} - \frac{1}{2n+1} \right)
\sum_{k=1}^{n} T_k = \frac{1}{2} \sum_{k=1}^n \frac{1}{2k-1} - \frac{1}{2k+1}
Now, try expanding that sum out, it doesn't have a common difference, but see what happens when you write that series out in...