\(\text{The formula seems to be }\\ \sum \limits_{k=1}^n \dfrac{1}{(2k-1)(2k+1)} = \dfrac{n}{2n+1}\)
Let's see if we can prove this by induction
\(P_1:\dfrac{1}{1\cdot 3} = \dfrac{1}{2(1)+1} = True\\ \text{Assume $P_n$ and prove $P_n \Rightarrow P_{n+1}$}\\~\\ \text{Let $S_n = \sum \limits_{k=1}^n \dfrac{1}{(2k-1)(2k+1)}$}\\ S_{n+1} = S_n + \dfrac{1}{(2n+1)(2n+3)} =\\ \dfrac{n}{2n+1}+\dfrac{1}{(2n+1)(2n+3)} = \\ \dfrac{1}{2n+1}\left(n + \dfrac{1}{2n+3}\right) = \\\)
\(\dfrac{1}{2n+1}\cdot \dfrac{2n^2+3n+1}{2n+3} = \\ \dfrac{n+1}{2n+3} = \dfrac{n+1}{2(n+1)+1}\\ \text{and thus $P_n \Rightarrow P_{n+1}$}\)
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