Help me again(HW PROB WARNING)

\(\text{The formula seems to be }\\ \sum \limits_{k=1}^n \dfrac{1}{(2k-1)(2k+1)} = \dfrac{n}{2n+1}\)

Let's see if we can prove this by induction

\(P_1:\dfrac{1}{1\cdot 3} = \dfrac{1}{2(1)+1} = True\\ \text{Assume $P_n$ and prove $P_n \Rightarrow P_{n+1}$}\\~\\ \text{Let $S_n = \sum \limits_{k=1}^n \dfrac{1}{(2k-1)(2k+1)}$}\\ S_{n+1} = S_n + \dfrac{1}{(2n+1)(2n+3)} =\\ \dfrac{n}{2n+1}+\dfrac{1}{(2n+1)(2n+3)} = \\ \dfrac{1}{2n+1}\left(n + \dfrac{1}{2n+3}\right) = \\\)

\(\dfrac{1}{2n+1}\cdot \dfrac{2n^2+3n+1}{2n+3} = \\ \dfrac{n+1}{2n+3} = \dfrac{n+1}{2(n+1)+1}\\ \text{and thus $P_n \Rightarrow P_{n+1}$}\)

What is the value of the sum\( \dfrac {1}{1\cdot 3} + \dfrac {1}{3\cdot 5} + \dfrac {1}{5\cdot 7} + \dfrac {1}{7\cdot 9} + \cdots + \dfrac {1}{199\cdot 201}\)?

Express your answer as a fraction in simplest form.

\(\begin{array}{rcll} && \dfrac{1}{1*3} + \dfrac{1}{3*5} + \dfrac{1}{5*7}+ \dfrac{1}{7*9}+\ldots+\dfrac{1}{199*201} \\ &=& \dfrac{1}{1*3} + \dfrac{1}{3*5} + \dfrac{1}{5*7}+ \dfrac{1}{7*9}+\ldots+\dfrac{1}{(2n-1)(2n+1)} \\ \hline && \dfrac{1}{(2n-1)(2n+1)} = \dfrac12\left( \dfrac{1}{2n-1} - \dfrac{1}{2n+1} \right) \\ && \dfrac{1}{1*3} = \dfrac12\left( \dfrac{1}{1} - \dfrac{1}{3} \right) \\ && \dfrac{1}{3*5} = \dfrac12\left( \dfrac{1}{3} - \dfrac{1}{5} \right) \\ && \dfrac{1}{5*7} = \dfrac12\left( \dfrac{1}{5} - \dfrac{1}{7} \right) \\ && \dfrac{1}{7*9} = \dfrac12\left( \dfrac{1}{7} - \dfrac{1}{9} \right) \\ && \ldots \\ && \dfrac{1}{199*201} = \dfrac12\left( \dfrac{1}{199} - \dfrac{1}{201} \right) \\ \hline &=& \dfrac12\left( \dfrac{1}{1} - \dfrac{1}{3} \right) + \dfrac12\left( \dfrac{1}{3} - \dfrac{1}{5} \right) + \dfrac12\left( \dfrac{1}{5} - \dfrac{1}{7} \right) + \dfrac12\left( \dfrac{1}{7} - \dfrac{1}{9} \right)+\ldots+\dfrac12\left( \dfrac{1}{199} - \dfrac{1}{201} \right) \\ &=& \dfrac12\left( \dfrac{1}{1} - \underbrace{\dfrac{1}{3} + \dfrac{1}{3}}_{=0} - \underbrace{\dfrac{1}{5}+\dfrac{1}{5}}_{=0} - \underbrace{\dfrac{1}{7} + \dfrac{1}{7}}_{=0} - \underbrace{\dfrac{1}{9}+ \dfrac{1}{9}}_{=0} +\ldots- \underbrace{\dfrac{1}{199}+\dfrac{1}{199}}_{=0} - \dfrac{1}{201} \right) \\ &=& \dfrac12\left( \dfrac{1}{1} - \dfrac{1}{201} \right) \\ &=& \dfrac12\left( 1 - \dfrac{1}{201} \right) \\ &=& \dfrac12\left( \dfrac{201-1}{201} \right) \\ &=& \dfrac12\left( \dfrac{200}{201} \right) \\ &=& \dfrac{100}{201} \\ \end{array}\)