idk

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\begin{itemize}

  \item
  dla każdego n $\in N_+$ \[\sum_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}\]
  \item KB: dla n=1\\
  L: $\frac{1}{i(i+1)} = \frac{1}{i(1+1)} = \frac{1}{2}$\\
  P: $\frac{n}{n+1} = \frac{1}{1+1}$
  \item Ki: Jeśli
  $\frac{1}{1+...+i(i+1)} = \frac{n}{n+1}$ to $\frac{n}{n+1} + \frac{1}{(n+1)(n+2)} = \frac{n+1}{n+2}$\\
  1+...+$\frac{1}{n(n+1)}=\frac{n}{n+1}$\\
  P: $\frac{n+1}{n+2}$\\
  L: $\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}=\frac{n(n+2)+1}{(n+1)(n+2)} = \frac{n^2+2n+1}{(n+1)(n+2)} = \frac{n+1}{n+2}$\\
  L = P $\square$
\end{itemize}


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