\frac{1^2}{1 \cdot 3} + \frac{2^2}{3 \cdot 5} + \dots \frac{n^2}{(2n-1) \cdot (2n+1)} + \frac{(n+1)^2}{(2(n+1)-1) \cdot (2(n+1)+1)} = \frac{1^2}{1 \cdot 3} + \frac{2^2}{3 \cdot 5} + \dots \frac{n^2}{(2n-1) \cdot (2n+1)} + \frac{(n+1)^2}{(2n+1) \cdot (2n+3)} = \textnormal{pretpostavka da vrijedi za $n$} = \\
\frac{n(n+1)}{2(2n+1)} + \frac{(n+1)^2}{(2n+1) \cdot (2n+3)}