[tex]1+3+5+...+(3n-1) = \\ \\ = \sum\limits_{k=1}^{?} (2k-1) \\ \\ 2k-1 = 3n-1 \Rightarrow 2k = 3n-1+1 \Rightarrow k = \dfrac{3n}{2}\\ \\ = \sum\limits_{k=1}^{\frac{3n}{2}} (2k-1) = 2\sum\limits_{k=1}^{\frac{3n}{2}}k - \sum\limits_{k=1}^{\frac{3n}{2}}1 = \\ \\ = 2\cdot \dfrac{\frac{3n}{2}\cdot (\frac{3n}{2}+1)}{2} - \dfrac{3n}{2} = \\ \\ = \dfrac{3n}{2}\cdot \Big(\dfrac{3n}{2}+1\Big) - \dfrac{3n}{2} = \\ \\ = \dfrac{3n}{2}\cdot \Big(\dfrac{3n}{2}+1-1\Big) = \\ \\ = \dfrac{3n}{2}\cdot \dfrac{3n}{2} [/tex]
[tex]= \left(\dfrac{3n}{2}\right)^2[/tex]
