I wanna transform $$(n+1)^2+3 * \left (\frac{1}{2} * \left ( -n*(n+3) + 5 *3^n - 3 \right ) \right )$$
to
$$\frac{1}{2}*\left (-(n+1)*((n+1) + 3) + 5 *3^{n+1}-3) \right )$$
I already tried so many things and ended up with: $n^2+n+(n+1)+\frac{3}{2}*-n^2+\frac{3}{2}*(-n*3)+\frac{1}{2} * 5*3^{n+1}-\frac{9}{2}$ but im stuck :(
Your second expression has unbalanced parentheses: $$ \frac{1}{2}*\color{green}{\big(}-\color{blue}{(}n+1\color{blue}{)}*\color{blue}{(}\color{orange}{(}n+1\color{orange}{)} + 3\color{blue}{)} + 5 *3^{n+1}-3\color{green}{\big)} \color{red}{\big)}\ . $$ If I ignore the final red parenthesis and write this out as a sum of powers of $\ n\ $ plus a multiple of a power of $3$, I get $$ -\frac{n^2}{2}-\frac{5n}{2}-\frac{7}{2}+\frac{5\cdot3^{n+1}}{2}\ . $$ Doing the same thing with your first and third expressions, I get the same result.