Let ${\left\{{\mathrm{a}}_{\mathrm{n}}\right\}}_{\mathrm{n}=0}^{\mathrm{\infty }}$ be a sequence such that ${\mathrm{a}}_0={\mathrm{a}}_1=0$ and ${\mathrm{\alpha }}_{\mathrm{n}+2}=3{\mathrm{a}}_{\mathrm{n}+1}-2{\mathrm{a}}_{\mathrm{n}}+1,\mathrm{\forall }\mathrm{n}⩾0$. Then ${\mathrm{a}}_{25}{\mathrm{a}}_{23}-2{\mathrm{a}}_{25}\cdot {\mathrm{a}}_{22}-2{\mathrm{a}}_{23}{\mathrm{a}}_{24}+4{\mathrm{a}}_{22}{\mathrm{a}}_{24}$ is equal to:

$\begin{array}{l}{\mathrm{a}}_0=0,{\mathrm{a}}_1=0\\ {\mathrm{a}}_{\mathrm{n}+2}-{\mathrm{a}}_{\mathrm{n}+1}=2\left({\mathrm{a}}_{\mathrm{n}+1}-{\mathrm{a}}_{\mathrm{n}}\right)+1\\ \mathrm{n}=0,{\mathrm{a}}_2-{\mathrm{a}}_1=2\left({\mathrm{a}}_1-{\mathrm{a}}_0\right)+1\\ \mathrm{n}=1,{\mathrm{a}}_3-{\mathrm{a}}_2=2\left({\mathrm{a}}_2-{\mathrm{a}}_1\right)+1\\ \mathrm{n}=2,{\mathrm{a}}_4-{\mathrm{a}}_3=2\left({\mathrm{a}}_3-{\mathrm{a}}_2\right)+1\\ \mathrm{n}=\mathrm{n},{\mathrm{a}}_{\mathrm{n}+2}-{\mathrm{a}}_{\mathrm{n}+1}=2\left({\mathrm{a}}_{\mathrm{n}+1}-{\mathrm{a}}_{\mathrm{n}}\right)+1\\ \left({\mathrm{a}}_{\mathrm{n}+2}-{\mathrm{a}}_1\right)-2\left({\mathrm{a}}_{\mathrm{n}+1}-{\mathrm{a}}_0\right)-(\mathrm{n}+1)=0\\ {\mathrm{a}}_{\mathrm{n}+2}=2{\mathrm{a}}_{\mathrm{n}+1}+(\mathrm{n}+1)\\ \mathrm{n}\to \mathrm{n}-2\\ {\mathrm{a}}_{\mathrm{n}}-2{\mathrm{a}}_{\mathrm{n}-1}=\mathrm{n}-1\end{array}$
Now ${\mathrm{a}}_{25}{\mathrm{a}}_{23}-2{\mathrm{a}}_{25}{\mathrm{a}}_{22}-2{\mathrm{a}}_{23}{\mathrm{a}}_{24}+4{\mathrm{a}}_{22}{\mathrm{a}}_{24}$
$=\left({\mathrm{a}}_{25}-2{\mathrm{a}}_{24}\right)\left({\mathrm{a}}_{23}-2{\mathrm{a}}_{22}\right)=\left(24\right)\left(22\right)=528$