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Q.
Let $\left\{a_{ n }\right\}_{ n =0}^{\infty}$ be a sequence such that $a_0=a_1=0$ and $a_{ n +2}=3 a_{ n +1}-2 a_{ n }+1, \forall n \geq 0$.
Then $a_{25} a_{23}-2 a_{25} a_{22}-2 a_{23} a_{24}+4 a_{22} a_{24}$ is equal to:
$ a_0=0, a_1=0 $
$ a_{ n +2}=3 a_{ n +1}-2 a_{ n +1}: n \geq 0 $
$ a_{ n +2}-a_{ n +1}=2\left(a_{ n +1}- a _{ n }\right)+1 $
$ n =0 a_2-a_1=2\left(a_1-a_0\right)+1 $
$ n =1 a_3-a_2=2\left(a_2-a_1\right)+1 $
$ n =2 a_4-a_3=2\left(a_3-a_2\right)+1 $
$ n = n a_{ n +2}-a_{ n +1}=2\left(a_{ n +1}-a_{ n }\right)+1$
$ \left(a_{ n +2}-a_1\right)-2\left(a_{ n +1}-a_0\right)-( n +1)=0$
$ a_{ n +2}=2 a_{ n +1}+( n +1)$
$ n \rightarrow n -2$
$ a_{ n }-2 a_{ n -1}= n -1$
Now $a _{25} a _{23}-2 a _{25} a _{22}-2 a _{23} a _{24}+4 a _{22} a _{24}$
$=\left(a_{25}-2 a_{24}\right)\left(a_{23}-2 a_{22}\right)=(24)(22)=528$