Question

Let ${\left\{a_n\right\}}_{n=0}^{\infty }$ be a sequence such that $a_0=a_1=0$ and $a_{n+2}=3a_{n+1}-2a_n+1,\forall n\ge 0$. Then $a_{25}{a}_{23}-2a_{25}{a}_{22}-2a_{23}{a}_{24}+4a_{22}{a}_{24}$ is equal to:

• A. 483
• B. 528
• C. 575
• D. 624

Answer 1

Answer: (B)

$a_0=0,a_1=0$
$a_{n+2}=3a_{n+1}-2a_{n+1}:n\ge 0$
$a_{n+2}-a_{n+1}=2\left(a_{n+1}-a_n\right)+1$
$n=0a_2-a_1=2\left(a_1-a_0\right)+1$
$n=1a_3-a_2=2\left(a_2-a_1\right)+1$
$n=2a_4-a_3=2\left(a_3-a_2\right)+1$
$n=n{a}_{n+2}-a_{n+1}=2\left(a_{n+1}-a_n\right)+1$
$\left(a_{n+2}-a_1\right)-2\left(a_{n+1}-a_0\right)-(n+1)=0$
$a_{n+2}=2a_{n+1}+(n+1)$
$n\to n-2$
$a_n-2a_{n-1}=n-1$
Now $a_{25}{a}_{23}-2a_{25}{a}_{22}-2a_{23}{a}_{24}+4a_{22}{a}_{24}$
$=\left(a_{25}-2a_{24}\right)\left(a_{23}-2a_{22}\right)=(24)(22)=528$