Let {a_{n}}_{n=0}^{infty} be a sequence such | vedclass

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(B) Given the recurrence relation: $a_{n+2} = 3a_{n+1} - 2a_{n} + 1$ with $a_{0} = 0, a_{1} = 0$.Rearranging the terms: $a_{n+2} - a_{n+1} = 2(a_{n+1}

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$528$

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(B) Given the recurrence relation: $a_{n+2} = 3a_{n+1} - 2a_{n} + 1$ with $a_{0} = 0, a_{1} = 0$.Rearranging the terms: $a_{n+2} - a_{n+1} = 2(a_{n+1} - a_{n}) + 1$.Let $b_{n} = a_{n+1} - a_{n}$. Then $b_{n+1} = 2b_{n} + 1$.Since $b_{0} = a_{1} - a_{0} = 0$,we have $b_{1} = 2(0) + 1 = 1$,$b_{2} = 2(1) + 1 = 3$,$b_{3} = 2(3) + 1 = 7$,and in general $b_{n} = 2^{n} - 1$.Thus,$a_{n} = \sum_{k=0}^{n-1} b_{k} = \sum_{k=0}^{n-1} (2^{k} - 1) = (2^{n} - 1) - n$.We want to compute $X = a_{25}a_{23} - 2a_{25}a_{22} - 2a_{23}a_{24} + 4a_{22}a_{24}$.Factoring the expression: $X = a_{25}(a_{23} - 2a_{22}) - 2a_{24}(a_{23} - 2a_{22}) = (a_{25} - 2a_{24})(a_{23} - 2a_{22})$.Using $a_{n} = 2^{n} - 1 - n$,we have $a_{n} - 2a_{n-1} = (2^{n} - 1 - n) - 2(2^{n-1} - 1 - (n-1)) = 2^{n} - 1 - n - 2^{n} + 2 + 2n - 2 = n - 1$.Therefore,$a_{25} - 2a_{24} = 25 - 1 = 24$ and $a_{23} - 2a_{22} = 23 - 1 = 22$.$X = 24 	imes 22 = 528$.

Let $\{a_{n}\}_{n=0}^{\infty}$ be a sequence such that $a_{0}=0, a_{1}=0$ and $a_{n+2}=3a_{n+1}-2a_{n}+1$ for all $n \geq 0$. Then the value of $a_{25}a_{23}-2a_{25}a_{22}-2a_{23}a_{24}+4a_{22}a_{24}$ is equal to:

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