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(B) Given the equation $x^2-x-1=0$,the roots $\alpha$ and $\beta$ satisfy $\alpha^2 = \alpha + 1$ and $\beta^2 = \beta + 1$.Given $S_n = 2023 \alpha^n

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$S_{12}=S_{11}+S_{10}$

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(B) Given the equation $x^2-x-1=0$,the roots $\alpha$ and $\beta$ satisfy $\alpha^2 = \alpha + 1$ and $\beta^2 = \beta + 1$.Given $S_n = 2023 \alpha^n + 2024 \beta^n$.Consider $S_{n-1} + S_{n-2} = (2023 \alpha^{n-1} + 2024 \beta^{n-1}) + (2023 \alpha^{n-2} + 2024 \beta^{n-2})$.$= 2023 \alpha^{n-2}(\alpha + 1) + 2024 \beta^{n-2}(\beta + 1)$.Since $\alpha + 1 = \alpha^2$ and $\beta + 1 = \beta^2$,we have:$= 2023 \alpha^{n-2}(\alpha^2) + 2024 \beta^{n-2}(\beta^2) = 2023 \alpha^n + 2024 \beta^n = S_n$.Thus,$S_n = S_{n-1} + S_{n-2}$.For $n=12$,we get $S_{12} = S_{11} + S_{10}$.

If $\alpha, \beta$ are the roots of the equation $x^2-x-1=0$ and $S_n=2023 \alpha^n+2024 \beta^n$,then

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